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1963 TMTA High School Algebra I Contest Problem 40

Revision as of 11:02, 2 February 2021 by Rubixmaster21 (talk | contribs) (→‎Problem)

Problem

If $64x^{3}-8y^{3}$ is divided by $4x-2y$, the quotient will be:

$\text{(A)} \quad 4x^{2}-2y^{2} \quad \text{(B)} \quad 4x^{2}+2y^{2} \quad \text{(C)} \quad 16x^{2}+8xy+4y^{2}$

$\text{(D)} \quad 16x^{2}-8xy+4y^{2} \quad \text{(E)} \quad \text{none of these}$

Solution

We can factor $64x^{3}-8y^{3}$ as \[(4x)^{3}-(2y)^{3}=(4x-2y)(16x^{2}+8xy+4y^{2})\] by using difference of cubes.

Then $\frac{64x^{3}-8y^{3}}{4x-2y}$ is equal to \[\frac{(4x-2y)(16x^{2}+8xy+4y^{2})}{4x-2y} = 16x^{2}+8xy+4y^{2},\] or $\boxed{\text{(C)}}$.

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