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

Mar 1, 2017

$6 {a}^{2} + 31 a b + 14 {b}^{2} = \left(2 a + b\right) \left(3 a + 14 b\right)$

#### Explanation:

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

To factor $6 {a}^{2} + 31 a b + 14 {b}^{2}$, one must split $31$ in two parts whose product is $6 \times 14 = 84$. 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 $6 {a}^{2} + 31 a b + 14 {b}^{2}$ as

$6 {a}^{2} + 28 a b + 3 a b + 14 {b}^{2}$

= $2 a \left(3 a + 14 b\right) + b \left(3 a + 14 b\right)$

= $\left(2 a + b\right) \left(3 a + 14 b\right)$