 # 10.02 Solving quadratic equations by factoring

Lesson

We have looked at many different ways to factor polynomials.

Recall that we can factor by:

• Using the greatest common factor
• Using knowledge of the difference of two squares or perfect squares
• Varied strategies for trinomials such as splitting and grouping in pairs

Once we can factor, we can solve equations algebraically to find the unknown value(s). There is a great benefit to factoring quadratics in order to solve them, and in order to understand why, we need to think about zero.

The property of $0$0 is very special. The only way two things that are being multiplied can have a product of $0$0, is if one, or both of those things are $0$0 themselves.

So if we have two factors, like $a$a and $b$b, and we multiply them together so that they equal $0$0, then one of those factors ($a$a or $b$b) must be $0$0. A written solution to a question like this would be similar to the following:

If $a\times b=0$a×b=0 then $a=0$a=0 or $b=0$b=0. This is known as the zero product property.

Zero product property

The product of any number and $0$0 is $0$0. That is:

If $ab=0$ab=0, then $a=0$a=0 or $b=0$b=0

#### Worked examples

##### Question 1

Solve $\left(x-2\right)\left(x+5\right)=0$(x2)(x+5)=0.

Think: This quadratic equation is already in factored form, so we can go straight to using the zero product property to find the solutions.

DoUsing the zero product property, either

 $\left(x-2\right)$(x−2) $=$= $0$0 or $\left(x+5\right)$(x+5) $=$= $0$0 $x$x $=$= $2$2 $x$x $=$= $-5$−5

So the solutions to the equation $\left(x-2\right)\left(x+5\right)=0$(x2)(x+5)=0 are $x=2$x=2 or $x=-5$x=5.

##### question 2

What are the solutions to the equation $x^2-25=0$x225=0?

Think: Notice that the left-hand side of this equation is actually a difference of two squares polynomial, and so we can rewrite it first in factored form.

Do: Let's first factor the equation, then apply the zero product property to find the solutions.

 $x^2-25$x2−25 $=$= $0$0 $\left(x-5\right)\left(x+5\right)$(x−5)(x+5) $=$= $0$0

Now we know from the zero product property that either $\left(x-5\right)=0$(x5)=0 or $\left(x+5\right)=0$(x+5)=0, and so either $x=5$x=5 or $x=-5$x=5.

##### question 3

Solve the equation $\left(2x-8\right)^2=0$(2x8)2=0.

Think: This perfect square quadratic equation is also already in a form that can make use of the zero product property.

Do: To make it clear, we'll rewrite the equation using the fact that $\left(\editable{}\right)^2=\editable{}\times\editable{}$()2=×.

 $\left(2x-8\right)^2$(2x−8)2 $=$= $0$0 $\left(2x-8\right)\left(2x-8\right)$(2x−8)(2x−8) $=$= $0$0

What does this last line mean? The zero product property tells us that either $\left(2x-8\right)=0$(2x8)=0 or $\left(2x-8\right)=0$(2x8)=0. But these are the same factor! This is actually an example of a double root, and we find the answer by considering the single factor $2x-8=0$2x8=0 by itself, which gives $x=4$x=4.

#### Practice questions

##### Question 4

Solve for the two possible values of $x$x:

$\left(x-7\right)\left(x-6\right)=0$(x7)(x6)=0

1. Write all solutions on the same line, separated by commas.

##### Question 5

Solve $3y-15y^2=0$3y15y2=0

1. Write all solutions on the same line, separated by commas.

##### Question 6

Solve the following equation by first factoring the left hand side of the equation.

$4x^2-17x+15=0$4x217x+15=0

1. Write all solutions on the same line, separated by commas.

### Outcomes

#### A1:A-SSE.B.3a

Factor a quadratic expression to reveal the zeros of the function it defines

#### A1:A-REI.B.4

Solve quadratic equations in one variable.