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#?

1 Answer
Sep 15, 2016

Answer:

Assuming one typo in the question...

#8a^2x^5y^4+10a^3x^3y^4+14a^2x^3y^7 = 2a^2x^3y^4(4x^2+5a+y^3)#

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 #14a^2x^3y^7#, though even that does not look very reasonable...

Consider:

#8a^2x^5y^4+10a^3x^3y^4+14a^2x^3y^7#

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

Hence we find:

#8a^2x^5y^4+10a^3x^3y^4+14a^2x^3y^7 = 2a^2x^3y^4(4x^2+5a+y^3)#

This does not factor any further.