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# 9.4: Polynomial Equations in Factored Form

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the zero-product property.
• Find greatest common monomial factors.
• Solve simple polynomial equations by factoring.

## Introduction

In the last few sections, we learned how to multiply polynomials by using the Distributive Property. All the terms in one polynomial had to be multiplied by all the terms in the other polynomial. In this section, you’ll start learning how to do this process in reverse. The reverse of distribution is called factoring.

The total area of the figure above can be found in two ways.

We could find the areas of all the small rectangles and add them: $ab+ac+ad+ae+2a$.

Or, we could find the area of the big rectangle all at once. Its width is $a$ and its length is $b+c+d+e+2$, so its area is $a(b+c+d+e+2)$.

Since the area of the rectangle is the same no matter what method we use, those two expressions must be equal.

$ab+ac+ad+ae+2a = a(b+c+d+e+2)$

To turn the right-hand side of this equation into the left-hand side, we would use the distributive property. To turn the left-hand side into the right-hand side, we would need to factor it. Since polynomials can be multiplied just like numbers, they can also be factored just like numbers—and we’ll see later how this can help us solve problems.

## Find the Greatest Common Monomial Factor

You will be learning several factoring methods in the next few sections. In most cases, factoring takes several steps to complete because we want to factor completely. That means that we factor until we can’t factor any more.

Let’s start with the simplest step: finding the greatest monomial factor. When we want to factor, we always look for common monomials first. Consider the following polynomial, written in expanded form:

$ax+bx+cx+dx$

A common factor is any factor that appears in all terms of the polynomial; it can be a number, a variable or a combination of numbers and variables. Notice that in our example, the factor $x$ appears in all terms, so it is a common factor.

To factor out the $x$, we write it outside a set of parentheses. Inside the parentheses, we write what’s left when we divide each term by $x$:

$x(a+b+c+d)$

Let’s look at more examples.

Example 1

Factor:

a) $2x+8$

b) $15x-25$

c) $3a+9b+6$

Solution

a) We see that the factor 2 divides evenly into both terms: $2x + 8 = 2(x) + 2(4)$

We factor out the 2 by writing it in front of a parenthesis: $2( \ )$

Inside the parenthesis we write what is left of each term when we divide by 2: $2(x + 4)$

b) We see that the factor of 5 divides evenly into all terms: $15x-25= 5(3x)-5(5)$

Factor out the 5 to get: $5(3x-5)$

c) We see that the factor of 3 divides evenly into all terms: $3a + 9b + 6 = 3(a) + 3(3b) + 3(2)$

Factor 3 to get: $3(a + 3b + 2)$

Example 2

Find the greatest common factor:

a) $a^3-3a^2+4a$

b) $12a^4-5a^3+7a^2$

Solution

a) Notice that the factor $a$ appears in all terms of $a^3-3a^2+4a$, but each term has $a$ raised to a different power. The greatest common factor of all the terms is simply $a$.

So first we rewrite $a^3-3a^2+4a$ as $a(a^2) + a(-3a) + a(4)$.

Then we factor out the $a$ to get $a(a^2 - 3a + 4)$.

b) The factor $a$ appears in all the terms, and it’s always raised to at least the second power. So the greatest common factor of all the terms is $a^2$.

We rewrite the expression $12a^4-5a^3+7a^2$ as $(12a^2 \cdot a^2) - (5a \cdot a^2) + (7 \cdot a^2)$

Factor out the $a^2$ to get $a^2(12a^2 - 5a + 7)$.

Example 3

Factor completely:

a) $3ax+9a$

b) $x^3y+xy$

c) $5x^3y-15x^2y^2+25xy^3$

Solution

a) Both terms have a common factor of 3, but they also have a common factor of $a$. It’s simplest to factor these both out at once, which gives us $3a(x+3)$.

b) Both $x$ and $y$ are common factors. When we factor them both out at once, we get $xy(x^2+1)$.

c) The common factors are 5, $x$, and $y$. Factoring out $5xy$ gives us $5xy(x^2-3xy+5xy^2)$.

## Use the Zero-Product Property

The most useful thing about factoring is that we can use it to help solve polynomial equations.

For example, consider an equation like $2x^2 + 5x - 42 = 0$. There’s no good way to isolate $x$ in this equation, so we can’t solve it using any of the techniques we’ve already learned. But the left-hand side of the equation can be factored, making the equation $(x + 6)(2x - 7)=0$.

How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at least one of those numbers must be zero. This is called the Zero-Product Property.

What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on the left-hand side must equal zero. So we can find the two solutions by setting each factor equal to zero and solving each equation separately.

Setting the factors equal to zero gives us:

$(x + 6) = 0 && \text{OR} && (2x - 7) = 0$

Solving both of those equations gives us:

$& x + 6 = 0 && && 2x - 7 =0\\& \underline{\underline{x = -6}} && \text{OR} && 2x = 7\\& && && \underline{\underline{x = \frac{7}{2}}}$ Notice that the solution is $x = -6$ OR $x = \frac{7}{2}$. The OR means that either of these values of $x$ would make the product of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct.

$& Check: x = - 6; && Check: x = \frac{7}{2}\\& (x + 6)(2x - 7)= && (x + 6)(2x - 7)=\\& (-6 +6)(2(-6) -7)= && \left ( \frac{7}{2} + 6 \right ) \left (2 \cdot \frac{7}{2} - 7 \right ) =\\& (0)(-19) = 0 && \left (\frac{19}{2} \right ) (7 - 7) = \\& && \left (\frac{19}{2} \right ) (0) = 0$

Both solutions check out.

Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into simpler separate steps. When we can’t factor a polynomial, the problem becomes harder and we must use other methods that you will learn later.

As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero. For example, if you multiplied two numbers and the answer was nine, that wouldn’t mean that one or both of the numbers must be nine. In order to use the property, the factored polynomial must be equal to zero.

Example 4

Solve each equation:

a) $(x - 9)(3x + 4)=0$

b) $x(5x - 4) = 0$

c) $4x (x+6)(4x - 9)=0$

Solution

Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler equations separately

a) $(x - 9)(3x + 4) = 0$ can be split up into two linear equations:

$& x - 9 = 0 && && 3x + 4 = 0\\& \underline{\underline{x = 9}} && \text{or} && 3x = -4\\& && && \underline{\underline{x = - \frac{4}{3}}}$

b) $x(5x - 4) = 0$ can be split up into two linear equations:

$& && && 5x - 4 = 0\\& \underline{\underline{x = 0}} && \text{or} && 5x = 4\\& && && \underline{\underline{x = \frac{4}{5}}}$

c) $4x(x + 6)(4x - 9) =0$ can be split up into three linear equations:

$& 4x = 0 && && x + 6 = 0 && && 4x - 9 =0\\& x= \frac{0}{4} && \text{or} && x = -6 && \text{or} && 4x = 9\\& \underline{\underline{x = 0}} && && && && \underline{\underline{x = \frac{9}{4}}}$

## Solve Simple Polynomial Equations by Factoring

Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we can use the Zero-Product Property to solve polynomials in factored form—now we can use that knowledge to solve polynomials by factoring them first. Here are the steps:

a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero.

b) Factor the polynomial completely.

c) Use the zero-product rule to set each factor equal to zero.

d) Solve each equation from step 3.

e) Check your answers by substituting your solutions into the original equation

Example 5

Solve the following polynomial equations.

a) $x^2 - 2x =0$

b) $2x^2 = 5x$

c) $9x^2 y - 6xy = 0$

Solution

a) $x^2 - 2x = 0$

Rewrite: this is not necessary since the equation is in the correct form.

Factor: The common factor is $x$, so this factors as $x(x-2)=0$.

Set each factor equal to zero:

$x = 0 && \text{or} && x - 2 = 0$

Solve:

$\underline{x = 0} && \text{or} && \underline{x = 2}$

Check: Substitute each solution back into the original equation.

$x & = 0 \Rightarrow (0)^2 - 2(0) = 0 && \text{works out}\\x & = 2 \Rightarrow (2)^2 - 2(2) = 4 - 4 = 0 && \text{works out}$

Answer: $x = 0, x = 2$

b) $2x^2 = 5x$

Rewrite: $2x^2 = 5x \Rightarrow 2x^2 - 5x = 0$

Factor: The common factor is $x$, so this factors as $x(2x - 5)=0$.

Set each factor equal to zero:

$x = 0 && \text{or} && 2x - 5 = 0$

Solve:

$& \underline{x = 0} && \text{or} && 2x = 5\\& &&&& \underline{x = \frac{5}{2}}$

Check: Substitute each solution back into the original equation.

$x & = 0 \Rightarrow 2(0)^2 = 5(0) \Rightarrow 0 = 0 && \text{works out}\\x & = \frac{5}{2} \Rightarrow 2 \left ( \frac{5}{2} \right )^2 = 5 \cdot \frac{5}{2} \Rightarrow 2 \cdot \frac{25}{4} = \frac{25}{2} \Rightarrow \frac{25}{2} = \frac{25}{2} && \text{works out}$

Answer: $x = 0, x =\frac{5}{2}$

c) $9x^2y - 6xy = 0$

Rewrite: not necessary

Factor: The common factor is $3xy$, so this factors as $3xy(3x - 2)=0$.

Set each factor equal to zero:

$3 = 0$ is never true, so this part does not give a solution. The factors we have left give us:

$x = 0 && \text{or} && y = 0 && \text{or} && 3x - 2 = 0$

Solve:

$& \underline{x = 0} && \text{or} && \underline{y = 0} && \text{or} && 3x = 2\\& &&&& \underline{x = \frac{2}{3}}$

Check: Substitute each solution back into the original equation.

$& x = 0 \Rightarrow 9(0)y - 6(0)y = 0 - 0 = 0 && \text{works out}\\& y = 0 \Rightarrow 9x^2 (0) - 6x(0) = 0 - 0 = 0 && \text{works out}\\& x = \frac{2}{3} \Rightarrow 9 \cdot \left ( \frac{2}{3} \right)^2 y - 6 \cdot \frac{2}{3} y = 9 \cdot \frac{4}{9} y - 4y = 4y - 4y = 0 && \text{works out}$

Answer: $x = 0, y = 0, x = \frac{2}{3}$

## Review Questions

Factor out the greatest common factor in the following polynomials.

1. $2x^2 - 5x$
2. $3x^3 - 21x$
3. $5x^6 + 15x^4$
4. $4x^3 + 10x^2 - 2x$
5. $-10x^6 + 12x^5 - 4x^4$
6. $12xy + 24xy^2 + 36xy^3$
7. $5a^3 - 7a$
8. $3y + 6z$
9. $10a^3 - 4ab$
10. $45y^{12} + 30y^{10}$
11. $16xy^2 z + 4x^3 y$
12. $2a - 4a^2 + 6$
13. $5xy^2 - 10xy + 5y^2$

Solve the following polynomial equations.

1. $x(x + 12) = 0$
2. $(2x + 1)(2x - 1) = 0$
3. $(x - 5)(2x + 7)(3x - 4) = 0$
4. $2x(x + 9)(7x - 20) = 0$
5. $x(3 + y) = 0$
6. $x(x - 2y) = 0$
7. $18y - 3y^2 = 0$
8. $9x^2 = 27x$
9. $4a^2 + a = 0$
10. $b^2 - \frac{5}{3}b = 0$
11. $4x^2 = 36$
12. $x^3 - 5x^2 = 0$

## Date Created:

Feb 23, 2012

Feb 12, 2015
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