Mathematics » Factoring and Factorisation I » Quadratic Equations

# Solving Quadratic Equations By Factoring

## Solving 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, $$a{x}^{2}+bx+c=0$$. Then we factor the expression on the left.

### Example: How to Solve a Quadratic Equation by Factoring

Solve: $${x}^{2}+2x-8=0$$.

### Solve a quadratic equation by factoring.

1. Write the quadratic equation in standard form, $$a{x}^{2}+bx+c=0$$.
3. Use the Zero Product Property.
4. Solve the linear equations.
5. Check.

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

## Example

Solve: $$2{y}^{2}=13y+45$$.

### Solution

 $$\phantom{\rule{4.65em}{0ex}}2{y}^{2}=13y+45$$ Write the quadratic equation in standard form. $$2{y}^{2}-13y-45=0$$ Factor the quadratic expression. $$\phantom{\rule{0.7em}{0ex}}\left(2y+5\right)\left(y-9\right)=0$$ Use the Zero Product Property to set each factor to 0. $$\phantom{\rule{3.46em}{0ex}}2y+5=0$$ $$y-9=0$$ Solve each equation. $$\phantom{\rule{5.68em}{0ex}}y=-\frac{5}{2}$$ $$y=9$$ Check your answers. This image shows the steps for solving the equation 2 y squared = 13 y + 45. The first step is writing the equation in standard quadratic form, 2 y squared – 13 y – 45 = 0. The second step is to factor the quadratic expression, (2 y + 5)(y – 9) = 0. The third step is to use the zero product property to set each factor equal to 0, 2 y + 5 = 0 or y – 9 = 0. Solve each equation, y = −5/2 or y = 9. Finally, check the answers by substituting them back into the original equation.

## Example

Solve: $$5{x}^{2}-13x=7x$$.

### Solution

 $$5{x}^{2}-13x=7x$$ Write the quadratic equation in standard form. $$5{x}^{2}-20x=0\phantom{\rule{0.5em}{0ex}}$$ Factor the left side of the equation. $$5x\left(x-4\right)=0\phantom{\rule{0.5em}{0ex}}$$ Use the Zero Product Property to set each factor to 0. $$5x=0\phantom{\rule{0.5em}{0ex}}$$ $$x-4=0$$ Solve each equation. $$x=0\phantom{\rule{0.5em}{0ex}}$$ $$x=4$$ Check your answers. This image shows the steps for solving the equation 5 x squared – 13 x = 7 x. The first step is writing the equation in standard quadratic form, 5 x squared – 20 x = 0. The second step is to factor the quadratic expression, 5 x(x – 4)= 0. The third step is to use the zero product property to set each factor equal to 0, 5 x = 0 or x − 4 = 0. Solve each equation, x = 0 or x = 4. Finally, check the answers by substituting them back into the original equation.

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

## Example

Solve: $$144{q}^{2}=25$$.

### Solution

$$\begin{array}{cccc}\begin{array}{}\text{Write the quadratic equation in standard form.}\hfill \\ \text{Factor. It is a difference of squares.}\hfill \end{array}\hfill & & & \begin{array}{ccc}\hfill 144{q}^{2}& =\hfill & 25\hfill \\ \hfill 144{q}^{2}-25& =\hfill & 0\hfill \\ \hfill \left(12q-5\right)\left(12q+5\right)& =\hfill & 0\hfill \end{array}\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to}\phantom{\rule{0.2em}{0ex}}0.\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\begin{array}{}\\ \\ \hfill 12q-5& =\hfill & 0\hfill & & & \hfill 12q+5& =\hfill & 0\hfill \\ \hfill 12q& =\hfill & 5\hfill & & & \hfill 12q& =\hfill & -5\hfill \\ \hfill q& =\hfill & \frac{5}{12}\hfill & & & \hfill q& =\hfill & -\frac{5}{12}\hfill \end{array}\hfill \\ \text{Check your answers.}\hfill & & & \end{array}$$

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.

## Example

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

### Solution

$$\begin{array}{cccc}\begin{array}{}\text{Multiply the binomials.}\hfill \\ \text{Write the quadratic equation in standard form.}\hfill \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & \begin{array}{ccc}\hfill \left(3x-8\right)\left(x-1\right)& =\hfill & 3x\hfill \\ \hfill 3{x}^{2}-11x+8& =\hfill & 3x\hfill \\ \hfill 3{x}^{2}-14x+8& =\hfill & 0\hfill \\ \hfill \left(3x-2\right)\left(x-4\right)& =\hfill & 0\hfill \end{array}\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \text{Solve each equation.}\hfill \\ \end{array}\hfill & & & \phantom{\rule{3.3em}{0ex}}\begin{array}{}\\ \hfill 3x-2& =\hfill & 0\hfill & & & \hfill x-4& =\hfill & 0\hfill \\ \hfill 3x& =\hfill & 2\hfill & & & \hfill x& =\hfill & 4\hfill \\ \hfill x& =\hfill & \frac{2}{3}\hfill \end{array}\hfill \\ \text{Check your answers.}\hfill & & & \text{The check is left to you!}\hfill \end{array}$$

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.

## Example

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

### Solution

$$\begin{array}{cccc}\begin{array}{}\text{Bring all the terms to one side so that the other side is zero.}\hfill \\ \text{Factor the greatest common factor first.}\hfill \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & \begin{array}{ccc}\hfill 9{m}^{3}+100m& =\hfill & 60{m}^{2}\hfill \\ \hfill 9{m}^{3}-60{m}^{2}+100m& =\hfill & 0\hfill \\ \hfill m\left(9{m}^{2}-60m+100\right)& =\hfill & 0\hfill \\ \hfill m\left(3m-10\right)\left(3m-10\right)& =\hfill & 0\hfill \end{array}\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to}\phantom{\rule{0.2em}{0ex}}0.\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & \phantom{\rule{0.5em}{0ex}}\begin{array}{ccccccccccccc}\hfill m& =\hfill & 0\hfill & & & \hfill 3m-10& =\hfill & 0\hfill & & & \hfill 3m-10& =\hfill & 0\hfill \\ \hfill m& =\hfill & 0\hfill & & & \hfill m& =\hfill & \frac{10}{3}\hfill & & & \hfill m& =\hfill & \frac{10}{3}\hfill \end{array}\hfill \\ \text{Check your answers.}\hfill & & & \text{The check is left to you.}\hfill \end{array}$$

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.

## Example

Solve: $$4{x}^{2}=16x+84$$.

### Solution

$$\begin{array}{cccc}\begin{array}{}\text{Write the quadratic equation in standard form.}\hfill \\ \text{Factor the greatest common factor first.}\hfill \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & \phantom{\rule{1em}{0ex}}\begin{array}{ccc}\hfill 4{x}^{2}& =\hfill & 16x+84\hfill \\ \hfill 4{x}^{2}-16x-84& =\hfill & 0\hfill \\ \hfill 4\left({x}^{2}-4x-21\right)& =\hfill & 0\hfill \\ \hfill 4\left(x-7\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & \phantom{\rule{0.4em}{0ex}}\begin{array}{ccccccccccccc}\hfill 4& \ne \hfill & 0\hfill & & & \hfill x-7& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill 4& \ne \hfill & 0\hfill & & & \hfill x& =\hfill & 7\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill \\ \text{Check your answers.}\hfill & & & \text{The check is left to you.}\hfill \end{array}$$

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