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

1 Answer
Shwetank Mauria
Mar 1, 2017

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)

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