6.7: Solving Factorable Polynomial Equations

section 6.7 Learning Objectives

6.7: Solving Factorable Quadratic Equations

Use factoring techniques and the Principle of Zero Products to solve polynomial equations
Expand and then factor expressions to solve

Not all of the techniques we use for solving linear equations will apply to solving all polynomial equations. In this section, we will introduce a method for solving polynomial equations that combines factoring and the zero product principle.

The Principle of Zero Products

The number zero

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 $2$ and $5$ ? Could they be $9$ and $1$ ? No! When the result (answer) from multiplying two numbers is zero, that means that at least 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.

Principle of Zero Products

The Principle of Zero Products states that if the product of two factors is $0$ , then at least one of the factors is $0$ . Equivalently, if $ab=0$ , then either $a=0$ or $b=0$ , or both a and b are $0$ .

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.

Example 1

Solve:

$-t^2+t=0$

Show Solution

First, note that it is critical that the equation is equal to zero.

Since each term has a common factor of $t$ , we can factor and use the zero product principle. Recall that we typically choose to also factor out the negative in the case of a negative leading coefficient, so we will factor out $-t$ .

$-t^2+t=-t(t-1)$

Rewrite the polynomial equation using the factored form of the polynomial.

$\begin{array}{c}-t^2+t=0\\-t(t-1)=0\end{array}$

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

$\begin{array}{c}{\displaystyle\frac{-t}{-1}=\frac{0}{-1}}\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,t-1=0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+1}\,\,\,\,\,\underline{+1}\\ \,\,\, \,\,\,\,\,\,t=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}$

Answer

$t=0\text{ OR }t=1$

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 the next video, we show that you can use previously learned methods to factor a trinomial in order to solve a quadratic equation.

It will not always be the case that the given equation will have zero on one side. As shown in the next example, it is crucial to first move all terms to the same side, setting up an equation equal to zero.

Example 2

Solve: $s^2-4s=5$

Show Solution

First, move all the terms to one side. The goal is to try to use the zero product principle since that is the only tool we currently know for solving polynomial equations of degree larger than 1.

$\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}$

We now have all the terms on the same side and zero on the other side. The polynomial $s^2-4s-5$ factors nicely using the product and sum technique.

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

We now apply the zero product principle, separating our factors into two linear equations.

$\begin{array}{c}s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}$

$\begin{array}{c}s+1=0\\s=-1\end{array}$

Answer

$s=-1\text{ OR }s=5$

In Section 1.5, we introduced a technique for clearing fractions out of an equation. In the following “Think About It,” we combine this strategy with solving an equation by factoring.

thinK about it

Solve $y^2-5=-\frac{7}{2}y+\frac{5}{2}$

Show Solution

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 the LCD of $2$ to eliminate fractions:

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

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.

$\begin{array}{c}2y^2-10=-7y+5\\2y^2-10+7y-5=0\\2y^2+7y-15=0\end{array}$

We can now check whether this polynomial will factor. Using the AC-Method with the optional assistance of a table, we can list factors until we find two numbers with a product of $2\cdot-15=-30$ and a sum of 7.

Factors of $2\cdot-15=-30$	Sum of Factors
$1,-30$	$-29$
$-1,30$	$29$
$2,-15$	$-13$
$-2,15$	$13$
$3,-10$	$-7$
$-3,10$	$7$

We have found the factors that will produce the middle term we want, $-3,10$ . We now rewrite the middle term and proceed by grouping.

$2y^2-3y+10y-15=0$

$y(2y-3)+5(2y-3)$

$\left(2y-3\right)\left(y+5\right)=0$

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

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

It is worth noting that for all of these solving problems, you can always check to make sure your solutions are correct:

Check $y=\frac{3}{2}$ .

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

$y=\frac{3}{2}$ is indeed a solution.

Now check $y=-5$ .

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

$y=-5$ is also a solution, so we must have done something right!

Answer

$y=\frac{3}{2}\text{ OR }y=-5$

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

Be aware that we must be prepared to use any of the factoring techniques we learned throughout this chapter. The next example shows a difference of squares in the context of solving an equation.

Example 3

Solve $8x^2-50=0$ .

Show Solution

First notice that there is a GCF of 2 that we can factor out before applying the difference of squares.

$2(4x^2-25)=0$

$2(2x+5)(2x-5)$

Often students get confused by a constant like 2 in front. We could try setting this factor to zero, but 2 can never equal zero so this makes no sense. Another way of looking at this is that we are solving for the variable (in this problem, x) and there is no x in this factor, so we cannot obtain a solution from it.

However, we still have two other factors to apply the zero product principle to.

$2x+5=0$

$x=-\frac{5}{2}$

OR

$2x-5=0$

$x=\frac{5}{2}$

Answer

$x=-\frac{5}{2} \mbox{ OR } x=\frac{5}{2}$

Up to this point, every example has been a degree 2 polynomial. However, this process can be applied to any polynomial equation, as long as we know how to factor it. The next example shows how factoring can help us solve a degree 3 polynomial equation.

Example 4

Solve $3z^3-9z^2-30z=0$ .

Show Solution

Like the previous example, there is a GCF. However, this time the GCF includes a variable. After factoring out the common factor of $3z$ , we can then apply the product and sum method to factor the resulting trinomial.

$3z(z^2-3z-10)=0$

$3z(z-5)(z+2)=0$

Setting the first factor of $3z$ equal to zero will result in a solution!

$3z=0$

$z=0$

$z-5=0$

$z=5$

$z+2=0$

$z=-2$

Answer

$z=0 \mbox{ OR } z=5 \mbox{ OR } z=-2$

A major theme of this section is that the polynomial equation must equal to zero prior to factoring. Our last example stresses this one more time.

Example 5

Solve $(x-3)(x+4)=8$ .

Show Solution

An all-too common error is to set each factor equal to 8, but it is absolutely not true that if the product of two numbers is 8, one of them must equal 8. This is (one reason) 0 is such a special number.

So first, we must “unfactor” (multiply out the left side of) this problem so that we can move the 8 over and start fresh.

$(x-3)(x+4)=8$

$x^2+4x-3x-12=8$

$x^2+x-12=8$

$x^2+x-20=0$

Now that we have a polynomial equal to zero, we can factor and apply the zero product principle.

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

Setting each to zero yields

$x+5=0$

$x=-5$

$x-4=0$

$x=4$

Answer

$x=-5 \mbox{ OR } x=4$

Module 6: Factoring Polynomials