How do you factor 2x^2-3x-9?

1 Answer
Dec 6, 2016

2x^2-3x-9 = (2x+3)(x-3)

Explanation:

The rule to factorise any quadratic is to find two numbers such that

"product" = x^2 " coefficient "xx" constant coefficient"
"sum" \ \ \ \ \ \ = x " coefficient"

So for 2x^2-3x-9 we seek two numbers such that

"product" = (2)*(-9) = -18
"sum" \ \ \ \ \ \ = -3

So we look at the factors of -18. As the product is negative one of the factors must also be negative and the other positive, We compute their sum we get

{: ("factor1", "factor2", "sum"), (18,-1,17), (9,-2,7), (6,-3,3), (-18,1,-17),(-9,2,-7), (-6,3,-3) :}

So the factors we seek are color(blue)(-6) and color(green)(3)

Therefore we can factorise the quadratic as follows:

\ \ \ \ \ 2x^2-3x-9 = 2x^2 color(blue)(-6)x + color(green)(3)x -9
:. 2x^2-3x-9 = 2x(x-3) + 3(x -3)
:. 2x^2-3x-9 = (2x+3)(x-3)

This approach works for all quadratics (assuming it does factorise) , The middle step in the last section can usually be skipped with practice.