How do you factorize #49a ^ { 8} b ^ { 4} - 35a ^ { 2} b ^ { 5} + 63a ^ { 6} b ^ { 7}#?

1 Answer
George C.
May 29, 2017

#49a^8b^4-35a^2b^5+63a^6b^7= 7a^2b^4(7a^6-5b+9a^4b^3)#

Explanation:

Given:

#49a^8b^4-35a^2b^5+63a^6b^7#

Note that all of the terms are divisible by #7#, #a^2# and #b^4#, so by #7a^2b^4#. So we can separate that out as a factor to find:

#49a^8b^4-35a^2b^5+63a^6b^7#

#= (7a^2b^4)(7a^6)-(7a^2b^4)(5b)+(7a^2b^4)(9a^4b^3)#

#= 7a^2b^4(7a^6-5b+9a^4b^3)#

The remaining trinomial cannot be broken down into simpler factors.

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