# 1965 AHSME Problems/Problem 19

## Problem 19

If $x^4 + 4x^3 + 6px^2 + 4qx + r$ is exactly divisible by $x^3 + 3x^2 + 9x + 3$, the value of $(p + q)r$ is: $\textbf{(A)}\ - 18 \qquad \textbf{(B) }\ 12 \qquad \textbf{(C) }\ 15 \qquad \textbf{(D) }\ 27 \qquad \textbf{(E) }\ 45 \qquad$

## Solution 1

Let $f(x)=x^3+3x^2+9x+3$ and $g(x)=x^4+4x^3+6px^2+4qx+r$.

Let 3 roots of $f(x)$ be $r_1, r_2$ and $r_3$. As $f(x)|g(x)$ , 3 roots of 4 roots of $g(x)$ will be same as roots of $f(x)$. Let the 4th root of $g(x)$ be $r_4$. By vieta's formula

In $f(x)$ $r_1+r_2+r_3=-3$ $r_1r_2+r_2r_3+r_1r_3=9$ $r_1r_2r_3=-3$

In $g(x)$ $r_1+r_2+r_3+r_4=-4$ $=>r_4=-1$ $r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p$ $=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p$ $=>9+(-1)(-3)=6p$ $=>p=2$ $r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q$ $=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q$ $=>-3+(-1)(9)=-4q$ $=>q=3$ $r_1r_2r_3r_4=r$ $=>(-3)(-1)=r$ $=>r=3$

so $(p+q)r=\fbox{15}$

By ~Ahmed_Ashhab

## Solution 2

Notice that to obtain the $x^4$ term one must multiply $x^4+4x^3+6px^2+4qx+r$ by some linear function of the form $x-a$. Looking at the $x^3$ term, it is clear that $a$ must equal $1$. Therefore by multiplying $x^4+4x^3+6px^2+4qx+r$ by $x+1$, the product will be $x^4+4x^3+12x^2+12x+3$. Therefore $p=2$, $q=3$, $r=3$. Thus $(2+3)3=\fbox{15}$

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