Question

Factorise
32x^5 + 243.

Answer 1

i used f(x) = (2x)^5 + (3y)^5

then f(-3y/2)= (2x)^5 + (3y)^5 = 0

(factor thereom) using synthetic divison and factor

(-3y/2)

i got.. 32x^4 -48x^3y + 72x^2y^2 -108xy^3 + 162y^4

The answer is

(2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4)

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