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

3 Answers
Feb 16, 2017

Answer:

#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)#

Jul 12, 2017

Answer:

#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.

Apr 25, 2018

Answer:

# (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!