Polynomial Equation Word Problems (Mixed Operations)

These lessons help Algebra students learn how to write and solve polynomial equations for algebra word problems.

Related Pages Solving Challenging Word Problems Math Word Problems More Algebra Lessons

How To Solve Polynomial Equation Word Problem?

Example: A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

Polynomial Equation Word Problem

Example: A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air as follows: h = -16t 2 + 8t + 8 a) How long will it take the gymnast to reach the ground? b) When will the gymnast be 8 feet above the ground?

How To Solve Word Problems With Polynomial Equations?

  • The sum of a number and its square is 72. Find the number.
  • The area of a triangle is 44m 2 . Find the lengths of the legs if one of the legs is 3m longer than the other leg.
  • The top of a 15-foot ladder is 3 feet farther up a wall than the foo is from the bottom of the wall. How far is the ladder from the bottom of the wall?
  • A projectile is launched upward from ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground?

How To Write Polynomials For Word Problems?

Learn to write a polynomial for Word problems involving perimeter and area of rectangles and circles.

Learn How To Write And Solve Polynomial Equations

Learn to write and solve polynomial equations for special integers, consecutive integers.

Example 1: Find a number that is 56 less than its square. Let n be the number. Example 2: Find two consecutive odd integers whose sum is 130.

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Solving Polynomials

"Solving" means finding the "roots" ...

... a "root" (or "zero") is where the function is equal to zero :

In between the roots the function is either entirely above, or entirely below, the x-axis

Example: −2 and 2 are the roots of the function x 2 − 4

Let's check:

  • when x = −2, then x 2 − 4 = (−2) 2 − 4 = 4 − 4 = 0
  • when x = 2, then x 2 − 4 = 2 2 − 4 = 4 − 4 = 0

How do we solve polynomials? That depends on the Degree !

The first step in solving a polynomial is to find its degree.

The Degree of a Polynomial with one variable is ...

... the largest exponent of that variable.

When we know the degree we can also give the polynomial a name:

How To Solve

So now we know the degree, how to solve?

  • Read how to solve Linear Polynomials (Degree 1) using simple algebra.
  • Read how to solve Quadratic Polynomials (Degree 2) with a little work,
  • It can be hard to solve Cubic (degree 3) and Quartic (degree 4) equations,
  • And beyond that it can be impossible to solve polynomials directly.

So what do we do with ones we can't solve? Try to solve them a piece at a time!

If we find one root, we can then reduce the polynomial by one degree (example later) and this may be enough to solve the whole polynomial.

Here are some main ways to find roots.

1. Basic Algebra

We may be able to solve using basic algebra:

Example: 2x+1

2x+1 is a linear polynomial:

The graph of y = 2x+1 is a straight line

It is linear so there is one root.

Use Algebra to solve:

A "root" is when y is zero: 2x+1 = 0

Subtract 1 from both sides: 2x = −1

Divide both sides by 2: x = −1/2

And that is the solution:

(You can also see this on the graph)

We can also solve Quadratic Polynomials using basic algebra (read that page for an explanation).

2. By experience, or simply guesswork.

It is always a good idea to see if we can do simple factoring:

Example: x 3 +2x 2 −x

This is cubic ... but wait ... we can factor out "x":

x 3 +2x 2 −x = x(x 2 +2x−1)

Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.

Example: x 3 −8

Again this is cubic ... but it is also the " difference of two cubes ":

x 3 −8 = x 3 −2 3

And so we can turn it into this:

x 3 −8 = (x−2)(x 2 +2x+4)

There is a root at x=2, because:

(2−2)(2 2 +2×2+4) = (0) (2 2 +2×2+4)

And we can then solve the quadratic x 2 +2x+4 and we are done

3. Graphically.

Graph the polynomial and see where it crosses the x-axis.

Graphing is a good way to find approximate answers, and we may also get lucky and discover an exact answer.

Caution: before you jump in and graph it, you should really know How Polynomials Behave , so you find all the possible answers!

This is useful to know: When a polynomial is factored like this:

f(x) = (x−a)(x−b)(x−c)...

Then a, b, c, etc are the roots !

So Linear Factors and Roots are related, know one and we can find the other.

(Read The Factor Theorem for more details.)

Example: f(x) = (x 3 +2x 2 )(x−3)

We see "(x−3)", and that means that 3 is a root (or "zero") of the function.

Well, let us put "3" in place of x:

f(x) = (3 3 +2·3 2 )(3−3)

f(3) = (3 3 +2·3 2 )( 0 )

Yes! f(3)=0, so 3 is a root.

How to Check

Found a root? Check it!

Simply put the root in place of "x": the polynomial should be equal to zero.

Example: 2x 3 −x 2 −7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2 . We can check easily, just put "2" in place of "x":

f(2) = 2(2) 3 −(2) 2 −7(2)+2 = 16−4−14+2 = 0

Yes! f(2)=0 , so we have found a root!

How about where it crosses near −1.8 :

f(−1.8) = 2(−1.8) 3 −(−1.8) 2 −7(−1.8)+2 = −11.664−3.24+12.6+2 = −0.304

No, it isn't equal to zero, so −1.8 will not be a root (but it may be close!)

But we did discover one root, and we can use that to simplify the polynomial, like this

Example (continued): 2x 3 −x 2 −7x+2

So, f(2)=0 is a root ... that means we also know a factor:

(x−2) must be a factor of 2x 3 −x 2 −7x+2

Next, divide 2x 3 −x 2 −7x+2 by (x−2) using Polynomial Long Division to find:

2x 3 −x 2 −7x+2 = (x−2)(2x 2 +3x−1)

So now we can solve 2x 2 +3x−1 as a Quadratic Equation and we will know all the roots.

That last example showed how useful it is to find just one root. Remember:

If we find one root, we can then reduce the polynomial by one degree and this may be enough to solve the whole polynomial.

How Far Left or Right

When trying to find roots, how far left and right of zero should we go?

There is a way to tell, and there are a few calculations to do, but it is all simple arithmetic. Read Bounds on Zeros for all the details.

Have We Got All The Roots?

There is an easy way to know how many roots there are. The Fundamental Theorem of Algebra says:

A polynomial of degree n ... ... has n roots (zeros) but we may need to use complex numbers

So: number of roots = the degree of polynomial .

Example: 2x 3 + 3x − 6

The degree is 3 (because the largest exponent is 3), and so:

There are 3 roots.

But Some Roots May Be Complex

Yes, indeed, some roots may be complex numbers (ie have an imaginary part), and so will not show up as a simple "crossing of the x-axis" on a graph.

But there is an interesting fact:

Complex Roots always come in pairs !

So we either get no complex roots, or 2 complex roots, or 4 , etc... Never an odd number.

Which means we automatically know this:

Positive or Negative Roots?

There is also a special way to tell how many of the roots are negative or positive called the Rule of Signs that you may like to read about.

Multiplicity of a Root

Sometimes a factor appears more than once. We call that Multiplicity :

  • Multiplicity is how often a certain root is part of the factoring.

Example: f(x) = (x−5) 3 (x+7)(x−1) 2

This could be written out in a more lengthy way like this:

f(x) = (x−5)(x−5)(x−5)(x+7)(x−1)(x−1)

(x−5) is used 3 times, so the root "5" has a multiplicity of 3 , likewise (x+7) appears once and (x−1) appears twice. So:

  • the root +5 has a multiplicity of 3
  • the root −7 has a multiplicity of 1 (a "simple" root)
  • the root +1 has a multiplicity of 2

Q: Why is this useful? A: It makes the graph behave in a special way!

When we see a factor like (x-r) n , "n" is the multiplicity, and

  • even multiplicity just touches the axis at "r" (and otherwise stays one side of the x-axis)
  • odd multiplicity crosses the axis at "r" (changes from one side of the x-axis to the other)

We can see it on this graph:

Example: f(x) = (x−2) 2 (x−4) 3

(x−2) has even multiplicity , so it just touches the axis at x=2

(x−4) has odd multiplicity , so it crosses the axis at x=4

  • We can directly solve polynomials of Degree 1 (linear) and 2 (quadratic)
  • For Degree 3 and up, graphs can be helpful
  • Know how far left or right the roots may be
  • Know how many roots (the same as its degree)
  • Estimate how many may be complex, positive or negative

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Mathematics LibreTexts

6.5: Polynomial Equations

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

Use the Zero Product Property

  • Solve quadratic equations by factoring
  • Solve equations with polynomial functions
  • Solve applications modeled by polynomial equations

Are you ready?

Before you get started, take this readiness quiz.

  • Solve: \(5y−3=0\). If you missed this problem, review [link] .
  • Factor completely: \(n^3−9n^2−22n\). If you missed this problem, review [link] .
  • If \(f(x)=8x−16\), find \(f(3)\) and solve \(f(x)=0\). If you missed this problem, review [link] .

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

POLYNOMIAL EQUATION

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form ax+b=c.ax+b=c.

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

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

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get n2+n.n2+n.

The general form of a quadratic equation is \(ax^2+bx+c=0\), with \(a\neq 0\). (If \(a=0\), then \(0·x^2=0\) and we are left with no quadratic term.)

QUADRATIC EQUATION

An equation of the form \(ax^2+bx+c=0\) is called a quadratic equation.

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

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

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, then 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

If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.

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

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

Solve: \((5n−2)(6n−1)=0\).

The equation is open parentheses 5n minus 2 close parentheses open parentheses 6n minus 1 close parentheses equals 0. The product equals zero, so at least one factor must equal zero. Step 1 is set each factor equal to zero. So, 5n minus 2 equals 0 and 6n minus 1 equals 0.

Example \(\PageIndex{2}\)

Solve: \((3m−2)(2m+1)=0\).

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

Example \(\PageIndex{3}\)

Solve: \((4p+3)(4p−3)=0\).

\(p=−\frac{3}{4},\space p=\frac{3}{4}\)

USE THE ZERO PRODUCT PROPERTY.

  • Set each factor equal to zero.
  • Solve the linear equations.

Solve Quadratic Equations by Factoring

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we be sure to start with the quadratic equation in standard form , \(ax^2+bx+c=0\). Then we factor the expression on the left.

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

The equation is 2 y squared equals 13y plus 45. Step 1 is to write it in standard form a x squared plus bx plus c. So we have 2 y squared minus 13y minus 45 equals 0.

Example \(\PageIndex{5}\)

Solve: \(3c^2=10c−8\).

\(c=2,\space c=\frac{4}{3}\)

Example \(\PageIndex{6}\)

Solve: \(2d^2−5d=3\).

\(d=3,\space d=−12\)

SOLVE A QUADRATIC EQUATION BY FACTORING.

  • Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
  • Factor the quadratic expression.
  • Use the Zero Product Property.
  • Check. Substitute each solution separately into the original equation.

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

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?

Example \(\PageIndex{7}\)

Solve: \(169q^2=49\).

\(\begin{array} {ll} &169x^2=49 \\ \text{Write the quadratic equation in standard form.} &169x^2−49=0 \\ \text{Factor. It is a difference of squares.} &(13x−7)(13x+7)=0 \\ \text{Use the Zero Product Property to set each factor to }0. & \\ \text{Solve each equation.} &\begin{array} {ll} 13x−7=0 &13x+7=0 \\ 13x=7 &13x=−7 \\ x=\frac{7}{13} &x=−\frac{7}{13} \end{array} \end{array}\)

We leave the check up to you.

Example \(\PageIndex{8}\)

Solve: \(25p^2=49\).

\(p=\frac{7}{5},p=−\frac{7}{5}\)

Example \(\PageIndex{9}\)

Solve: \(36x^2=121\).

\(x=\frac{11}{6},x=−\frac{11}{6}\)

In the next example, the left side of the equation 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.

Example \(\PageIndex{10}\)

Solve: \((3x−8)(x−1)=3x\).

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

Example \(\PageIndex{11}\)

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

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

Example \(\PageIndex{12}\)

Solve: \((k+1)(k−1)=8\).

\(k=3,\space k=−3\)

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

Example \(\PageIndex{13}\)

Solve: \(3x^2=12x+63\).

\(\begin{array} {ll} &3x^2=12x+63 \\ \text{Write the quadratic equation in standard form.} &3x^2−12x−63=0 \\ \text{Factor the greatest common factor first.} &3(x^2−4x−21)=0 \\ \text{Factor the trinomial.} &3(x−7)(x+3)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {lll} 3\neq 0 &x−7=0 &x+3=0 \\ 3\neq 0 &x=7 &x=−3 \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)

Example \(\PageIndex{14}\)

Solve: \(18a^2−30=−33a\).

\(a=−\frac{5}{2},a=\frac{2}{3}\)

Example \(\PageIndex{15}\)

Solve: \(123b=−6−60b^2\)

\(b=−2,\space b=−\frac{1}{20}\)

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 greater than two by using the Zero Product Property, just like we solved quadratic equations.

Example \(\PageIndex{16}\)

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

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

Example \(\PageIndex{17}\)

Solve: \(8x^3=24x^2−18x\).

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

Example \(\PageIndex{18}\)

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

\(y=0,\space y=14\)

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

Example \(\PageIndex{19}\)

For the function \(f(x)=x^2+2x−2\),

ⓐ find \(x\) when \(f(x)=6\) ⓑ find two points that lie on the graph of the function.

ⓐ \(\begin{array} {ll} &f(x)=x^2+2x−2 \\ \text{Substitute }6\text{ for }f(x). &6=x^2+2x−2 \\ \text{Put the quadratic in standard form.} &x^2+2x−8=0 \\ \text{Factor the trinomial.} &(x+4)(x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &x−2=0 \\ x=−4 &\text{or} &x=2 \end{array} \\ \text{Check:} & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ \begin{array} {lll} \quad &\hspace{3mm} f(x)=x^2+2x−2 &f(x)=x^2+2x−2 \\ \quad &f(−4)=(−4)^2+2(−4)−2 &f(2)=2^2+2·2−2 \\ \quad &f(−4)=16−8−2 &f(2)=4+4−2 \\ \quad &f(−4)=6\checkmark &f(2)=6\checkmark \end{array} & \end{array} \) ⓑ Since \(f(−4)=6\) and \(f(2)=6\), the points \((−4,6)\) and \((2,6)\) lie on the graph of the function.

Example \(\PageIndex{20}\)

For the function \(f(x)=x^2−2x−8\),

ⓐ find \(x\) when \(f(x)=7\) ⓑ Find two points that lie on the graph of the function.

ⓐ \(x=−3\) or \(x=5\) ⓑ \((−3,7)\space (5,7)\)

Example \(\PageIndex{21}\)

For the function \(f(x)=x^2−8x+3\),

ⓐ find \(x\) when \(f(x)=−4\) ⓑ Find two points that lie on the graph of the function.

ⓐ \(x=1\) or \(x=7\) ⓑ \((1,−4)\space (7,−4)\)

The Zero Product Property also helps us determine where the function is zero. A value of \(x\) where the function is \(0\), is called a zero of the function .

ZERO OF A FUNCTION

For any function \(f\), if \(f(x)=0\), then \(x\) is a zero of the function .

When \(f(x)=0\), the point \((x,0)\) is a point on the graph. This point is an \(x\)- intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.

Example \(\PageIndex{22}\)

For the function \(f(x)=3x^2+10x−8\), find

ⓐ the zeros of the function, ⓑ any \(x\)-intercepts of the graph of the function ⓒ any \(y\)-intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0. \(\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Substitute }0\text{ for}f(x). &0=3x^2+10x−8 \\ \text{Factor the trinomial.} &(x+4)(3x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &3x−2=0 \\ x=−4 &\text{or} &x=\frac{2}{3} \end{array} \end{array}\) ⓑ An \(x\)-intercept occurs when \(y=0\). Since \(f(−4)=0\) and \(f(\frac{2}{3})=0\), the points \((−4,0)\) and \((\frac{2}{3},0)\) lie on the graph. These points are \(x\)-intercepts of the function. ⓒ A \(y\)-intercept occurs when \(x=0\). To find the \(y\)-intercepts we need to find \(f(0)\). \(\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Find }f(0)\text{ by substituting }0\text{ for }x. &f(0)=3·0^2+10·0−8 \\ \text{Simplify.} &f(0)=−8 \end{array} \) Since \(f(0)=−8\), the point \((0,−8)\) lies on the graph. This point is the \(y\)-intercept of the function.

Example \(\PageIndex{23}\)

For the function \(f(x)=2x^2−7x+5\), find

ⓐ the zeros of the function ⓑ any \(x\)-intercepts of the graph of the function ⓒ any \(y\)-intercepts of the graph of the function.

ⓐ \(x=1\) or \(x=\frac{5}{2}\) ⓑ \((1,0),\space (\frac{5}{2},0)\) ⓒ \((0,5)\)

Example \(\PageIndex{24}\)

For the function \(f(x)=6x^2+13x−15\), find

ⓐ \(x=−3\) or \(x=\frac{5}{6}\) ⓑ \((−3,0),\space (\frac{5}{6},0)\) ⓒ \((0,−15)\)

Solve Applications Modeled by Polynomial 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 polynomial 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 algebraic equation.
  • Solve the equation using appropriate 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 polynomial equation.

Example \(\PageIndex{25}\)

The product of two consecutive odd integers is 323. Find the integers.

\(\begin{array} {ll} \textbf{Step 1. Read }\text{the problem.} & \\ \textbf{Step 2. Identify }\text{what we are looking for.} &\text{We are looking for two consecutive integers.} \\ \textbf{Step 3. Name}\text{ what we are looking for.} &\text{Let } n=\text{ the first integer.} \\ &n+2= \text{ next consecutive odd integer} \\ \begin{array} {l} \textbf{Step 4. Translate }\text{into an equation. Restate the}\hspace{20mm} \\ \text{problem in a sentence.} \end{array} &\begin{array} {l} \text{The product of the two consecutive odd} \\ \text{integers is }323. \end{array} \\ &\quad n(n+2)=323 \\ \textbf{Step 5. Solve }\text{the equation.} n^2+2n=323 \\ \text{Bring all the terms to one side.} &n^2+2n−323=0 \\ \text{Factor the trinomial.} &(n−17)(n+19)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve the equations.} \end{array} &\begin{array} {ll} n−17=0 \hspace{10mm}&n+19=0 \\ n=17 &n=−19 \end{array} \end{array} \) There are two values for \(n\) that are solutions to this problem. So there are two sets of consecutive odd integers that will work.

\(\begin{array} {ll} \text{If the first integer is } n=17 \hspace{60mm} &\text{If the first integer is } n=-19 \\ \text{then the next odd integer is} &\text{then the next odd integer is} \\ \hspace{53mm} n+2 &\hspace{53mm} n+2 \\ \hspace{51mm} 17+2 &\hspace{51mm} -19+2 \\ \hspace{55mm} 19 &\hspace{55mm} -17 \\ \hspace{51mm} 17,19 &\hspace{51mm} -17,-19 \\ \textbf{Step 6. Check }\text{the answer.} & \\ \text{The results are consecutive odd integers} & \\ \begin{array} {ll} 17,\space 19\text{ and }−19,\space −17. & \\ 17·19=323\checkmark &−19(−17)=323\checkmark \end{array} & \\ \text{Both pairs of consecutive integers are solutions.} & \\ \textbf{Step 7. Answer }\text{the question} &\text{The consecutive integers are }17, 19\text{ and }−19,−17. \end{array} \)

Example \(\PageIndex{26}\)

The product of two consecutive odd integers is 255. Find the integers.

\(−15,−17\) and \(15, 17\)

Example \(\PageIndex{27}\)

The product of two consecutive odd integers is 483 Find the integers.

\(−23,−21\) and \(21, 23\)

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 positive results.

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

Example \(\PageIndex{28}\)

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

Example \(\PageIndex{29}\)

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

The width is 5 feet and length is 6 feet.

Example \(\PageIndex{30}\)

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

The length of the patio is 12 feet and the width 15 feet.

In the next example, we will use the Pythagorean Theorem \((a^2+b^2=c^2)\). This formula gives the relation between the legs and the hypotenuse of a right triangle.

Figure shows a right triangle with the shortest side being a, the second side being b and the hypotenuse being c.

We will use this formula to in the next example.

Example \(\PageIndex{31}\)

A boat’s sail is in the shape of a right triangle as shown. 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 sail.

Figure shows a right triangle with the shortest side being x, the second side being x minus 7 and the hypotenuse being 17.

Example \(\PageIndex{32}\)

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

5 feet and 12 feet

Example \(\PageIndex{33}\)

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 the other leg. Find the lengths of the hypotenuse and the other leg.

24 feet and 25 feet

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

Example \(\PageIndex{34}\)

Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function \(h(t)=−16t^2+64t+80\) models the height, \(h\), of the ball above the ground as a function of time, \(t\). Find:

ⓐ the zeros of this function which tell us when the ball hits the ground ⓑ when the ball will be 80 feet above the ground ⓒ the height of the ball at \(t=2\) seconds.

ⓐ The zeros of this function are found by solving \(h(t)=0\). This will tell us when the ball will hit the ground. \(\begin{array} {ll} &h(t)=0 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=0 \\ \text{Factor the GCF, }−16. &−16(t^2−4t−5)=0 \\ \text{Factor the trinomial.} &−16(t−5)(t+1)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.} \end{array} &\begin{array} {ll} t−5=0 &t+1=0 \\ t=5 &t=−1 \end{array} \end{array} \)

The result \(t=5\) tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result \(t=−1\) is discarded.

ⓑ The ball will be 80 feet above the ground when \(h(t)=80\). \(\begin{array} {ll} &h(t)=80 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=80 \\ \text{Subtract 80 from both sides.} &−16t^2+64t=0 \\ \text{Factor the GCF, }−16t. &−16t(t−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.}\end{array} &\begin{array} {ll} −16t=0 &t−4=0 \\ t=0 &t=4 \end{array} \\ &\text{The ball will be at 80 feet the moment Dennis} \\ &\text{tosses the ball and then 4 seconds later, when} \\ &\text{the ball is falling.} \end{array} \)

ⓒ To find the height ball at \(t=2\) seconds we find \(h(2)\). \(\begin{array} {ll} &h(t)=−16t^2+64t+80 \\ \text{To find }h(2)\text{ substitute }2\text{ for }t. &h(2)=−16(2)^2+64·2+80 \\ \text{Simplify.} &h(2)=144 \\ &\text{After 2 seconds, the ball will be at 144 feet.} \end{array}\)

Example \(\PageIndex{35}\)

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function \(h(t)=−16t^2+48t+160\) models the height, \(h\), of the rock above the ocean as a function of time, \(t\). Find:

ⓐ the zeros of this function which tell us when the rock will hit the ocean ⓑ when the rock will be 160 feet above the ocean. ⓒ the height of the rock at \(t=1.5\) seconds.

ⓐ 5 ⓑ 0;3 ⓒ 196

Example \(\PageIndex{36}\)

Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function \(h(t)=−16t^2+32t+128\) models the height, \(h\), of the penny above the ocean as a function of time, \(t\). Find:

ⓐ the zeros of this function which is when the penny will hit the ocean ⓑ when the penny will be 128 feet above the ocean. ⓒ the height the penny will be at \(t=1\) seconds which is when the penny will be at its highest point.

ⓐ 4 ⓑ 0;2 ⓒ 144

Access this online resource for additional instruction and practice with quadratic equations.

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  • How do you solve polynomials equations?
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Multiplying Polynomials Word Problems - Examples & Practice - Expii

Multiplying polynomials word problems - examples & practice, explanations (3).

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Polynomial Word Problems

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As of right now, you should know how to add , subtract , and multiply polynomials.

So when we look at word problems they'll only be asking you to add, subtract, or multiply polynomials.

The trickier part is figuring out which they want you to use. Fortunately, the problems usually deal with shapes which can help clue us in on what to use. If it references perimeter, chances are you need to use addition or subtraction. If references area or volume, chances are you need to use multiplication.

Let's look at an example.

Linda has a rectangular garden that's 3x−5 feet by 2x+8 feet. She wants to put a fence around the perimeter of the garden. If each foot of fence costs 5 dollars, how much will the total fencing around the garden cost?

Step 1: Draw a Diagram

Seriously, don't skip this step. Diagrams are a great way to help visualize the problem and keep things in order. Plus you get to make a quick doodle.

polynomial problem solving floor plan

Image Source: Expii

Here's Linda's garden. Each side is either 3x−5 feet or 2x+8 feet.

Step 2: Figure out What to Use

We want to find the price of the fencing that will surround the garden. To do this, we first need to figure out how much fencing there is.

This is a perimeter problem. Which means, we'll need to add up the sides.

Step 3: Set up the Problem

The last trickier step, is to set the problem up . First, we need to find the perimeter of the rectangle. perimeter=(3x−5)+(2x+8)+(3x−5)+(2x+8) Then, we multiply this by the price per foot of fencing, which is 5 dollars. 5(perimeter)=5[(3x−5)+(2x+8)+(3x−5)+(2x+8)]

Step 4: Solve

In my opinion, this is the easiest step of word problems. We already have it all set up, and finally we just add together some polynomials which we already know how to do. 5[(3x−5)+(2x+8)+(3x−5)+(2x+8)]=5[3x−5+2x+8+3x−5+2x+8]=5[(3x+2x+3x+2x)+(−5+8−5+8)]=5[10x+6]=50x+30 It will cost 50x+30 dollars to put fencing around this garden.

Related Lessons

Word problems involving binomials are often related to basic geometry (like the concept of area , for example).

An important thing to remember is that word problems are just presenting you with familiar information in a new way (i.e., in words). You already know how to handle these problems , so don't let the fact that they're word problems throw you off.

Let's look at an example:

The length of the rectangle below is (2x-3) and the width is (x+7). Find the area of the rectangle in terms of x.

Image source: by Hannah Bonville

First, let's remind ourselves of how to find area. What's the formula for finding the area of the shape above?

Area=length+width

Area=length×width

Area=(2×length)+(2×width)

Area=length/width

(Video) Polynomials 06 Multiply Polynomials Word Problem

by Mister Zuidema

polynomial problem solving floor plan

This video by Mister Zuidema walks through a word problem with polynomials.

Let's look at it.

Sally has a photo which is 20 cm tall and 32 cm wide. She wants to put a frame around the photo which is the same width all around. What is the expression for the area of the framed photo.

First, we draw out the diagram of the photo and the frame. We don't know the width of the frame, so we label it x. We want the area of the whole frame, so we need to write an expression that is equal to the length times the height of the frame. We know the photo is 32 cm wide and the frame is x wide on each side. So, the length of the frame is (32+2x). We find the height of the frame similarly. We know the height of the photo is 20 and each side of the frame is x, so the height is (20+2x). Now, the area is the length times the width, so we multiply binomials. In the video, he uses a different way of multiplying, but I will be using FOIL . We see,

(32+2x)×(20+2x)=(32)(20)+(32)(2x)+(2x)(20)+(2x)(2x)=640+64x+40x+4x2=640+104x+4x2

The expression that represents the area of the whole frame is 640+104x+4x2 which is equal to 4x2+104x+640

Polynomial Word Problems

Solution of exercise solved polynomial word problems, solution of exercise 1, solution of exercise 2, solution of exercise 3, solution of exercise 4, solution of exercise 5, solution to exercise 6, solution to exercise 7, solution to exercise 8, solution to exercise 9.

A polynomial is an expression which consists of two or more than two algebraic expressions.  In a polynomial expression, the same variable has different powers. If the polynomial is added to another polynomial, the resulting expression is also a polynomial. The same goes with the operations of addition, subtraction, multiplication and division.

In this article, we will see how to find the unknown constants, and how to multiply and divide the polynomials. Below are some of the examples of polynomial word problems which you will find quite useful in understanding polynomials and their attributes when they are added, subtracted, multiplied or divided.

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polynomial problem solving floor plan

Set the factors equal to zero:

polynomial problem solving floor plan

Add both expressions together to get

polynomial problem solving floor plan

Take 4 on the left side of the equation:

polynomial problem solving floor plan

Subtract 3 from both sides of the equation to get the final answer:

polynomial problem solving floor plan

The area of the rectangle = length x width

polynomial problem solving floor plan

We know that the amount of revenue generated is equal to the:

Number of items sold x Price per item

Multiply these two expressions together:

polynomial problem solving floor plan

The formula for area of the rectangle = length x width

Hence, to find the width of the rectangle, we need to divide the area by the length:

polynomial problem solving floor plan

Use the polynomial long division method to solve the above expression:

polynomial problem solving floor plan

Since the formula for the distance is speed x time, hence we can easily derive formula of speed from this formula of distance:

polynomial problem solving floor plan

Put the values in the questions in the above formula to get the speed:

polynomial problem solving floor plan

Use the polynomial long division method to find the answer.

polynomial problem solving floor plan

The total amount of profit is calculated by the formula:

Profit = Price per item x Number of items sold

Hence, we will find the profit by multiplying the price of the single shirt with the total number of shirts sold.

polynomial problem solving floor plan

Did you like this article? Rate it!

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

Dividing Algebraic Fractions

Operations with monomials, algebraic identities, polynomial roots, factoring polynomials, dividing polynomials, algebraic fractions, adding or subtracting algebraic fractions, binomial theorem, adding polynomials, reducing algebraic fractions to a common denominator, multiplying polynomials, algebraic expressions, remainder theorem, multiplying algebraic fractions, binimial theorem worksheet, polynomials, ruffini’s rule, binomial formula, polynomial formulas, algebra formulas, algebraic fractions worksheet, polynomial worksheet, factoring polynomials worksheet, binomial worksheet, monomial worksheets, cancel reply.

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Polynomial Equations

Learning objectives.

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

Use the Zero Product Property

  • Solve quadratic equations by factoring
  • Solve equations with polynomial functions
  • Solve applications modeled by polynomial equations

Before you get started, take this readiness quiz.

5y-3=0.

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

ax+b=c.

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

{x}^{2}+5x+6=0\phantom{\rule{3em}{0ex}}3{y}^{2}+4y=10\phantom{\rule{3em}{0ex}}64{u}^{2}-81=0\phantom{\rule{3em}{0ex}}n\left(n+1\right)=42

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

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, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

a·b=0,

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

\left(5n-2\right)\left(6n-1\right)=0.

  • Set each factor equal to zero.
  • Solve the linear equations.

Solve Quadratic Equations by Factoring

a{x}^{2}+bx+c=0.

  • Factor the quadratic expression.
  • Use the Zero Product Property.
  • Check. Substitute each solution separately into the original equation.

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

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?

169{q}^{2}=49.

We leave the check up to you.

25{p}^{2}=49.

In the next example, the left side of the equation 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.

\left(3x-8\right)\left(x-1\right)=3x.

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

3{x}^{2}=12x+63.

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 greater than two by using the Zero Product Property, just like we solved quadratic equations.

9{m}^{3}+100m=60{m}^{2}.

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

f\left(x\right)={x}^{2}+2x-2,

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

f\left(x\right)=0,

ⓐ the zeros of the function,

ⓑ any x -intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0.

\begin{array}{cccccc}& & & & & f\left(x\right)=3{x}^{2}+10x-8\hfill \\ \text{Substitute 0 for}\phantom{\rule{0.2em}{0ex}}f\left(x\right).\hfill & & & & & \phantom{\rule{1.2em}{0ex}}0=3{x}^{2}+10x-8\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(x+4\right)\left(3x-2\right)=0\hfill \\ \begin{array}{c}\text{Use the zero product property.}\hfill \\ \text{Solve.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccc}\hfill x+4& =\hfill & 0\hfill & & & \text{or}\hfill & & & \hfill 3x-2& =\hfill & 0\hfill \\ \hfill x& =\hfill & -4\hfill & & & \text{or}\hfill & & & \hfill x& =\hfill & \frac{2}{3}\hfill \end{array}\hfill \end{array}

ⓐ the zeros of the function

x=\frac{5}{2}

Solve Applications Modeled by Polynomial 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 polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

  • 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 algebraic equation.
  • Solve the equation using appropriate 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 polynomial equation.

The product of two consecutive odd integers is 323. Find the integers.

\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{We are looking for two consecutive integers.}\hfill \\ \mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{Let}\phantom{\rule{0.2em}{0ex}}n=\text{the first integer.}\hfill \\ & & & & & n+2=\phantom{\rule{0.2em}{0ex}}\text{next consecutive odd integer}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation. Restate the}\hfill \\ \text{problem in a sentence.}\hfill \end{array}\hfill & & & & & \begin{array}{c}\text{The product of the two consecutive odd}\hfill \\ \text{integers is 323.}\hfill \end{array}\hfill \\ & & & & & \phantom{\rule{3.47em}{0ex}}n\left(n+2\right)=323\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill & & & & & \phantom{\rule{3.57em}{0ex}}{n}^{2}+2n=323\hfill \\ \text{Bring all the terms to one side.}\hfill & & & & & \phantom{\rule{0.85em}{0ex}}{n}^{2}+2n-323=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(n-17\right)\left(n+19\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property.}\hfill \\ \text{Solve the equations.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccc}\hfill n-17& =\hfill & 0\hfill & & & & \hfill n+19& =\hfill & 0\hfill \\ \hfill n& =\hfill & 17\hfill & & & & \hfill n& =\hfill & -19\hfill \end{array}\hfill \end{array}

There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work.

\begin{array}{cccccc}\text{If the first integer is}\phantom{\rule{0.2em}{0ex}}n=17\hfill & & & & & \text{If the first integer is}\phantom{\rule{0.2em}{0ex}}n=-19\hfill \\ \text{then the next odd integer is}\hfill & & & & & \text{then the next odd integer is}\hfill \\ \phantom{\rule{11em}{0ex}}n+2\hfill & & & & & \phantom{\rule{11em}{0ex}}n+2\hfill \\ \phantom{\rule{10.5em}{0ex}}17+2\hfill & & & & & \phantom{\rule{10em}{0ex}}\text{−}19+2\hfill \\ \phantom{\rule{11.5em}{0ex}}19\hfill & & & & & \phantom{\rule{11em}{0ex}}\text{−}17\hfill \\ \phantom{\rule{10.5em}{0ex}}17,19\hfill & & & & & \phantom{\rule{10em}{0ex}}\text{−}17,-19\hfill \\ \mathbf{\text{Step 6. Check}}\phantom{\rule{0.2em}{0ex}}\text{the answer.}\hfill & & & & & \\ \text{The results are consecutive odd integers}\hfill & & & & & \\ 17,19\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{−}19,-17.\hfill & & & & & \\ 17·19=323✓\phantom{\rule{4em}{0ex}}\text{−}19\left(\text{−}17\right)=323✓\hfill & & & & & \\ \text{Both pairs of consecutive integers are solutions.}\hfill & & & & & \\ \mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question}\hfill & & & & & \text{The consecutive integers are 17, 19 and}\phantom{\rule{0.2em}{0ex}}\text{−}19,-17.\hfill \end{array}

The product of two consecutive odd integers is 255. Find the integers.

-15,-17

The product of two consecutive odd integers is 483 Find the integers.

-23,-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 positive results.

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

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

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

The width is 5 feet and length is 6 feet.

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

The length of the patio is 12 feet and the width 15 feet.

\left({a}^{2}+{b}^{2}={c}^{2}\right).

We will use this formula to in the next example.

A boat’s sail is in the shape of a right triangle as shown. 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 sail.

Figure shows a right triangle with the shortest side being x, the second side being x minus 7 and the hypotenuse being 17.

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

5 feet and 12 feet

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 the other leg. Find the lengths of the hypotenuse and the other leg.

24 feet and 25 feet

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

h\left(t\right)=-16{t}^{2}+64t+80

ⓐ the zeros of this function which tell us when the ball hits the ground

ⓑ when the ball will be 80 feet above the ground

t=2

ⓐ the zeros of this function which tell us when the rock will hit the ocean

ⓑ when the rock will be 160 feet above the ocean.

t=1.5

ⓐ 5 ⓑ 0;3 ⓒ 196

h\left(t\right)=-16{t}^{2}+32t+128

ⓐ the zeros of this function which is when the penny will hit the ocean

ⓑ when the penny will be 128 feet above the ocean.

t=1

ⓐ 4 ⓑ 0;2 ⓒ 144

Access this online resource for additional instruction and practice with quadratic equations.

  • Beginning Algebra & Solving Quadratics with the Zero Property

Key Concepts

  • Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

a,b,c\phantom{\rule{0.2em}{0ex}}\text{are real numbers and}\phantom{\rule{0.2em}{0ex}}a\ne 0

Section Exercises

Practice makes perfect.

In the following exercises, solve.

\left(3a-10\right)\left(2a-7\right)=0

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

f\left(x\right)=9{x}^{2}-4

Solve Applications Modeled by Quadratic Equations

The product of two consecutive odd integers is 143. Find the integers.

-13,-11

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

-14,-12

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

-4

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The height of the carport is five feet less than twice its length. Find the height and the length of the carport.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

h\left(t\right)=-16{t}^{2}+32t

ⓐ the zeros of this function which tells us when the rocket will hit the ground. ⓑ the time the rocket will be 16 feet above the ground.

h\left(t\right)=-16{t}^{2}+32t+48

Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Answers will vary.

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve quadratic equations by using the zero product property, solve quadratic equations by factoring and solve applications modeled by quadratic equations.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Greatest common factor and factor by grouping.

Find the Greatest Common Factor of Two or More Expressions

In the following exercises, find the greatest common factor.

12{a}^{2}{b}^{3},15a{b}^{2}

Factor the Greatest Common Factor from a Polynomial

In the following exercises, factor the greatest common factor from each polynomial.

35y+84

Factor by Grouping

In the following exercises, factor by grouping.

ax-ay+bx-by

Factor Trinomials

{x}^{2}+bx+c

In the following exercises, factor completely using trial and error.

{x}^{3}+5{x}^{2}-24x

In the following exercises, factor.

2{x}^{2}+9x+4

Factor using substitution

In the following exercises, factor using substitution.

{x}^{4}-13{x}^{2}-30

Factor Special Products

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

25{x}^{2}+30x+9

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

81{r}^{2}-25

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

{a}^{3}-125

General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

24{x}^{3}+44{x}^{2}

In each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

f\left(x\right)=64{x}^{2}-49

The product of two consecutive numbers is 399. Find the numbers.

-21

The area of a rectangular shaped patio 432 square feet. The length of the patio is 6 feet more than its width. Find the length and width.

A ladder leans against the wall of a building. The length of the ladder is 9 feet longer than the distance of the bottom of the ladder from the building. The distance of the top of the ladder reaches up the side of the building is 7 feet longer than the distance of the bottom of the ladder from the building. Find the lengths of all three sides of the triangle formed by the ladder leaning against the building.

The lengths are 8, 15, and 17 ft.

Chapter Practice Test

80{a}^{2}+120{a}^{3}

In the following exercises, solve

5{a}^{2}+26a=24

The product of two consecutive integers is 156. Find the integers.

The area of a rectangular place mat is 168 square inches. Its length is two inches longer than the width. Find the length and width of the placemat.

The width is 12 inches and the length is 14 inches.

t=4

Intermediate Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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  11. Polynomial

    A polynomial is a function in one or more variables that consists of a sum of variables raised to nonnegative, integral powers and multiplied by coefficients from a predetermined set (usually the set of integers; rational, real or complex numbers; but in abstract algebra often an arbitrary field ). Note that a constant is also a polynomial.

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    Polynomial equations of degree one are linear equations are of the form ax+b=c.ax+b=c. We are now going to solve polynomial equations of degree two. A polynomial equation of degree two is called a quadratic equation. Listed below are some examples of quadratic equations: x2 + 5x + 6 = 0 3y2 + 4y = 10 64u2 − 81 = 0 n(n + 1) = 42.

  15. PDF Chapter 9 Polynomials Unit Plan

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  18. Polynomial Word Problems

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  19. Lesson Plan: Polynomial Functions

    Lesson Plan: Polynomial Functions Mathematics. Lesson Plan: Polynomial Functions. Start Practising. This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to identify, write, and evaluate a one-variable polynomial function and state its degree and leading coefficient.

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  23. Lesson PLAN Polynomial Function

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