Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

8.9.2 • Using Factored Form and the Zero Product Property to Solve Quadratic Equations

Activity

To solve the equation $n^{2} - 2 n = 99$ , Tyler wrote out the following steps. Analyze Tyler's work by explaining what Tyler did in each step.

Original equation $n^{2} - 2 n = 99$

1.

Step 1 - $n^{2} - 2 n - 99 = 0$

Compare your answer:

Step 1 - Subtract 99 from each side.

2.

Step 2 - $(n - 11) (n + 9) = 0$

Compare your answer:

Step 2 - Rewrite the left side in factored form.

3.

Step 3 - $n - 11 = 0$ or $n + 9 = 0$

Compare your answer:

Step 3 - Apply the zero product property, which leads to two separate equations.

4.

Step 4 - $n = 11$ and $n = - 9$

Compare your answer:

Step 4 - Solve each equation by performing the same operation on each side.

For questions 5 – 9, solve each equation by rewriting it in factored form and using the zero product property .

5.

$x^{2} + 8 x + 15 = 0$

Compare your answer:

$x = - 3$ and $x = - 5$ because $(x + 3) (x + 5) = 0$ .

6.

$x^{2} - 8 x + 12 = 5$

Compare your answer:

$x = 7$ and $x = 1$ because $(x - 7) (x - 1) = 0$ .

7.

$x^{2} - 10 x - 11 = 0$

Compare your answer:

$x = 11$ and $x = - 1$ because $(x - 11) (x + 1) = 0$ .

8.

$49 - x^{2} = 0$

Compare your answer:

$x = 7$ and $x = - 7$ because $(7 - x) (7 + x) = 0$ .

9.

$(x + 4) (x + 5) - 30 = 0$

Compare your answer:

$x = - 10$ and $x = 1$ because $(x + 10) (x - 1) = 0$ .

Are you ready for more?

Extending Your Thinking

1.

Solve this equation and explain or show your reasoning.

$(x^{2} - x - 20) (x^{2} + 2 x - 3) = (x^{2} + 2 x - 8) (x^{2} - 8 x + 15)$

Compare your answer:

$x = 5$ , $x = - 4$ , and $x = \frac{9}{7}$

Rewriting each quadratic expression in factored form gives:

$(x - 5) (x + 4) (x - 1) (x + 3) = (x + 4) (x - 2) (x - 5) (x - 3)$

Step 1 - Writing the equation so that it equals zero yields:

$(x - 5) (x + 4) (x - 1) (x + 3) - (x + 4) (x - 2) (x - 5) (x - 3) = 0$

Step 2 - Notice that the factors $(x - 5)$ and $(x + 4)$ appear in each of the terms. This means we can factor those binomials to the front of our expression. This means our expression is a list of factors and not terms.

$(x - 5) (x + 4) [(x - 1) (x + 3) - (x - 2) (x - 3)] = 0$

Step 3 - Now, using the zero product property, we set each of the factors equal to zero and solve.

$(x - 5) = 0$ , so $x = 5$ .

$(x + 4) = 0$ , so $x = - 4$ .

$[(x - 1) (x + 3) - (x - 2) (x - 3)] = 0$

Solving this equation will take a little more work!

$x^{2} + 2 x - 3 - (x^{2} - 5 x + 6) = 0$

$x^{2} + 2 x - 3 - x^{2} + 5 x - 6 = 0$

$7 x - 9 = 0$

$7 x = 9$

$x = \frac{9}{7}$

Solving the linear equation gives $x = \frac{9}{7}$ .

Self Check

Solve the equation by rewriting it in factored form and then using the zero product property.

$x^2+ 3x − 59 = −5$

$x=-9$ and $x=-6$
$x=-9$ and $x=6$
$x=9$ and $x=6$
$x=9$ and $x=-6$

Additional Resources

Using Factored Form and the Zero Product Property to Solve Quadratic Equations

We have seen quadratic equations in many different forms. Let’s look at an example in which we find the factored form and then use the zero product property to solve.

Example 1

Solve $x^{2} + 3 x - 50 = - 10$ .

Step 1 - Set the equation equal to 0 by adding 10 to both sides.

$x^{2} + 3 x - 40 = 0$

Step 2 - Find the factored form.

Which numbers have a product of –40 and a sum of +3?

$(x + 8) (x - 5) = 0$

Step 3 - Use the zero product property to solve.

Set each factor equal to 0 and solve.

$(x + 8) = 0 (x - 5) = 0$

$x = - 8 x = 5$

The solution is $x = - 8$ and $x = 5$ .

Example 2

Solve $x^{2} + 12 x + 21 = - 15$ .

Step 1 - Set the equation equal to 0 by adding 15 to both sides.

$x^{2} + 12 x + 36 = 0$

Step 2 - Find the factored form.

Which numbers have a product of 36 and a sum of 12?

$(x + 6) (x + 6) = 0$

Step 3 - Use the zero product property to solve.

Set each factor equal to 0 and solve. The factors are the same, so there is one solution.

$(x + 6) = 0 (x + 6) = 0$

$x = - 6 x = - 6$

The only solution is $x = - 6$ .

Try it

Using Factored Form and the Zero Product Property to Solve Quadratic Equations

1.

Solve $z^{2} - 5 z - 48 = - 12$ .

Here is how to solve the quadratic equations using the factored form and the zero product property:

$z^{2} - 5 z - 48 = - 12$

Step 1 - Set the equation equal to 0.

$z^{2} - 5 z - 36 = 0$

Step 2 - Find the factored form.

$(z - 9) (z + 4) = 0$

Step 3 - Use the zero product property to solve.

$z = 9$ and $z = - 4$

2.

Solve $p^{2} - 3 p - 108 = - 20$ .

Here is how to solve the quadratic equations using the factored form and the zero product property:

$p^{2} - 3 p - 108 = - 20$

Step 1 - Set the equation equal to 0.

$p^{2} - 3 p - 88 = 0$

Step 2 - Find the factored form.

$(p - 11) (p + 8) = 0$

Step 3 - Use the zero product property to solve.

$p = 11$ and $p = - 8$

8.9.2 Using Factored Form and the Zero Product Property to Solve Quadratic Equations

Activity

Are you ready for more?

Extending Your Thinking

Additional Resources

Using Factored Form and the Zero Product Property to Solve Quadratic Equations

Using Factored Form and the Zero Product Property to Solve Quadratic Equations