Question

How do you factor the quadratics `6x^2-x-2` and `x^2-7x+12` ?

Answer 1

`6x^2-x-2 = (2x+1)(3x-2)`

`x^2-7x+12 = (x-3)(x-4)`

Explanation:

In order to find the appropriate split of the middle terms in each of these quadratics, we can use an AC method:

Example 1

Given:

`6x^2-x-2`

look for a pair of factors of `AC=6*2 = 12` which differ by `B=1`. Note that we look for a suitable difference because the coefficient of the constant term is negative.

The pair `4, 3` works in that `4 * 3 = 12` and `4 - 3 = 1`.

Use this pair to split the middle term and factor by grouping:

`6x^2-x-2 = (6x^2-4x)+(3x-2)`

`color(white)(6x^2-x-2) = 2x(3x-2)+1(3x-2)`

`color(white)(6x^2-x-2) = (2x+1)(3x-2)`

So this quadratic has zeros `x=-1/2` and `x=2/3`

Example 2

Given:

`x^2-7x+12`

look for a pair of factors of `AC=1*12 = 12` with sum `B=7`. Note that we look for a suitable sum because the coefficient of the constant term is positive.

The pair `4, 3` works in that `4 * 3 = 12` and `4 + 3 = 7`.

Use this pair to split the middle term and factor by grouping:

`x^2-7x+12 = (x^2-4x)-(3x-12)`

`color(white)(x^2-7x+12) = x(x-4)-3(x-4)`

`color(white)(x^2-7x+12) = (x-3)(x-4)`

So this quadratic has zeros `x=3` and `x=4`