3.14 Solving quadratic equations by factoring

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.

 

 

Worked examples

Question 1

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

Do: Using 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 $\left(x-2\right)\left(x+5\right)=0$(x−2)(x+5)=0 are $x=2$x=2 or $x=-5$x=−5.

question 2

What are the solutions to the equation $x^2-25=0$x2−25=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 $\left(x-5\right)=0$(x−5)=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 $\left(2x-8\right)^2=0$(2x−8)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 $\left(2x-8\right)=0$(2x−8)=0 or $\left(2x-8\right)=0$(2x−8)=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$2x−8=0 by itself, which gives $x=4$x=4.

 

Practice questions

Question 4

Question 5

Question 6