# How do you factor 14n^2+23n-15?

May 7, 2016

$14 {n}^{2} + 23 n - 15 = \left(2 n - 1\right) \left(7 n + 15\right)$

#### Explanation:

Given: $14 {n}^{2} + 23 n - 15$

Use an AC method.

Find a pair of factors of $A C = 14 \cdot 15 = 210$ which differ by $B = 23$

The pair $30 , 7$ works.

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

$14 {n}^{2} + 23 n - 15$

$= 14 {n}^{2} + 30 n - 7 n - 15$

$= \left(14 {n}^{2} + 30 n\right) - \left(7 n + 15\right)$

$= 2 n \left(7 n + 15\right) - 1 \left(7 n + 15\right)$

$= \left(2 n - 1\right) \left(7 n + 15\right)$

$\textcolor{w h i t e}{}$
Alternatively, multiply by $4 \cdot 14 = 56$, make into a difference of squares, factor using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(28 n + 23\right)$ and $b = 37$, then divide by $56$:

$56 \left(14 {n}^{2} + 23 n - 15\right)$

$= 784 {n}^{2} + 1288 n - 840$

$= {\left(28 n\right)}^{2} + 2 \left(28 n\right) \left(23\right) - 840$

$= {\left(28 n + 23\right)}^{2} - {23}^{2} - 840$

$= {\left(28 n + 23\right)}^{2} - 529 - 840$

$= {\left(28 n + 23\right)}^{2} - 1369$

$= {\left(28 n + 23\right)}^{2} - {37}^{2}$

$= \left(\left(28 n + 23\right) - 37\right) \left(\left(28 n + 23\right) + 37\right)$

$= \left(28 n - 14\right) \left(28 n + 60\right)$

$= \left(14 \left(2 n - 1\right)\right) \left(4 \left(7 n + 15\right)\right)$

$= 56 \left(2 n - 1\right) \left(7 n + 15\right)$

So:

$14 {n}^{2} + 23 n - 15 = \left(2 n - 1\right) \left(7 n + 15\right)$