When Betty substitutes $x=1$ into the expression $a x^{3}-2 x+c$ its value is -5 . When she substitutes $x=4$ the expression has value 52 . One value of $x$ that makes the expression equal to zero is
(A) 2
(B) $\frac{5}{2}$
(C) 3
(D) $\frac{7}{2}$
(E) 4
When Betty substitutes $x=1$ into the expression $a x^{3}-2 x+c$ its value is -5 . When she substitutes $x=4$ the expression has value 52 . One value of $x$ that makes the expression equal to zero is
(A) 2
(B) $\frac{5}{2}$
(C) 3
(D) $\frac{7}{2}$
(E) 4

## Solution

When $x=1$, we are told that $a(1)^{3}-2(1)+c=-5$, or $a+c=-3$ (1). Similarly, when $x=4$, $a(4)^{3}-2(4)+c=52$, or $64 a+c=60$ (2). Subtracting equation (1) from equation (2) gives $63 a=63$, or $a=1$. Substituting $a=1$ into equation (1) gives $c=-4$. The original expression
$a x^{3}-2 x+c$ becomes $x^{3}-2 x-4$. By trial and error, using divisors of 4 , when $x=2$ we get $2^{3}-2(2)-4=0$.

ANSWER: (A)