Solving Polynomial Equations With a Greatest Common Factor

Learning Objectives

Factor the greatest common monomial out of a polynomial
Solve a polynomial in factored form by setting it equal to zero

In this section we will apply factoring a monomial from a polynomial to solving polynomial equations. Recall that not all of the techniques we use for solving linear equations will apply to solving polynomial equations, so we will be using the zero product principle to solve for a variable.

We will begin with an example where the polynomial is already equal to zero.

Example

Solve:

$-t^2+t=0$

Show Solution

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

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

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

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

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.

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

Answer

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

Notice how we were careful with signs in the last example. Even though one of the terms was negative, we factored out the positive common term of t. In the next example we will see what to do when the polynomial you are working with is not set equal to zero. Int eh followign video, we present more examples of solving quadratic equations by factoring.

Example

Solve: $6t=3t^2-12t$

Show Solution

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.

$\begin{array}{c}\,\,\,\,\,\,\,6t=3t^2-12t\\\underline{-6t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-6t}\\\,\,\,\,\,\,\,0=3t^2-18t\\\end{array}$

We now have all the terms on the right side, and zero on the other side. Each term has a common factor of 3t, so we can factor and use the zero product principle. If you are uncertain how we found the common factor 3t, review the section on greatest common factor.

Rewrite each term as the product of the GCF and the remaining terms. Note how we leave the negative sign on the 6 when we factor 3t out of -18t.

$\begin{array}{c}3t^2=3t\left(t\right)\\-18t=3t\left(-6\right)\end{array}$

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

$\begin{array}{c}0=3t^2-18t\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=3t\left(t\right)+3t\left(-6\right)\\0=3t\left(t-6\right)\end{array}$

Solve the two equations.

$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=t-6\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=3t\\\,\,\,\,\,\,\,\,\,\underline{+6}\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{0}{3}=\frac{3t}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,6=t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=t\end{array}$

Answer

$t=6\text{ OR }t=0$

The video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.

We will work through one more example that is similar to the ones above, except this example has fractions and the greatest common monomial is negative.

Example

Solve $\frac{1}{2}y=-4y-\frac{1}{2}y^2$

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. We will work through the second way.

First, we will find a common denominator, then factor fractions.

$\begin{array}{l}\frac{1}{2}y=-4y-\frac{1}{2}y^2\\0=-\frac{1}{2}y-4y-\frac{1}{2}y^2\end{array}$

2 is a common denominator for $-\frac{1}{2}y\text{ and }-4y$

$\frac{2}{2}\cdot{-4y}=-\frac{8y}{2}$

Rewrite the equation with the common denominator, then combine like terms.

$\begin{array}{l}0=-\frac{1}{2}y-4y-\frac{1}{2}y^2\\\text{ }\\0=-\frac{1}{2}y-\frac{8y}{2}-\frac{1}{2}y^2\\\text{}\\0=-\frac{9}{2}y-\frac{1}{2}y^2\end{array}$

Find the greatest common factor of the terms of the polynomial:

Factors of $-\frac{9}{2}y$

$-\frac{1}{2}\cdot{3}\cdot{3}\cdot{y}$

Factors of $-\frac{1}{2}y^2$

$-\frac{1}{2}\cdot{y}\cdot{y}$

Both terms have $-\frac{1}{2}\text{ and }y$ in common.

Rewrite each term as the product of the GCF and the remaining terms.

$-\frac{9}{2}y=-\frac{1}{2}y\left(3\cdot{3}\right)=-\frac{1}{2}y\left(9\right)$

$-\frac{1}{2}y^2=-\frac{1}{2}y\left(y\right)$

Rewrite the polynomial equation using the factored terms in place of the original terms. Pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. $\left(9+y\right)$

$\begin{array}{l}0=-\frac{9}{2}y-\frac{1}{2}y^2\\\text{}\\-\frac{1}{2}y\left(9\right)-\frac{1}{2}y\left(y\right)\\\text{}\\-\frac{1}{2}y\left(9+y\right)\end{array}$

Solve the two equations.

$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=9+y\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=-\frac{1}{2}y\\\,\,\,\,\,\,\,\,\,\underline{-9}\,\,\,\,\,\,\underline{-9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,-9=y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=y\end{array}$

Answer

$y=-9\text{ OR }y=0$

Wow! In the last example, we used many skills to solve one equation. Let’s summarize them:

We needed a common denominator to combine the like terms $-4y\text{ and }-\frac{1}{2}y$ , after we moved all the terms to one side of the equation
We found the GCF of the terms $-\frac{9}{2}y\text{ and }-\frac{1}{2}y^2$
We used the GCF to factor the polynomial $-\frac{9}{2}y-\frac{1}{2}y^2$
We used the zero product principle to solve the polynomial equation $0=-\frac{1}{2}y\left(9+y\right)$

Sometimes solving an equation requires the combination of many algebraic principles and techniques. The last facet of solving the polynomial equation $\frac{1}{2}y=-4y-\frac{1}{2}y^2$ that we should talk about is negative signs.

We found that the GCF $-\frac{1}{2}y$ contained a negative coefficient. This meant that when we factored it out of all the terms in the polynomial, we were left with two positive factors, 9 and y. This explains why we were left with $\left(9+y\right)$ as one of the factors of our final product.

In the following video we present another example of solving a quadratic polynomial with fractional coefficients using factoring and the zero product principle.

If the GCF of a polynomial is negative, pay attention to the signs that are left when you factor it from the terms of a polynomial.

In the next unit, we will learn more factoring techniques that will allow you to be able to solve a wider variety of polynomial equations such as $3x^2-x=2$ .

Summary

In this section we practiced using the zero product principle as a method for solving polynomial equations. We found that the techniques we used to solve linear equations did not give us the correct answer when used to solve a polynomial equation. We also found that a polynomial can be rewritten as a product by factoring out the greatest common factor. We used both factoring and the zero product principle to solve second degree polynomials.

Objective 3