How to factorize #6a^2+31ab+14b^2#?

1 Answer
Mar 1, 2017

Answer:

#6a^2+31ab+14b^2=(2a+b)(3a+14b)#

Explanation:

Observe that it is a homogeneous quadratic polynomial and hence is equivalent to a quadratic polynomial.

To factor #6a^2+31ab+14b^2#, one must split #31# in two parts whose product is #6xx14=84#. Observe that #7# is a factor of #84# and hence one of them must be a multiple of #7#.

Thus the two factors, whose product is #84#, sum is #31# and one of whom is multiple of #7#, are #28# and #3#.

Hence, we can write #6a^2+31ab+14b^2# as

#6a^2+28ab+3ab+14b^2#

= #2a(3a+14b)+b(3a+14b)#

= #(2a+b)(3a+14b)#