Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 40"

Latest revision as of 12:42, 2 February 2021

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

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

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

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 $\boxed{\text{(C)}}$ .

@@ Line 16: / Line 16: @@
 <cmath>\frac{(4x-2y)(16x^{2}+8xy+4y^{2})}{4x-2y} = 16x^{2}+8xy+4y^{2},</cmath>
 or <math>\boxed{\text{(C)}}</math>.
+== See Also ==
+{{Succession box
+|header=[[1963 TMTA High School Mathematics Contests]] ([[1963 TMTA High School Algebra I Contest Problems|Problems]])
+|before=[[1963 TMTA High School Algebra I Contest Problem 39 | Problem 39]]
+|title=[[TMTA High School Mathematics Contest Past Problems/Solutions]]
+|after=Last Problem
+}}

1963 TMTA High School Mathematics Contests (Problems)
Preceded by Problem 39	TMTA High School Mathematics Contest Past Problems/Solutions	Followed by Last Problem