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

##### 3 Answers

#### Explanation:

Given:

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

Note that all of the terms are of degree

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

If we ignore the terms involving

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

If we ignore the terms involving

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

If we ignore the terms involving

#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)#

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#### Explanation:

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

2) I used

#### Explanation:

Here is another way to factorise the poly.

So, if we subst.

**respected George C.** and **Cem**

**Sentin** have **already derived!**

**Enjoy Maths!**