Question

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

Answer 1

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

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` to make the arithmetic simpler, not forgetting to divide through by `4` at the end.

`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 `4` we get:

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

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Why did I multiply by `4` to start?

The middle term of the original quadratic expression is odd, so if you do not multiply through by some multiple of `4` first then you end up doing arithmetic with `1/2`'s and `1/4`'s. It was not necessary to multiply through by anything else to avoid fractions since the leading term `9x^2 = (3x)^2` was already a perfect square.

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)`