1964 AHSME Problems/Problem 25

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Problem 25

The set of values of $m$ for which $x^2+3xy+x+my-m$ has two factors, with integer coefficients, which are linear in $x$ and $y$, is precisely:

$\textbf{(A)}\ 0, 12, -12\qquad \textbf{(B)}\ 0, 12\qquad \textbf{(C)}\ 12, -12\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 0$

Solution

Since there are only $3$ candidate values for $m$, we test $0, 12, -12$.

If $m=0$, then the expression is $x^2 + 3xy + x$. The term $x$ appears in each monomial, giving $x(x + 3y + 1)$, which has two factors that are linear in $x$ and $y$ with integer coefficients. This eliminates $C$ and $D$.

We next test $m=12$. This gives $x^2 + 3xy + x + 12y - 12$. The linear factors will be of the form $(Ax + By + C)(Dx + Ey + F)$. We will match coefficients from the skeleton form to the expression that is to be factored.

Matching the $y^2$ coefficient gives $BE = 0$, and WLOG we can let $B=0$ to give $(Ax + C)(Dx + Ey + F)$.

Matching the $x^2$ coefficients gives $AD = 1$. Since $A, D$ are integers, we either have $(A, D) = (1, 1), (-1, -1)$. WLOG we can pick the former, since if $(a)(b)$ is a factorization, so is $(-a)(-b)$.

We now have $(x + C)(x + Ey + F)$. Matching the $xy$ term gives $E =3$. We now have $(x+C)(x + 3y + F)$. Matching the $y$ term gives $3C = 12$, which leads to $C = 4$. Finally, matching constants leads to $4F = -12$ and $F = -3$ for a provisional factorization of $(x + 4)(x + 3y - 3)$. We've matched every coefficient except for $x$, and finding the $x$ term by selectively multiplying leads to $4x + -3x = x$, which matches, so this is a real factorization.

We now must check $m=-12$. The expression is $x^2+3xy+x-12y+12m$, and the skeleton once again is $(Ax + By + C)(Dx + Ey + F)$. The beginning three steps matching $y^2, x^2, xy$ are the same, leading to $(x + C)(x + 3y + F)$. Matching the $y$ term gives $3C = -12$, or $C = -4$. The skeleton becomes $(x -4)(x + 3y + F)$. The constant locks in $F = -3$ for a provisional factorization of $(x-4)(x + 3y - 3)$. However, the $x$ term does not match up, because it is $-4x + -3x = -7x$, when it needs to be $x$.

Thus, $m=0, 12$ work and $m=-12$ does not, giving answer $\boxed{\textbf{(B)}}$.

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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