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

Dec 8, 2017

$6 {x}^{2} - x - 2 = \left(2 x + 1\right) \left(3 x - 2\right)$

${x}^{2} - 7 x + 12 = \left(x - 3\right) \left(x - 4\right)$

#### 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:

$6 {x}^{2} - x - 2$

look for a pair of factors of $A C = 6 \cdot 2 = 12$ 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 $4 \cdot 3 = 12$ and $4 - 3 = 1$.

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

$6 {x}^{2} - x - 2 = \left(6 {x}^{2} - 4 x\right) + \left(3 x - 2\right)$

$\textcolor{w h i t e}{6 {x}^{2} - x - 2} = 2 x \left(3 x - 2\right) + 1 \left(3 x - 2\right)$

$\textcolor{w h i t e}{6 {x}^{2} - x - 2} = \left(2 x + 1\right) \left(3 x - 2\right)$

So this quadratic has zeros $x = - \frac{1}{2}$ and $x = \frac{2}{3}$

Example 2

Given:

${x}^{2} - 7 x + 12$

look for a pair of factors of $A C = 1 \cdot 12 = 12$ 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 $4 \cdot 3 = 12$ and $4 + 3 = 7$.

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

${x}^{2} - 7 x + 12 = \left({x}^{2} - 4 x\right) - \left(3 x - 12\right)$

$\textcolor{w h i t e}{{x}^{2} - 7 x + 12} = x \left(x - 4\right) - 3 \left(x - 4\right)$

$\textcolor{w h i t e}{{x}^{2} - 7 x + 12} = \left(x - 3\right) \left(x - 4\right)$

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