# How do you factor 7n^2-13n-2?

Jul 6, 2015

$7 \left(n - 2\right) \left(n + \frac{1}{7}\right)$

#### Explanation:

7n^2 - 13n - 2 = 0 \Rightarrow n = (13 ± sqrt{225})/14

${n}_{1} = \frac{13 + 15}{14} = 2$

${n}_{2} = \frac{13 - 15}{14} = - \frac{1}{7}$

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

Jul 6, 2015

Alternatively, use a variant of the AC Method to find:

$7 {n}^{2} - 13 n - 2 = \left(7 n + 1\right) \left(n - 2\right)$

#### Explanation:

Given $7 {n}^{2} - 13 n - 2$, let $A = 7$, $B = 13$ and $C = 2$.

Notice that the sign of the constant term is negative, so look for a pair of factors of $A C = 7 \cdot 2 = 14$, whose difference is $B = 13$.

It should not take long to spot that the pair $B 1 = 1 , B 2 = 14$ works.

Next, for each of the pairs $\left(A , B 1\right)$ and $\left(A , B 2\right)$ divide by the HCF (highest common factor) to find a pair of coefficients of a factor, choosing signs as appropriate:

$\left(7 , 1\right) \to \left(7 , 1\right) \to \left(7 n + 1\right)$
$\left(7 , 14\right) \to \left(1 , 2\right) \to \left(n - 2\right)$

We choose $+$ for the pair $\left(A , B 1\right)$ and $-$ for the pair $\left(A , B 2\right)$ to get $\left(+ B 1\right) + \left(- B 2\right) = - 13$ - the coefficient of the middle term.