How do you factor #36x^2y^6 - 16#?

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Answer 1 · 2018-03-10T14:13:54

# 4 * (3xy^3 + 2) *(3xy^3 - 2)#

#36x^2y^6-16#

= #((2^2+3^2*x^2) * y^6) - 16) #

Because #(a^2 - b^2)# can be factored into # (a+b) * (a-b)#

From the simplified expression above, we can factorise the numbers as it is.

Simply half the index numbers as you would normally when you square root a number with an index.

Therefore:
#([(3xy^3 + 2) *(3xy^3 - 2)] - 16)#

We can also square root the 16

So
# 4 * (3xy^3 + 2) *(3xy^3 - 2)#