# How do you factor 14m² +19mn - 3n²?

May 17, 2015

Factor this expression like factoring a trinomial in x. Call m = x.

$f \left(x\right) = 14 {x}^{2} + 19 n . x - 3 {n}^{2.}$ = (x - p)(x - q).
I use the new AC Method (Google, Yahoo Search) to factor trinomials

Converted trinomial:$f ' \left(x\right) = {x}^{2} + 19 n - 42 {n}^{2} =$(x - p')(x - q')
with $\left(a . c = - 42 {n}^{2}\right)$.
Compose factor pairs of $\left(- 42 {n}^{2}\right)$ -> (-n, 42n)(-2n, 21n).
Then, p' = -2n and q' = 21n.
We get: $p = \frac{p '}{a} = \frac{- 2 n}{14} = \frac{- n}{7}$, and $q = \frac{q '}{a} = \frac{21 n}{14} = \frac{3 n}{2}$

Factored form: $f \left(x\right) = \left(x - \frac{n}{7}\right) \left(x + \frac{3 n}{2}\right) = \left(7 x - n\right) \left(2 x + 3 n\right)$

Check by developing and replace x by m

$f \left(x\right) = 14 {m}^{2} + 21 n . m - 2 n . m - 3 {n}^{2} = 14 {m}^{2} + 19 n . m + 3 {n}^{2.}$
Correct..

May 17, 2015

Because the quadratic $14 {m}^{2} + 19 m n - 3 {n}^{2}$ is homogenous (the sums of the powers of $m$ and $n$ are the same in each term), this is directly analogous to factoring the quadratic $14 {x}^{2} + 19 x - 3$.

You can consider $x = \frac{m}{n}$, factor the quadratic in $x$ then multiply each linear term by $n$.

$14 {x}^{2} + 19 x - 3$ is of the form $a {x}^{2} + b x + c$ with $a = 14$, $b = 19$ and $c = - 3$.

The discriminant is given by the formula:

$\Delta = {b}^{2} - 4 a c$

$= {19}^{2} - \left(4 \times 14 \times - 3\right) = 361 + 168 = 529 = {23}^{2}$

The roots of $14 {x}^{2} + 19 x - 3 = 0$ are given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 19 \pm 23}{28}$

So one root is $x = \frac{- 19 + 23}{28} = \frac{4}{28} = \frac{1}{7}$

and the other is $x = \frac{- 19 - 23}{28} = - \frac{42}{28} = - \frac{3}{2}$

Multiplying the first of these by $7$ we get:

$7 x = 1$ showing us that $\left(7 x - 1\right)$ is a factor.

Multiplying the second by $2$ we get:

$2 x = - 3$ showing us that $\left(2 x + 3\right)$ is a factor.

We therefore find that:

$14 {x}^{2} + 19 x - 3 = \left(7 x - 1\right) \left(2 x + 3\right)$

and hence that:

$14 {m}^{2} + 19 m n - 3 {n}^{2} = \left(7 m - n\right) \left(2 m + 3 n\right)$