Question

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

Answer 1

`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.