Question

How do you factor 14m² +19mn - 3n²?

Answer 1

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

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

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

Factored form: `f(x) = (x - n/7)(x + (3n)/2) = (7x - n)(2x + 3n)`

Check by developing and replace x by m

`f(x) = 14m^2 + 21n.m - 2n.m - 3n^2 = 14m^2 + 19n.m + 3n^2.`
Correct..

Answer 2

Because the quadratic `14m^2+19mn-3n^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 `14x^2+19x-3`.

You can consider `x=m/n`, factor the quadratic in `x` then multiply each linear term by `n`.

`14x^2+19x-3` is of the form `ax^2+bx+c` with `a=14`, `b=19` and `c=-3`.

The discriminant is given by the formula:

`Delta = b^2-4ac`

`= 19^2-(4xx14xx-3) = 361 + 168 = 529 = 23^2`

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

`x = (-b+-sqrt(Delta))/(2a) = (-19+-23)/28`

So one root is `x = (-19+23)/28 = 4/28 = 1/7`

and the other is `x = (-19-23)/28 = -42/28 = -3/2`

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

`7x = 1` showing us that `(7x-1)` is a factor.

Multiplying the second by `2` we get:

`2x = -3` showing us that `(2x+3)` is a factor.

We therefore find that:

`14x^2+19x-3 = (7x-1)(2x+3)`

and hence that:

`14m^2+19mn-3n^2 = (7m-n)(2m+3n)`