How do you factorize #36x ^ { 8} y ^ { 6} - 25#?

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Answer 1 · 2017-07-15T14:17:37

See a solution process below:

This is a special case of polynomials where:

#a^2 - b^2 = (a + b)(a - b)#

Substitute :

#6x^4y^3# for #a#

#5# for #b# gives:

#(6x^4y^3)^2 - 5^2 = (6x^4y^3 + 5)(6x^4y^3 - 5)#

#36x^8y^6 - 25 = (6x^4y^3 + 5)(6x^4y^3 - 5)#