# 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 $\boxed{\text{B}}$.

## 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