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

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

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Answer 1 · 2017-02-16T00:29:58

$3 a^{2} + 4 a b + b^{2} - 2 a c - c^{2} = (3 a + b + c) (a + b - c)$ $3a^2+4ab+b^2-2ac-c^2 = (3a+b+c)(a+b-c)$

Explanation:

Given:

$3 a^{2} + 4 a b + b^{2} - 2 a c - c^{2}$ $3a^2+4ab+b^2-2ac-c^2$

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

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

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

$b^{2} - c^{2} = (b - c) (b + c)$ $b^2-c^2 = (b-c)(b+c)$

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

$3 a^{2} + 4 a b + b^{2} = (3 a + b) (a + b)$ $3a^2+4ab+b^2 = (3a+b)(a+b)$

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

$3 a^{2} - 2 a c - c^{2} = (3 a + c) (a - c)$ $3a^2-2ac-c^2 = (3a+c)(a-c)$

These various linear binomial factors can be combined as follows:

$(b + c), (3 a + b), (3 a + c) \to (3 a + b + c)$ $(b+c), (3a+b), (3a+c) rarr (3a+b+c)$

$(b - c), (a + b), (a - c) \to (a + b - c)$ $(b-c), (a+b), (a-c) rarr (a+b-c)$

Hence we find the factorisation:

$3 a^{2} + 4 a b + b^{2} - 2 a c - c^{2} = (3 a + b + c) (a + b - c)$ $3a^2+4ab+b^2-2ac-c^2 = (3a+b+c)(a+b-c)$

Answer 2 · 2017-07-12T13:28:18

$3 a^{2} + 4 a b + b^{2} - 2 a c - c^{2}$ $3a^2+4ab+b^2-2ac-c^2$

= $(4 a^{2} + 4 a b + b^{2}) - (a^{2} + 2 a c + c^{2})$ $(4a^2+4ab+b^2)-(a^2+2ac+c^2)$

= ${(2 a + b)}^{2} - {(a + c)}^{2}$ $(2a+b)^2-(a+c)^2$

= $(2 a + b + a + c) \cdot (2 a + b - a - c)$ $(2a+b+a+c)*(2a+b-a-c)$

= $(3 a + b + c) \cdot (a + b - c)$ $(3a+b+c)*(a+b-c)$

Explanation:

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

2) I used $u^{2} - v^{2} = (u + v) \cdot (u - v)$ $u^2-v^2=(u+v)*(u-v)$ identity.

Answer 3 · 2018-04-25T18:58:28

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

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

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