# How do you factor completely #9x^2+9x-10#?

##### 1 Answer

#### Answer:

#### Explanation:

One method involves completing the square, then using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

First multiply through by

#4(9x^2+9x-10)#

#=36x^2+36x-40#

#=(6x+3)^2-9-40#

#=(6x+3)^2-7^2#

#=((6x+3)-7)((6x+3)+7)#

#=(6x-4)(6x+10)#

#=(2(3x-2))(2(3x+5))#

#=4(3x-2)(3x+5)#

Dividing both ends by

#9x^2+9x-10 = (3x-2)(3x+5)#

Why did I multiply by

The middle term of the original quadratic expression is odd, so if you do not multiply through by some multiple of

Without premultiplying, it looks like this:

#9x^2+9x-10#

#=(3x+3/2)^2-9/4-10#

#=(3x+3/2)^2-49/4#

#=(3x+3/2)^2-(7/2)^2#

#=((3x+3/2)-7/2)((3x+3/2)+7/2)#

#=(3x-2)(3x+5)#