How do you find the product #(8-10a)^2#?

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Answer 1 · 2017-06-30T10:16:56

See a solution process below:

This is a special for of a quadratic where:

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

Substituting:

#8# for #a#

#10a# for #b#

Gives:

#(8 - b)^2 = 8^2 - (2 * 8 * 10a) + (10a)^2 =#

#64 - 160a + 100a^2#

Or, in standard polynomial form:

#100a^2 - 160a + 64#