Question

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

Answer 1

`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)`