# How do you factor #9q ^ { 2} - 7q - 18#?

##### 3 Answers

Well,

What we need to do is find two numbers that multiply to

*let's leave the sign out of it for now*

............

Well, that didn't work. We should check to make sure we even *can* factor this.

To do that, we need to graph the equation and see the roots (

graph{y=9x^2-7x-18}

So this does have

However, we can still find solutions, using the *quadratic formula*:

We can't simplify

So, we have two solutions:

**number #1#**

**number #2#**

can't be factored.

#### Explanation:

Since D is not a perfect square, this trinomial can't be factored.

#### Explanation:

The difference of squares identity can be written:

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

Use this with

#36(9q^2-7q-18) = 324q^2-252q-648#

#color(white)(36(9q^2-7q-18)) = (18q)^2-2(18q)7+49-697#

#color(white)(36(9q^2-7q-18)) = (18q-7)^2-(sqrt(697))^2#

#color(white)(36(9q^2-7q-18)) = ((18q-7)-sqrt(697))((18q-7)+sqrt(697))#

#color(white)(36(9q^2-7q-18)) = (18q-7-sqrt(697))(18q-7+sqrt(697))#

So:

#9q^2-7q-18 = 1/36(18q-7-sqrt(697))(18q-7+sqrt(697))#

Note that