How do you factor completely 8 a^2 x^5 y^4 + 10 a^3 x^3 y^4 + 14 a^2 + x^3 y^7?

Sep 15, 2016

Assuming one typo in the question...

$8 {a}^{2} {x}^{5} {y}^{4} + 10 {a}^{3} {x}^{3} {y}^{4} + 14 {a}^{2} {x}^{3} {y}^{7} = 2 {a}^{2} {x}^{3} {y}^{4} \left(4 {x}^{2} + 5 a + {y}^{3}\right)$

Explanation:

This question has no really useful answer in the form given in that it has no simpler factors.

I suspect that there is a typo in the question, whereby the last two terms should be a single term $14 {a}^{2} {x}^{3} {y}^{7}$, though even that does not look very reasonable...

Consider:

$8 {a}^{2} {x}^{5} {y}^{4} + 10 {a}^{3} {x}^{3} {y}^{4} + 14 {a}^{2} {x}^{3} {y}^{7}$

All of the terms are divisible by $2$, ${a}^{2}$, ${x}^{3}$ and ${y}^{4}$.

Hence we find:

$8 {a}^{2} {x}^{5} {y}^{4} + 10 {a}^{3} {x}^{3} {y}^{4} + 14 {a}^{2} {x}^{3} {y}^{7} = 2 {a}^{2} {x}^{3} {y}^{4} \left(4 {x}^{2} + 5 a + {y}^{3}\right)$

This does not factor any further.