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How to Factor Algebraic Equations

Last Updated: September 30, 2022 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 664,061 times.

In mathematics, factoring is the act of finding the numbers or expressions that multiply together to make a given number or equation. Factoring is a useful skill to learn for the purpose of solving basic algebra problems; the ability to competently factor becomes almost essential when dealing with quadratic equations and other forms of polynomials. Factoring can be used to simplify algebraic expressions to make solving simpler. Factoring can even give you the ability to eliminate certain possible answers much more quickly than you would be able to by solving manually. [1] X Research source

Factoring Numbers and Basic Algebraic Expressions

Step 1 Understand the definition of factoring when applied to single numbers.

  • Another way to think of this is that a given number's factors are the numbers by which it is evenly divisible .
  • The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

Step 2 Understand that variable expressions can also be factored.

  • We can even go as far as to factor 12x multiple times . In other words, we don't have to stop with 3(4x) or 2(6x) - we can factor 4x and 6x to give 3(2(2x) and 2(3(2x), respectively. Obviously, these two expressions are equal.

Step 3 Apply the distributive property of multiplication to factor algebraic equations.

  • Let's try an example problem. To factor the algebraic equation 12 x + 6, first, let's try to find the greatest common factor of 12x and 6. 6 is the biggest number that divides evenly into both 12x and 6, so we can simplify the equation to 6(2x + 1).
  • This process also applies to equations with negatives and fractions. x/2 + 4, for instance, can be simplified to 1/2(x + 8), and -7x + -21 can be factored to -7(x + 3).

Factoring Quadratic Equations

Step 1 Ensure the equation is in quadratic form (ax2 + bx + c = 0).

  • For example, let's consider the algebraic equation. 5x 2 + 7x - 9 = 4x 2 + x - 18 can be simplified to x 2 + 6x + 9 = 0, which is in the quadratic form.
  • Equations with greater powers of x, like x 3 , x 4 , etc. can't be quadratic equations. They are cubic equations, quartic equations, and so on, unless the equation can be simplified to eliminate these terms of x above the power of 2.

Step 2 In quadratic equations where a = 1, factor to (x+d )(x+e), where d × e = c and d + e = b.

  • For example, let's consider the quadratic equation x 2 + 5x + 6 = 0. 3 and 2 multiply together to make 6 and also add up to make 5, so we can simplify this equation to (x + 3)(x + 2).
  • If the quadratic equation is in the form x 2 -bx+c, your answer is in this form: (x - _)(x - _).
  • If it is in the form x 2 +bx+c, your answer looks like this: (x + _)(x + _).
  • If it is in the form x 2 -bx-c, you answer is in the form (x + _)(x - _).
  • Note: the numbers in the blanks can be fractions or decimals. For example, the equation x 2 + (21/2)x + 5 = 0 factors to (x + 10)(x + 1/2).

Step 3 If possible, factor by inspection.

  • Let's consider an example problem. 3x 2 - 8x + 4 at first seems intimidating. However, once we realize that 3 only has two factors (3 and 1), it becomes easier, because we know that our answer must be in the form (3x +/- _)(x +/- _). In this case, adding a -2 to both blank spaces gives the correct answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x add to -8x. -2 × -2 = 4, so we can see that the factored terms in parentheses multiply to become the original equation.

Step 4 Solve by completing the square.

  • For example, the equation x 2 + 6x + 9 fits this form. 3 2 is 9 and 3 × 2 is 6. So, we know that the factored form of this equation is (x + 3)(x + 3), or (x + 3) 2 .

Step 5 Use factors to solve quadratic equations.

  • Let's return to the equation x 2 + 5x + 6 = 0. This equation factored to (x + 3)(x + 2) = 0. If either of the factors equals 0, the entire equation equals 0, so our possible answers for x are the numbers that make (x + 3) and (x + 2) equal 0. These numbers are -3 and -2, respectively.

Step 6 Check your answers - some of them may be extraneous!

  • (-2) 2 + 5(-2) + 6 = 0
  • 4 + -10 + 6 = 0
  • 0 = 0. This is correct, so -2 is a valid answer.
  • (-3) 2 + 5(-3) + 6 = 0
  • 9 + -15 + 6 = 0
  • 0 = 0. This is also correct, so -3 is also a valid answer.

Factoring Other Forms of Equations

Step 1 If the equation is in the form a2-b2, factor it to (a+b)(a-b).

  • For example, the equation 9x 2 - 4y 2 = (3x + 2y)(3x - 2y).

Step 2 If the equation is in the form a2+2ab+b2, factor it to (a+b)2.

  • The equation 4x 2 + 8xy + 4y 2 can be re-written as 4x 2 + (2 × 2 × 2)xy + 4y 2 . We can now see that it's in the correct form, so we can say with confidence that our equation factors to (2x + 2y) 2

Step 3 If the equation is in the form a3-b3, factor it to (a-b)(a2+ab+b2).

  • For instance, 8x 3 - 27y 3 factors to (2x - 3y)(4x 2 + ((2x)(3y)) + 9y 2 )

Community Q&A

Community Answer

Video . By using this service, some information may be shared with YouTube.

  • If you have a trinomial in the form x 2 +bx+ (b/2) 2 , the factored form is (x+(b/2)) 2 . (You may have this situation while completing the square.) Thanks Helpful 1 Not Helpful 2
  • a 2 -b 2 is factorable, a 2 +b 2 isn't factorable. Thanks Helpful 1 Not Helpful 1
  • Remember how to factor constants- it might help. Thanks Helpful 0 Not Helpful 1

how to solve equations with factoring

Things You'll Need

  • Math Book (if necessary)

You Might Also Like

Solve Systems of Algebraic Equations Containing Two Variables

  • ↑ https://www.basic-mathematics.com/what-is-factoring.html
  • ↑ https://www.mathsisfun.com/algebra/factoring.html
  • ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-grouping/a/factoring-by-grouping
  • ↑ https://www.purplemath.com/modules/factquad.htm
  • ↑ https://brownmath.com/alge/polysol.htm
  • ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratics-solve-factoring/a/solving-quadratic-equations-by-factoring
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut19_radeq.htm

About This Article

David Jia

To factor a basic algebraic equation, start by looking for the largest factor that all the numbers in the equation have in common. For instance, if your equation is 6x + 2 = 0, the largest common factor that can be divided evenly into both terms on the left side of the equation is 2. Divide each term by the largest common factor, then rewrite the expression in the form a(b + c), where a is the largest common factor. After factoring, our example equation would become 2(3x + 1) = 0. If the equation contains variables that are common factors in multiple terms, you can factor those out as well. For instance, in the equation 4x² – 2x = 0, each term contains the common factors 2 and x. To factor the left side of the equation, divide each term by those factors to get 2x(2x – 1) = 0. If you want to learn how to factor quadratic equations, keep reading the article! Did this summary help you? Yes No

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Factoring in Algebra

Numbers have factors :

And expressions (like x 2 +4x+3 ) also have factors:

Factoring (called " Factorising " in the UK) is the process of finding the factors :

Factoring: Finding what to multiply together to get an expression.

It is like "splitting" an expression into a multiplication of simpler expressions.

Example: factor 2y+6

Both 2y and 6 have a common factor of 2:

So we can factor the whole expression into:

2y+6 = 2(y+3)

So 2y+6 has been "factored into" 2 and y+3

Factoring is also the opposite of Expanding :

Common Factor

In the previous example we saw that 2y and 6 had a common factor of 2

But to do the job properly we need the highest common factor , including any variables

Example: factor 3y 2 +12y

Firstly, 3 and 12 have a common factor of 3 .

So we could have:

3y 2 +12y = 3(y 2 +4y)

But we can do better!

3y 2 and 12y also share the variable y .

Together that makes 3y :

  • 3y 2 is 3y × y
  • 12y is 3y × 4

3y 2 +12y = 3y(y+4)

Check: 3y(y+4) = 3y × y + 3y × 4 = 3y 2 +12y

More Complicated Factoring

Factoring can be hard .

The examples have been simple so far, but factoring can be very tricky.

Because we have to figure what got multiplied to produce the expression we are given!

Experience Helps

With more experience factoring becomes easier.

Example: Factor 4x 2 − 9

Hmmm... there don't seem to be any common factors.

But knowing the Special Binomial Products gives us a clue called the "difference of squares" :

Because 4x 2 is (2x) 2 , and 9 is (3) 2 ,

So we have:

4x 2 − 9 = (2x) 2 − (3) 2

And that can be produced by the difference of squares formula:

(a+b)(a−b) = a 2 − b 2

Where a is 2x, and b is 3.

So let us try doing that:

(2x+3)(2x−3) = (2x) 2 − (3) 2 = 4x 2 − 9

So the factors of 4x 2 − 9 are (2x+3) and (2x−3) :

Answer: 4x 2 − 9 = (2x+3)(2x−3)

How can you learn to do that? By getting lots of practice, and knowing "Identities"!

Remember these Identities

Here is a list of common "Identities" (including the "difference of squares" used above).

It is worth remembering these, as they can make factoring easier.

There are many more like those, but those are the most useful ones.

The factored form is usually best.

When trying to factor, follow these steps:

  • "Factor out" any common terms
  • See if it fits any of the identities, plus any more you may know
  • Keep going till you can't factor any more

There are also Computer Algebra Systems (called "CAS") such as Axiom, Derive, Macsyma, Maple, Mathematica, MuPAD, Reduce and others that can do factoring.

More Examples

Experience does help, so here are more examples to help you on the way:

Example: w 4 − 16

An exponent of 4? Maybe we could try an exponent of 2:

w 4 − 16 = (w 2 ) 2 − 4 2

Yes, it is the difference of squares

w 4 − 16 = (w 2 + 4)(w 2 − 4)

And "(w 2 − 4)" is another difference of squares

w 4 − 16 = (w 2 + 4)(w+ 2)(w− 2)

That is as far as I can go (unless I use imaginary numbers)

Example: 3u 4 − 24uv 3

Remove common factor "3u":

3u 4 − 24uv 3 = 3u(u 3 − 8v 3 )

Then a difference of cubes:

3u 4 − 24uv 3 = 3u(u 3 − (2v) 3 )

= 3u(u−2v)(u 2 +2uv+4v 2 )

That is as far as I can go.

Example: z 3 − z 2 − 9z + 9

Try factoring the first two and second two separately:

z 2 (z−1) − 9(z−1)

Wow, (z-1) is on both, so let us use that:

(z 2 −9)(z−1)

And z 2 −9 is a difference of squares

(z−3)(z+3)(z−1)

Now get some more experience:

Module 2: Linear and Quadratic Equations

Quadratic equations and factoring, learning outcomes.

  • Recognize a quadratic equation
  • Use the zero product principle to solve a quadratic equation that can be factored

When a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form [latex]ax^{2}+bx+c=0[/latex] is called a quadratic equation .  When solving polynomials where the highest degree is degree 2, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[/latex], where a , b , and c are real numbers and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] is in standard form.

Quadratic Equation

A quadratic equation is an equation containing a second-degree polynomial; for example,

where a , b , and c are real numbers, and if [latex]a\ne 0[/latex], it is in standard form.

Often the easiest method of solving a quadratic equation is by  factoring . Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. Note that we will not spend a lot of time explaining how to factor in this section. We cover factoring in an earlier module of this course and you can sharpen your skills there .

Solving by factoring depends on the Principle of Zero Products. What if we told you that we multiplied two numbers together and got an answer of zero? What could you say about the two numbers? Could they be [latex]2[/latex] and [latex]5[/latex]? Could they be [latex]9[/latex] and [latex]1[/latex]? No! When the result (answer) from multiplying two numbers is zero, that means that one of them  had  to be zero. This idea is called the zero product principle, and it is useful for solving polynomial equations that can be factored. You can further review the   Principle of Zero Products here .

Principle of Zero Products

The Principle of Zero Products states that if the product of two numbers is [latex]0[/latex], then at least one of the factors is [latex]0[/latex]. If [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or both a and b are [latex]0[/latex].

In this section, we will show several examples of solving quadratic equations using factoring.  For each example, we will be using the same general technique:

How to solve a quadratic equation using factoring

  • Make sure your quadratic equation is written in standard form, that is, [latex]a{x}^{2}+bx+c=0[/latex] where a , b , and c are real numbers, and [latex]a\ne 0[/latex].
  • Factor the quadratic expression on the left-hand side of the equation.
  • Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Let us start with a simple example. We will factor a GCF from a binomial and apply the principle of zero products to solve a polynomial equation.  ( You can review how to factor out a greatest common factor here )

[latex]-t^2+t=0[/latex]

Each term has a common factor of [latex]t[/latex], so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms.

[latex]\begin{array}{c}-t^2=t\left(-t\right)\\t=t\left(1\right)\end{array}[/latex]

Rewrite the polynomial equation using the factored terms in place of the original terms.

[latex]\begin{array}{c}-t^2+t=0\\t\left(-t\right)+t\left(1\right)\\t\left(-t+1\right)=0\end{array}[/latex]

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.

[latex]\begin{array}{c}t=0\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,-t+1=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\underline{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-t=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-t}{-1}=\frac{-1}{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}[/latex]

Therefore, [latex]t=0\text{ OR }t=1[/latex].

In the following video, we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation.

In our next example, we will solve a quadratic equation that has a leading coefficient of 1.  In the quadratic equation [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], is [latex]1[/latex]. Notice that if we compare this equation to the standard form of a quadratic equation, then [latex]a=1[/latex], [latex]b=1[/latex], and [latex]c=-b[/latex].  For a review on how to factor a quadratic polynomial with a leading coefficient of 1, you can visit this page.

Factor and solve the equation: [latex]{x}^{2}+x - 6=0[/latex].

This equation is already in standard form, so we will proceed with factoring the left side of the equation.

To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals [latex]1[/latex]. Begin by looking at the possible factors of [latex]-6[/latex].

The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to [latex]1[/latex], so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex].

If we graph the equation [latex]y={x}^{2}+x - 6[/latex], we will get the parabola in the figure below (NOTE: we will learn how to graph quadratic equations in greater detail when we learn about functions ). The solutions to the equation [latex]{x}^{2}+x - 6=0[/latex] are the x- intercepts of [latex]y={x}^{2}+x - 6[/latex]. Recall that x-intercepts are where the y values are zero, therefore the points [latex](-3,0)[/latex] and [latex](2,0)[/latex] represent the places where the parabola crosses the x-axis.  This matches up with our solution to the equation [latex]{x}^{2}+x - 6=0[/latex] because it shows that when [latex]x=2[/latex] and [latex]x=-3[/latex], our equation equals [latex]0[/latex]

Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 7 to 7. The function x squared plus x minus six equals zero is graphed, with the x-intercepts (-3,0) and (2,0), plotted as well.

In the following video, we provide more examples of factoring to solve quadratic equations where the leading coefficient is equal to 1.

We all know that it is rare to be given an equation to solve that has zero on one side, so let us try an example where we first have to move all the terms of the equation to the left-hand side.

Solve: [latex]s^2-4s=5[/latex]

First, move all the terms to one side. The goal is to try and see if we can use the zero product principle since that is the only tool we know for solving polynomial equations.

[latex]\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}[/latex]

We now have all the terms on the left side and zero on the other side. The polynomial [latex]s^2-4s-5[/latex] factors nicely which makes this equation a good candidate for the zero product principle.

[latex]\begin{array}{c}s^2-4s-5=0\\\left(s+1\right)\left(s-5\right)=0\end{array}[/latex]

We separate our factors into two linear equations using the principle of zero products.

[latex]\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}[/latex]

[latex]\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array}[/latex]

Therefore, [latex]s=-1\text{ OR }s=5[/latex].

In our next example, we solve a quadratic equation with a leading coefficient that is not equal to 1. Recall that when the leading coefficient is not [latex]1[/latex], we factor a quadratic equation using a method called grouping, which requires four terms.  You can review how to factor by grouping here.

Solve the quadratic equation: [latex]4{x}^{2}+15x+9=0[/latex].

We can see that this equation is already in standard for, so we will proceed with factoring.

Because this equation has a leading coefficient not equal to 1, we will factor by grouping:

multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the b term, [latex]15x[/latex], with two terms using [latex]3[/latex] and [latex]12[/latex] as coefficients of x . Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

[latex]\begin{array}{lll}&\left(4x+3\right)\left(x+3\right)=0 & \hfill \\ \left(4x+3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=-\frac{3}{4} & & x=-3 \hfill \end{array}[/latex]

The solutions are [latex]x=-\frac{3}{4}[/latex], [latex]x=-3[/latex].

Coordinate plane with the x-axis ranging from negative 6 to 2 with every other tick mark labeled and the y-axis ranging from negative 6 to 2 with each tick mark numbered. The equation: four x squared plus fifteen x plus nine is graphed with its x-intercepts: (-3/4,0) and (-3,0) plotted as well.

Notice that when we graph the equation  [latex]y=4{x}^{2}+15x+9[/latex] we can see that the x-intercepts are [latex](-3,0)[/latex] and [latex](-\frac{3}{4},0)[/latex]. This agrees with our solutions!

The example below shows a quadratic equation where neither side is originally equal to zero.  Remember, we cannot solve quadratic equations unless one side is equal to zero because this allows us to use the Zero Product Principle. (Note that the factoring sequence has been shortened.)

Solve [latex]5b^{2}+4=−12b[/latex] for b.

[latex]5b^{2}+4+12b=−12b+12b[/latex]

Combine like terms.

[latex]5b^{2}+12b+4=0[/latex]

Rewrite [latex]12b[/latex] as [latex]10b+2b[/latex].

[latex]5b^{2}+10b+2b+4=0[/latex]

Factor out [latex]5b[/latex] from the first pair and 2 from the second pair.

[latex]5b\left(b+2\right)+2\left(b+2\right)=0[/latex]

Factor out [latex]b+2[/latex].

[latex]\left(5b+2\right)\left(b+2\right)=0[/latex]

Apply the Zero Product Property.

[latex]5b+2=0\,\,\,\text{or}\,\,\,b+2=0[/latex]

Solve each equation.

[latex]b=-\frac{2}{5}\,\,\,\text{OR}\,\,\,b=−2[/latex]

[latex]b=-\frac{2}{5}\,\,\,\text{or}\,\,\,b=−2[/latex]

We will work through one more example that is similar to the one above, except this example has fractions, yay!

Solve [latex]y^2-5=-\frac{7}{2}y+\frac{5}{2}[/latex]

We can solve this in one of two ways.  One way is to eliminate the fractions like you may have done when solving linear equations, and the second is to find a common denominator and factor fractions. Eliminating fractions is easier, so we will show that way.

Start by multiplying the whole equation by [latex]2[/latex] to eliminate fractions:

[latex]\begin{array}{ccc}2\left(y^2-5=-\frac{7}{2}y+\frac{5}{2}\right)\\\,\,\,\,\,\,2(y^2)+2(-5)=2\left(-\frac{7}{2}y\right)+2\left(\frac{5}{2}\right)\\2y^2-10=-7y+5\end{array}[/latex]

Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products.

[latex]\begin{array}{c}2y^2-10=-7y+5\\2y^2-10+7y-5=0\\2y^2-15+7y=0\\2y^2+7y-15=0\end{array}[/latex]

We can now check whether this polynomial will factor. Using a table we can list factors until we find two numbers with a product of [latex]2\cdot-15=-30[/latex] and a sum of 7.

We have found the factors that will produce the middle term we want,[latex]-3,10[/latex]. We need to place the factors in a way that will lead to a middle term of [latex]7y[/latex]:

[latex]\left(2y-3\right)\left(y+5\right)=0[/latex]

Now we can set each factor equal to zero and solve:

[latex]\begin{array}{ccc}\left(2y-3\right)=0\text{ OR }\left(y+5\right)=0\\2y=3\text{ OR }y=-5\\y=\frac{3}{2}\text{ OR }y=-5\end{array}[/latex]

You can always check your work to make sure your solutions are correct:

Check [latex]y=\frac{3}{2}[/latex]

[latex]\begin{array}{ccc}\left(\frac{3}{2}\right)^2-5=-\frac{7}{2}\left(\frac{3}{2}\right)+\frac{5}{2}\\\frac{9}{4}-5=-\frac{21}{4}+\frac{5}{2}\\\text{ common denominator = 4}\\\frac{9}{4}-\frac{20}{4}=-\frac{21}{4}+\frac{10}{4}\\-\frac{11}{4}=-\frac{11}{4}\end{array}[/latex]

[latex]y=\frac{3}{2}[/latex] is indeed a solution, now check [latex]y=-5[/latex]

[latex]\begin{array}{ccc}\left(-5\right)^2-5=-\frac{7}{2}\left(-5\right)+\frac{5}{2}\\25-5=\frac{35}{2}+\frac{5}{2}\\20=\frac{40}{2}\\20=20\end{array}[/latex]

[latex]y=-5[/latex] is also a solution, so we must have done something right!

Therefore, [latex]y=\frac{3}{2}\text{ OR }y=-5[/latex].

In our last video, we show how to solve another quadratic equation that contains fractions.

The following video contains another example of solving a quadratic equation using factoring with grouping.

It is useful to remember that, if you factor out a constant, the constant will never equal [latex]0[/latex]. So it can essentially be ignored when solving. See the following example.

Solve for k: [latex]-2k^2+90=-8k[/latex]

We need to move all the terms to one side so we can use the zero product principle.

[latex]\begin{array}{l}-2k^2+90=-8k\\\underline{+8k}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+8k}\\-2k^2+8k+90=0\end{array}[/latex]

You will either need to try to factor out a [latex]-2[/latex], or use the method where we multiply [latex]-2\cdot{90}[/latex] and find factors that sum to [latex]8[/latex]. Each term is divisible by  [latex]2[/latex], so we can factor out [latex]-2[/latex].

[latex]-2\left(k^2-4k-45\right)=0[/latex]

Note how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.

Using the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because [latex]45[/latex] is negative and the only way to get a product that is negative is if one of the factors is negative.

[latex]-2\left(k-\,\,\,\right)\left(k+\,\,\,\right)=0[/latex]

We want our factors to have a product of [latex]-45[/latex] and a sum of [latex]-4[/latex]:

There are more factors that will give [latex]-45[/latex], but we have found the ones that sum to [latex]-4[/latex], so we will stop. Fill in the rest of the binomials with the factors we found.

[latex]-2\left(k-9\right)\left(k+5\right)=0[/latex]

Now we can set each factor equal to zero using the zero product rule.

[latex]-2=0[/latex] This solution is nonsense so we discard it.

[latex]k-9=0, k=9[/latex]

[latex]k+5=0, k=-5[/latex]

[latex]k=9\text{ and }k=-5[/latex]

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF) and for equations that have special factoring formulas as well, such as the difference of squares, which we will see later in this section.

In our next example, we will solve a quadratic equation that is written as a difference of squares.  For a review of difference of squares and other special cases, you can visit this page.

Solve the difference of squares equation using the zero-product property: [latex]{x}^{2}-9=0[/latex].

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

[latex]\begin{array}{lll}& {x}^{2}-9=0 & \hfill \\ \left(x - 3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=3 & & x=-3 \hfill \end{array}[/latex]

The solutions are [latex]x=3[/latex] and [latex]x=-3[/latex].

Sometimes, we may be given an equation that does not look like a quadratic at first glance. In our next examples we will solve a cubic polynomial equation where the GCF of each term is x and can be factored. The result is a quadratic equation that we can solve.

Solve the equation by factoring: [latex]-3{x}^{3}-5{x}^{2}-2x=0[/latex].

This equation does not look like a quadratic, as the highest power is [latex]3[/latex], not [latex]2[/latex]. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out [latex]-x[/latex] from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

[latex]\begin{array}{ccc}-x=0&\left(3x+2\right)=0&\left(x+1\right)=0 \hfill \\ x=0 & x=-\frac{2}{3} & x=-1 \hfill \end{array}[/latex]

The solutions are [latex]x=0[/latex], [latex]x=-\frac{2}{3}[/latex], and [latex]x=-1[/latex].

In this last video example, we solve a quadratic equation with a leading coefficient of -1 using a shortcut method of factoring and the zero product principle.

You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or both a and b are [latex]0[/latex]. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.

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  • Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
  • Ex 2: Quadratic Equation App - Find the Dimensions of a Rectangle Given Area (Factoring). Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/PvXsWZp588o . License : CC BY: Attribution
  • Ex: Solve a Quadratic Equation Using Factor By Grouping. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/04zEXaOiO4U . License : CC BY: Attribution
  • Ex: Factor and Solve Quadratic Equation - Trinomial a = -1. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/nZYfgHygXis . License : CC BY: Attribution
  • Unit 12: Factoring, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/resources/downloads/nroc-math-open-textbook-units-1-12-pdf-and-word-formats/ . License : CC BY: Attribution

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

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How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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4.1: Quadratic Equations and Solving by Factoring

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Learning Objectives

By the end of this section, you will be able to:

  • Solve quadratic equations by using the Zero Product Property
  • Solve quadratic equations factoring
  • Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

  • Solve: \(5y−3=0\). If you missed this problem, review Exercise 2.3.1 .
  • Solve: \(10a=0\). If you missed this problem, review Exercise 2.2.1 .
  • Combine like terms: \(12 x^{2}-6 x+4 x\). If you missed this problem, review Exercise 1.3.37 .
  • Factor \(n^{3}-9 n^{2}-22 n\) completely. If you missed this problem, review Exercise 7.3.10 .

We have already solved linear equations, equations of the form \(a x+b y=c\). In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations:

\[x^{2}+5 x+6=0 \quad 3 y^{2}+4 y=10 \quad 64 u^{2}-81=0 \quad n(n+1)=42\]

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get \(n^{2}+n\).

The general form of a quadratic equation is \(a x^{2}+b x+c=0\),with \(a \neq 0\).

QUADRATIC EQUATION

An equation of the form \(a x^{2}+b x+c=0\) is called a quadratic equation.

\[a, b, \text { and } c \text { are real numbers and } a \neq 0\]

To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Solve Quadratic Equations Using the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

ZERO PRODUCT PROPERTY

\(\text { If } a \cdot b=0, \text { then either } a=0 \text { or } b=0 \text { or both. }\)

We will now use the Zero Product Property , to solve a quadratic equation.

Exercise \(\PageIndex{1}\): How to Use the Zero Product Property to Solve a Quadratic Equation

Solve: \((x+1)(x-4)=0\)

This table gives the steps for solving (x + 1)(x – 4) = 0. The first step is to set each factor equal to 0. Since it is a product equal to 0, at least one factor must equal 0. x + 1 = 0 or x – 4 = 0.

Exercise \(\PageIndex{2}\)

Solve: \((x-3)(x+5)=0\)

\(x=3, x=-5\)

Exercise \(\PageIndex{3}\)

Solve: \((y-6)(y+9)=0\)

\(y=6, y=-9\)

We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.

Exercise \(\PageIndex{4}\)

Solve: \((5 n-2)(6 n-1)=0\)

Exercise \(\PageIndex{5}\)

Solve: \((3 m-2)(2 m+1)=0\)

\(m=\frac{2}{3}, m=-\frac{1}{2}\)

Exercise \(\PageIndex{6}\)

Solve: \((4 p+3)(4 p-3)=0\)

\(p=-\frac{3}{4}, p=\frac{3}{4}\)

Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.

Exercise \(\PageIndex{7}\)

Solve: \(3 p(10 p+7)=0\)

Exercise \(\PageIndex{8}\)

Solve: \(2 u(5 u-1)=0\)

\(u=0, u=\frac{1}{5}\)

Exercise \(\PageIndex{9}\)

Solve: \(w(2 w+3)=0\)

\(w=0, w=-\frac{3}{2}\)

It may appear that there is only one factor in the next example. Remember, however, that \((y-8)^{2}\) means \((y-8)(y-8)\).

Exercise \(\PageIndex{10}\)

Solve: \((y-8)^{2}=0\)

Exercise \(\PageIndex{11}\)

Solve: \((x+1)^{2}=0\)

Exercise \(\PageIndex{12}\)

Solve: \((v-2)^{2}=0\)

Solve Quadratic Equations by Factoring

Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we be sure to start with the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Then we factor the expression on the left.

Exercise \(\PageIndex{13}\)

Solve: \(x^{2}+2 x-8=0\)

This table gives the steps for solving the equation x squared + 2 x – 8 = 0. The first step is writing the equation in standard quadratic form, which it is.

Exercise \(\PageIndex{14}\)

Solve: \(x^{2}-x-12=0\)

\(x=4, x=-3\)

Exercise \(\PageIndex{15}\)

Solve: \(b^{2}+9 b+14=0\)

\(b=-2, b=-7\)

SOLVE A QUADRATIC EQUATION BY FACTORING.

  • Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\).
  • Factor the quadratic expression.
  • Use the Zero Product Property.
  • Solve the linear equations.

Before we factor, we must make sure the quadratic equation is in standard form.

Exercise \(\PageIndex{16}\)

Solve: \(2 y^{2}=13 y+45\)

Exercise \(\PageIndex{17}\)

Solve: \(3 c^{2}=10 c-8\)

\(c=0, c=\frac{4}{3}\)

Exercise \(\PageIndex{18}\)

Solve: \(2 d^{2}-5 d=3\)

\(d=3, d=-\frac{1}{2}\)

Exercise \(\PageIndex{19}\)

Solve: \(5 x^{2}-13 x=7 x\)

Exercise \(\PageIndex{20}\)

Solve: \(6 a^{2}+9 a=3 a\)

\(a=0, a=-1\)

Exercise \(\PageIndex{21}\)

Solve: \(45 b^{2}-2 b=-17 b\)

\(b=0, b=-\frac{1}{3}\)

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

Exercise \(\PageIndex{22}\)

Solve: \(144 q^{2}=25\)

\(\begin{array}{lrllrl} & 144 q^{2}&=&25 \\ \text { Write the quadratic equation in standard form. }& 144 q^{2}-25&=& 0 \\ \text { Factor. It is a difference of squares. } & (12 q-5)(12 q+5) & = & 0 \\ \text { Use the Zero Product Property to set each factor to } 0 . & 12 q-5&=&0 & 12 q+5&=&0 \\\text { Solve each equation. } & 12 q & = & 5 & 12 q&=&-5 \\ & q&=&\frac{5}{12} & q & =&-\frac{5}{12} \\ \text { Check your answers. }\end{array}\)

Exercise \(\PageIndex{23}\)

Solve: \(25 p^{2}=49\)

\(p=\frac{7}{5}, p=-\frac{7}{5}\)

Exercise \(\PageIndex{24}\)

Solve: \(36 x^{2}=121\)

\(x=\frac{11}{6}, x=-\frac{11}{6}\)

The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Exercise \(\PageIndex{25}\)

Solve: \((3 x-8)(x-1)=3 x\)

\(\begin{array}{ll} & (3 x-8)(x-1)=3 x \\ \text { Multiply the binomials. }& 3 x^{2}-11 x+8=3 x \\ \text { Write the quadratic equation in standard form. }& 3 x^{2}-14 x+8=0\\ \text { Factor the trinomial. }& (3 x-2)(x-4)=0\\\text { Use the Zero Product Property to set each factor to } 0 . & 3 x-2=0 \quad x-4=0 \\ \text { Solve each equation. } & 3 x=2 \quad x=4 \\ & x=\frac{2}{3} \\ \text { Check your answers. } & \text {The check is left to you! } \end{array}\)

Exercise \(\PageIndex{26}\)

Solve: \((2 m+1)(m+3)=12 m\)

\(m=1, m=\frac{3}{2}\)

Exercise \(\PageIndex{27}\)

Solve: \((k+1)(k-1)=8\)

\(k=3, k=-3\)

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.

Exercise \(\PageIndex{28}\)

Solve: \(9 m^{3}+100 m=60 m^{2}\)

\(\begin{array}{lrllrl} & 9 m^{3}+100 m&=&60 m^{2}\\ \text { Bring all the terms to one side so that the other side is zero. } & 9 m^{3}-60 m^{2}+100 m&=&0 \\ \text { Factor the greatest common factor first. } & m\left(9 m^{2}-60 m+100\right)&=&0 \\ \text { Factor the trinomial. } & m(3 m-10)(3 m-10)&=&0 \\ \text { Use the Zero Product Property to set each factor to 0. } & m&=&0 & 3 m-10&=&0 & 3 m-10&=&0 \\ \text { Solve each equation. } & m&=&0 & m&=&\frac{10}{3}& m&=&\frac{10}{3} \\ \text { Check your answers. } & \text { The check is left to you. } \end{array}\)

Exercise \(\PageIndex{29}\)

Solve: \(8 x^{3}=24 x^{2}-18 x\)

\(x=0, x=\frac{3}{2}\)

Exercise \(\PageIndex{30}\)

Solve: \(16 y^{2}=32 y^{3}+2 y\)

\(y=0, y=\frac{1}{4}\)

When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

Exercise \(\PageIndex{31}\)

Solve: \(4 x^{2}=16 x+84\)

\(\begin{array}{lrllrl} & 4 x^{2}&=&16 x+84\\ \text { Write the quadratic equation in standard form. }& 4 x^{2}-16 x-84&=&0 \\ \text { Factor the greatest common factor first. }& 4\left(x^{2}-4 x-21\right)&=&0 \\ \text { Factor the trinomial. } & 4(x-7)(x+3)&=&0 \\ \text { Use the Zero Product Property to set each factor to 0. } & 4&\neq&0 & x-7&=&0 & x +3&=&0 \\ \text { Solve each equation. } & 4&\neq&0 & x&=&7& x&=&-3 \\ \text { Check your answers. } & \text { The check is left to you. } \end{array}\)

Exercise \(\PageIndex{32}\)

Solve: \(18 a^{2}-30=-33 a\)

\(a=-\frac{5}{2}, a=\frac{2}{3}\)

Exercise \(\PageIndex{33}\)

Solve: \(123 b=-6-60 b^{2}\)

\(b=2, b=\frac{1}{20}\)

Solve Applications Modeled by Quadratic Equations

The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference.

USE A PROBLEM-SOLVING STRATEGY TO SOLVE WORD PROBLEMS

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose a variable to represent that quantity.
  • Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
  • Solve the equation using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a quadratic equation.

Exercise \(\PageIndex{34}\)

The product of two consecutive integers is \(132 .\) Find the integers.

\(\begin{array}{ll} \textbf { Step 1. Read} \text { the problem. } \\ \textbf { Step 2. Identify} \text { what we are looking for. } & \text { We are looking for two consecutive integers. } \\ \textbf { Step 3. Name} \text{ what we are looking for. } & \begin{array}{l}{\text { Let } n=\text { the first integer }} \\ {\space n+1=\text { the next consecutive integer }}\end{array} \\\textbf { Step 4. Translate} \text { into an equation. Restate the } & \text { The product of the two consecutive integers is } 132 . \\ \text { problem in a sentence. } \\ \text { Translate to an equation. } & \begin{array}{c}{\text { The first integer times the next integer is } 132 .} \\ {n(n+1)=132}\end{array} \\ \textbf { Step 5. Solve}\text { the equation. } & n^{2}+n=132 \\ \text { Bring all the terms to one side. } & n^{2}+n-132=0 \\ \text { Factor the trinomial. } & (n-11)(n+12)=0 \\ \text { Use the zero product property. } & n-11=0 \quad n+12=0 \\ \text { Solve the equations. } & n=11 \quad n=-12 \end{array}\)

Exercise \(\PageIndex{35}\)

The product of two consecutive integers is \(240 .\) Find the integers.

\(-15,-16\) and \(15,16\)

Exercise \(\PageIndex{36}\)

The product of two consecutive integers is \(420 .\) Find the integers.

\(-21,-20\) and \(20,21\)

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give 132.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

Exercise \(\PageIndex{37}\)

A rectangular garden has an area of 15 square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

Exercise \(\PageIndex{38}\)

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

55 feet and 66 feet

Exercise \(\PageIndex{39}\)

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

12 feet and 15 feet

In an earlier chapter, we used the Pythagorean Theorem \(\left(a^{2}+b^{2}=c^{2}\right)\). It gave the relation between the legs and the hypotenuse of a right triangle.

This figure is a right triangle.

We will use this formula to in the next example.

Exercise \(\PageIndex{40}\)

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle, as shown below. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the deck.

This figure is a right triangle. The vertical leg is labeled “x – 7”. the horizontal leg, the base, is labeled “x”. The hypotenuse is labeled “17”.

Exercise \(\PageIndex{41}\)

A boat’s sail is a right triangle. The length of one side of the sail is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the sail.

5 feet and 12 feet

Exercise \(\PageIndex{42}\)

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg.

24 feet and 25 feet

Key Concepts

  • Zero Product Property If \(a \cdot b=0\), then either a=0 or b=0 or both. See Example .

Calcworkshop

Solving Equations by Factoring 17+ Amazing Examples!

// Last Updated: January 20, 2020 - Watch Video //

Remember when we learned how to solve linear equations? Well, guess what, we were solving a polynomial!

Jenn (B.S., M.Ed.) of Calcworkshop® teaching factoring equations

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

While a linear equation is a first-degree polynomial, where we don’t need to factor before we solve, some of these same skills will undoubtedly help us to solve polynomial equations by factoring .

So we begin by focusing on the number zero!

When we have a product of two or more factors, the only way for the product to equal zero is for one or more of the factors to also equal zero.

In other words, for all real numbers a and b ,

To solve polynomial equations by factoring

The Zero-Product Property

We call this property, the Zero Product Theorem, as Wyzant so accurately states, we will use this important rule of zero when we are solving by factoring.

So, what are the steps for solving an equation by factoring?

  • Write the polynomial in standard form all on one side of the equation set equal to zero.
  • Factor the polynomial and write as a product of factors.
  • Set each factor equal to zero and solve for the variable using our SCAM technique.

In this lesson, we’re going to expand our skills to solve polynomials of all different degrees such as solving quadratic equations (degree 2 polynomial), cubic equations (degree 3 polynomials), and more.

Solve By Factoring (How-To) – Video

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COMMENTS

  1. Solving quadratic equations by factoring (article)

    Solving quadratics by factoring Google Classroom Learn how to solve quadratic equations like (x-1) (x+3)=0 and how to use factorization to solve other forms of equations. What you should be familiar with before taking this lesson Factoring using the Sum-Product pattern Factoring by grouping Factoring special products

  2. How To Solve Quadratic Equations By Factoring

    This algebra video tutorial explains how to solve quadratic equations by factoring in addition to using the quadratic formula. This video contains plenty o...

  3. 3 Ways to Factor Algebraic Equations

    Method 1 Factoring Numbers and Basic Algebraic Expressions Download Article 1 Understand the definition of factoring when applied to single numbers. Factoring is conceptually simple, but, in practice, can prove to be challenging when applied to complex equations.

  4. 6.6: Solving Equations by Factoring

    6.6: Solving Equations by Factoring Page ID Anonymous LibreTexts Table of contents Solving Quadratic Equations by Factoring

  5. Factoring quadratic expressions: how to walkthrough (video)

    Omkar A. Katta 12 years ago Yes, a and b can be equal. In the example you wrote, that expression is a special case. It is the square of a binomial. (x+10)* (x+10) = (x+10)^2. x+10 is a binomial that is being squared. I hope I helped. 3 comments ( 164 votes) Upvote Downvote Flag

  6. Solving a quadratic equation by factoring

    Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadr...

  7. How To Solve Quadratic Equations By Factoring

    This algebra introduction tutorial explains how to solve quadratic equations by factoring. It discusses how to factor the gcf - greatest common factor, trin...

  8. 1.1: Solve Polynomial Equations by Factoring

    Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF). Step 2: Determine the number of terms in the polynomial. Factor four-term polynomials by grouping (either GCF of pairs, or binomial square then difference of squares).

  9. Solving Quadratic Equations by Factoring Method

    Here is the complete solution. You should back-substitute to verify that [latex]x = 0 [/latex], [latex]x = - \,3 [/latex], and [latex]x = 3 [/latex] are the correct solutions. Example 5: Solve the quadratic equation below using the Factoring Method. The first thing I realize in this problem is that one side of the equation doesn't contain zero.

  10. Factoring Quadratics

    With the quadratic equation in this form: Step 1: Find two numbers that multiply to give ac (in other words a times c), and add to give b. Example: 2x2 + 7x + 3. ac is 2×3 = 6 and b is 7. So we want two numbers that multiply together to make 6, and add up to 7. In fact 6 and 1 do that (6×1=6, and 6+1=7)

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    Full pad Examples Frequently Asked Questions (FAQ) What is the sum of cubes formula? The sum of cubes formula is a³ + b³ = (a+b) (a² - ab + b²) What is the difference of squares formula? The difference of squares formula is a² - b² = (a+b) (a-b) What is the difference of cubes formula?

  12. 4.4: Solve Polynomial Equations by Factoring

    Solving Polynomial Equations by Factoring. In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property 20: \(a⋅b=0\) if and only if \(a=0\) or \(b=0\) The zero-product property is true for any number of factors that make up an equation. ...

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  14. Factoring two-variable quadratics (video)

    Helen White 8 years ago The middle term isn't a square so you can't do a difference of two squares. This equation should be in the form (x - cy) (x + dy).

  15. Factoring in Algebra

    The examples have been simple so far, but factoring can be very tricky. Because we have to figure what got multiplied to produce the expression we are given! It is like trying to find which ingredients went into a cake to make it so delicious. It can be hard to figure out! Experience Helps. With more experience factoring becomes easier.

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  17. 10.3: Solving Quadratic Equations by Factoring

    To solve quadratic equations by factoring, we must make use of the zero-factor property. Factoring Method. Set the equation equal to zero, that is, get all the nonzero terms on one side of the equal sign and 0 on the other. \(ax^2 + bx + c = 0\) Factor the quadratic expression.

  18. Quadratic Equations and Factoring

    How to solve a quadratic equation using factoring. Make sure your quadratic equation is written in standard form, that is, [latex]a{x}^{2}+bx+c=0[/latex] where a, b, and c are real numbers, and [latex]a\ne 0[/latex]. Factor the quadratic expression on the left-hand side of the equation.

  19. Factoring higher degree polynomials (video)

    The easiest way to solve this is to factor by grouping. To do that, you put parentheses around the first two terms and the second two terms. (x^3 - 4x^2) + (6x - 24). Now we take out the GCF from both equations and move it to the outside of the parentheses. x^2(x - 4) + 6(x - 4). As you can see, the sets of numbers inside the parentheses are ...

  20. Solving Equations

    Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions. ... Example 5: solve equations by factoring. Solve the following equation by ...

  21. 4.1: Quadratic Equations and Solving by Factoring

    Solve a quadratic equation by factoring To solve a quadratic equation by factoring: See Example. Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Factor the quadratic expression. Use the Zero Product Property. Solve the linear equations. Check. Use a problem solving strategy to solve word problems See Example. Read the ...

  22. Solving quadratic equations by factoring

    Solving quadratic equations by factoring, step by step, example. Learn how to solve quadratic equations by factoring, at http://MathMeeting.com.

  23. Solving Equations by Factoring (17+ Amazing Examples!)

    Write the polynomial in standard form all on one side of the equation set equal to zero. Factor the polynomial and write as a product of factors. Set each factor equal to zero and solve for the variable using our SCAM technique. In this lesson, we're going to expand our skills to solve polynomials of all different degrees such as solving ...