Question

How do you factorise `3a^2+4ab+b^2-2ac-c^2` ?

Answer 1

`3a^2+4ab+b^2-2ac-c^2 = (3a+b+c)(a+b-c)`

Explanation:

Given:

`3a^2+4ab+b^2-2ac-c^2`

Note that all of the terms are of degree `2`.

So if this factors into simpler polynomials then its factors are homogeneous of degree `1`.

If we ignore the terms involving `a`, then we are looking for a factorisation of `b^2-c^2`, which can be written:

`b^2-c^2 = (b-c)(b+c)`

If we ignore the terms involving `c`, then we find a factorisation:

`3a^2+4ab+b^2 = (3a+b)(a+b)`

If we ignore the terms involving `b`, then we find a factorisation:

`3a^2-2ac-c^2 = (3a+c)(a-c)`

These various linear binomial factors can be combined as follows:

`(b+c), (3a+b), (3a+c) rarr (3a+b+c)`

`(b-c), (a+b), (a-c) rarr (a+b-c)`

Hence we find the factorisation:

`3a^2+4ab+b^2-2ac-c^2 = (3a+b+c)(a+b-c)`

Answer 2

`3a^2+4ab+b^2-2ac-c^2`

=`(4a^2+4ab+b^2)-(a^2+2ac+c^2)`

=`(2a+b)^2-(a+c)^2`

=`(2a+b+a+c)*(2a+b-a-c)`

=`(3a+b+c)*(a+b-c)`

Explanation:

1) I added and subtracted a^2 for resembling difference of squares.

2) I used `u^2-v^2=(u+v)*(u-v)` identity.

Answer 3

`(a+b-c)(3a+b+c)`.

Explanation:

Here is another way to factorise the poly.

`ul(3a^2+4ab+b^2)-2ac-c^2`,

`=(3a+b)(a+b)-2ac-c^2`.

So, if we subst. `x=3a+b, and, y=a+b`, then, since,

`x-y=(3a+b)-(a+b)=2a`.

`:. 3a^2+4ab+b^2-2ac-c^2`,

`=(3a+b)(a+b)-2ac-c^2`,

`=xy-(x-y)c-c^2`,

`=ul(xy-cx)+ul(cy-c^2)`,

`=x(y-c)+c(y-c)`,

`=(y-c)(x+c)`,

`=(a+b-c)(3a+b+c)`, as respected George C. and Cem

Sentin have already derived!

Enjoy Maths!