1. ## [SOLVED] polynomial quesstion

hello all,
i hope this question is in the right board, anyhow, here is my problem:

if $\displaystyle x^3+px+q$ is divisible by both $\displaystyle x-3$ and $\displaystyle x+2$, find the values of p and q

thanks

2. don't worry guys, after some more work i got the answer myself, here is the working out for anybody interested:

$\displaystyle P(x) = x^3+px+q$
if both x-3 and x+2 are factors of P(x), then P(3) = 0 = P(-2)

therefore
$\displaystyle (3)^3+p3+q = (-2)^3+p2+q$
simplifying $\displaystyle 27+3p = -8-2p$
$\displaystyle -35 = 5p$
$\displaystyle p = -7$

therefore $\displaystyle P(x)=(x-3)(x+2)(x+k)$

now, $\displaystyle (k*-1) + (k*-6) =7$
$\displaystyle k=1$

therefore $\displaystyle P(x)=(x-3)(x+2)(x+1)$
expanding $\displaystyle P(x)=x^3-7x-6$

therefore $\displaystyle q=-6$

3. Originally Posted by str33tl0rd
hello all,
i hope this question is in the right board, anyhow, here is my problem:

if $\displaystyle x^3+px+q$ is divisible by both $\displaystyle x-3$ and $\displaystyle x+2$, find the values of p and q

thanks
Hello :There is the solution
I'have :

$\displaystyle \begin{array}{l} x^2 + px + q = \left( {x - 3} \right)\left( {x + 2} \right)\left( {ax + b} \right) \\ x^2 + px + q = ax^3 + \left( {b - a} \right)x^2 + \left( { - b - 6a} \right)x - 6b \\ \end{array}$
I'HAVE THE SYSTEM :
$\displaystyle \left\{ \begin{array}{l} a = 1 \\ b - a = 0 \\ - b - 6a = p \\ - 6b = q \\ \end{array} \right.$

$\displaystyle a = 1,b =1$
Subst :
CONCLUSION
$\displaystyle \left\{ \begin{array}{l} q = - 6 \\ p = - 7 \\ \end{array} \right.$

4. "if is divisible by both and , find the values of p and q."

Not so ELEGANT SOLUTION:

Notice that x^3 + px + q = (x - a)(x - b)(x - c) = (x - 3)(x + 2)(x - k)

that the coefficient of x^2 is ZERO, that means the sum of its roots is ZERO.

then, -(-3 + 2 - k) = 0, or

3 - 2 + k = 0,

k = -1; thus the 3 roots are, -2, -1 and 3.

p = (-2)(-1) + (-2)(3) + (-1)(3) = 2 - 6 - 3 = 2 - 9 = -7

q = -(-2)(-1)(3) = -6

x^3 -7x - 6 = (x - 3)(x + 2)(x + 1).