## Sunday, May 23, 2010

### GR9768.65: Polynomial

65.  Let p(x) be the polynomial $x^3 + ax^2 + bx + c$, where a, b, c are real constants. If p(-3) = p(2) = 0 and p'(-3) < 0, which of the following is a possible value of c?
(A)  -27
(B)  -18
(C)  -6
(D)  -3
(E)  -1/2

Solution:
Since $p^(x) = 3x^2 + 2ax + b$, we have $p^(-3) = 27 - 6a + b < 0$.
$p(-3) = -27 + 9a -3b + c = 0$ and $p(2) = 8 + 4a + 2b + c = 0$.
From these two equations, we see that $c = 27 - 9a + 3b = -8 - 4a - 2b$.
So $35 = 5a - 5b$, or $7 = a - b$. Thus, $b = a -7$.
But since $27 - 6a + b < 0$, we see that $27 - 6a + a - 7 = 20 -5a < 0$.
So $a < 4$. This implies $b = -7 +a < -7 + 4 = -3$.
Therefore,
$c = 27 - 9a + 3b < 27 - 9(4) + 3(-3) = 27 - 36 - 9 = -18$.
$c = -8 - 4a - 2b < -8 - 4(4) -2(-3) = -8 +4 + 6 = 2$.
Thus, $c$ must be strictly less than $-18$ so the possible value of $c$ is $-27$.
The answer is A.
-sg-

#### 3 comments:

1. a must be > than 4

2. you have 20-5a<0, a must be > 4 for this to be true, not <4

3. There is an easier way to approach this one. The problem already gives us two roots of p(x) , so we have

p(x) = (x+3)(x-2)(x+t) = x^3 + (t+1)x^2 + (t-6)x -6t

where
a = t+1
b = t-6
c = -6t

Thus

p'(x) = 3x^2 + 2(t+1)x + t-6
p'(-3) = -5t + 15 < 0

Equivalently,

(5c)/6 < -15

and so

c < -18