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-

a must be > than 4

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

ReplyDeleteThere is an easier way to approach this one. The problem already gives us two roots of

ReplyDeletep(x), so we havep(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

The polynomial goes to infinity as x grows and to - oo as x goes to -oo since it's a 3rd degree polynomials. It's decreasing in -3. Hence, t is strictly less than -3

ReplyDeleteP(x) = (x+t)(x+3)(x-2)

D = P(0) = -6t <-18