1988 AHSME Problems/Problem 19

Problem

Simplify

$\frac{bx(a^2x^2 + 2a^2y^2 + b^2y^2) + ay(a^2x^2 + 2b^2x^2 + b^2y^2)}{bx + ay}$

$\textbf{(A)}\ a^2x^2 + b^2y^2\qquad \textbf{(B)}\ (ax + by)^2\qquad \textbf{(C)}\ (ax + by)(bx + ay)\qquad\\ \textbf{(D)}\ 2(a^2x^2+b^2y^2)\qquad \textbf{(E)}\ (bx+ay)^2$

Solution 1

We can multiply each answer choice by $bx + ay$ and then compare with the numerator. This gives $\boxed{\text{B}}$ .

Solution 2

Expanding everything in the brackets, we get $\frac{ba^2x^3 + 2ba^2xy^2 + b^3xy^2 + a^3x^2y + 2ab^2x^2y + ab^2y^3}{bx + ay}$ . We can then group numbers up in pairs so they equal $n(bx + ay)$ :

$= \frac{ba^2x^3 + a^3x^2y + 2ab^2x^2y + 2ba^2xy^2 + b^3xy^2 + ab^2y^3}{bx+ay}$

$= \frac{bx + ay(a^2x^2) + bx + ay(2baxy) + bx + ay(b^2y^2)}{bx+ay}$

$= a^2x^2 + 2baxy + b^2y^2$

$= (ax + by)^2$

We get $\boxed{\text{B}}$ .

-ThisUsernameIsTaken

Solution 3

If you were out of time and your algebra isn't that good, you could just plug in some values for the variables and see which answer choice works.

Solution 4 (fastest)

After regrouping, the numerator becomes $(bx+ay)(a^2x^2+b^2y^2)+2bxa^2y^2+2ayb^2x^2$ . Factoring further, we get $(bx+ay)(a^2x^2+b^2y^2)+2bxay(bx+ay)$ . After dividing, we get $a^2x^2+b^2y^2+2bxay$ , which can be factored as $(ax+by)^2$ , so the answer is $\boxed{\text{B}}$ . -Pengu14

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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