1956 AHSME Problems/Problem 17

Problem 17

The fraction $\frac {5x - 11}{2x^2 + x - 6}$ was obtained by adding the two fractions $\frac {A}{x + 2}$ and $\frac {B}{2x - 3}$ . The values of $A$ and $B$ must be, respectively:

$\textbf{(A)}\ 5x,-11\qquad\textbf{(B)}\ -11,5x\qquad\textbf{(C)}\ -1,3\qquad\textbf{(D)}\ 3,-1\qquad\textbf{(E)}\ 5,-11$

This is essentially asking for the partial fraction decomposition of $\frac{5x-11}{2x^2 + x - 6}$ Looking at $A$ and $B$ , we can write $A(2x-3) + B(x+2) = 5x-11$ . Substituting $x= -2$ , we get $A(-4-3) + B(0) = -21 \Rrightarrow A = 3$ Substituting $x = \frac{3}{2}$ , we get $A(0) + B(\frac{7}{2}) = \frac{15}{2} - \frac{22}{2} \Rrightarrow B = -1$ Thus, our answer is $\boxed{D}$

~JustinLee2017

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1956 AHSME Problems/Problem 17

Problem 17

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