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

##### 1 Answer

#### 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 *differ* by *difference* because the coefficient of the constant term is negative.

The pair

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

**Example 2**

Given:

#x^2-7x+12#

look for a pair of factors of *sum* *sum* because the coefficient of the constant term is positive.

The pair

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