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

Feb 16, 2017

$3 {a}^{2} + 4 a b + {b}^{2} - 2 a c - {c}^{2} = \left(3 a + b + c\right) \left(a + b - c\right)$

#### Explanation:

Given:

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

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:

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

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

$3 {a}^{2} + 4 a b + {b}^{2} = \left(3 a + b\right) \left(a + b\right)$

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

$3 {a}^{2} - 2 a c - {c}^{2} = \left(3 a + c\right) \left(a - c\right)$

These various linear binomial factors can be combined as follows:

$\left(b + c\right) , \left(3 a + b\right) , \left(3 a + c\right) \rightarrow \left(3 a + b + c\right)$

$\left(b - c\right) , \left(a + b\right) , \left(a - c\right) \rightarrow \left(a + b - c\right)$

Hence we find the factorisation:

$3 {a}^{2} + 4 a b + {b}^{2} - 2 a c - {c}^{2} = \left(3 a + b + c\right) \left(a + b - c\right)$

Jul 12, 2017

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

=$\left(4 {a}^{2} + 4 a b + {b}^{2}\right) - \left({a}^{2} + 2 a c + {c}^{2}\right)$

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

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

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

#### Explanation:

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

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

Apr 25, 2018

$\left(a + b - c\right) \left(3 a + b + c\right)$.

#### Explanation:

Here is another way to factorise the poly.

$\underline{3 {a}^{2} + 4 a b + {b}^{2}} - 2 a c - {c}^{2}$,

$= \left(3 a + b\right) \left(a + b\right) - 2 a c - {c}^{2}$.

So, if we subst. $x = 3 a + b , \mathmr{and} , y = a + b$, then, since,

$x - y = \left(3 a + b\right) - \left(a + b\right) = 2 a$.

$\therefore 3 {a}^{2} + 4 a b + {b}^{2} - 2 a c - {c}^{2}$,

$= \left(3 a + b\right) \left(a + b\right) - 2 a c - {c}^{2}$,

$= x y - \left(x - y\right) c - {c}^{2}$,

$= \underline{x y - c x} + \underline{c y - {c}^{2}}$,

$= x \left(y - c\right) + c \left(y - c\right)$,

$= \left(y - c\right) \left(x + c\right)$,

$= \left(a + b - c\right) \left(3 a + b + c\right)$, as respected George C. and Cem