Cubic Polynomial With an Extra Variable

Question

I was trying so desperately to find the value of $c$. Let

$$P(x) = 2x^3+2x^2-2cx+4$$

$x+2$ is the factor of $P(x)$. So I started factoring it, going $x+2(2x^2-2x...$ and then got stuck at factoring $2cx$. It just seems impossible, because 4 does not have $c$ with it.

What I need here is the process with which to solve this problem, not an answer. My explanation book just tells me to graph it, saying it's the fastest way. But I don't want to.

Hint: Suppose that the polynomial $P(x)$ has factor $x+2$. Then $P(x)=(x+2)Q(x)$ for some polynomial $Q(x)$, and therefore $P(-2)=0$. — André Nicolas, Jul 01 '16 at 17:15
I flagged the answer for removal and so will respond on donotcopy's behalf. It's $2x^2$. — , Jul 01 '16 at 17:42

score 1 · Answer 1 · answered Jul 01 '16 at 17:44

1

Dividing $P(x)$ by $x+2$ you have the residu equal to zero which give you $c=1$ and you get finally $$P(x)=2(x+2)(x^2-x+1)$$

answered Jul 01 '16 at 17:44

Piquito

29,594

score 1 · Answer 2 · answered Jul 01 '16 at 18:14

If $x+2$ is a factor of $P(x)$, then we know that $$P(x) = (x+2)Q(x)$$ for some polynomial $Q(x)$. It follows that $x = -2$ is a root of $P(x)$, since $$P(-2) = (-2 + 2)Q(-2) = 0.$$ Consequently, $$0 = 2(-2)^3 + 2(-2)^2 -2c (-2) + 4,$$ from which the value of $c$ is trivially deduced.

Cubic Polynomial With an Extra Variable

2 Answers2