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4.02 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

$\left(x-2\right)$(x2) $=$= $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.

$x^2-25$x225 $=$= $0$0
$\left(x-5\right)\left(x+5\right)$(x5)(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=×.

$\left(2x-8\right)^2$(2x8)2 $=$= $0$0
$\left(2x-8\right)\left(2x-8\right)$(2x8)(2x8) $=$= $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

II.A.SSE.3.a

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

II.A.CED.1

Create equations and inequalities in one variable and use them to solve problems.

II.A.REI.4

Solve quadratic equations in one variable.

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