# Thread: Finding the k value of a cubic equation with factor theorem

1. ## Finding the k value of a cubic equation with factor theorem

Hi, I'm working with the factor theorem and need help with the following question:

If x^3 + 4x^2 + kx - 5 is divisible by (x+2), what is the value of k?

I thought of doing the following

f(-2)= (-2)^3 + 4(-2)^2 -2k - 5, and then replacing f(-2) with 0 since if f(x) is evenly divisible by d(x), then f(p) should equal 0, but obviously my thinking can't be correct since that would mean k = 0. I would be extremely grateful for any help. Thank you!

2. You could do synthetic division and put a k as the coefficient of x to arrive at $-5-2(k-4)=0$

Since you know the remainder should be 0. Now you have:
$-5-2(k-4)=0 \Longleftrightarrow -5-2k+8=0 \Longleftrightarrow 3-2k=0 \Longleftrightarrow k=\frac{3}{2}$

3. Originally Posted by rualin
You could do synthetic division and put a k as the coefficient of x to arrive at $-5-2(k-4)=0$

Since you know the remainder should be 0. Now you have:
$-5-2(k-4)=0 \Longleftrightarrow -5-2k+8=0 \Longleftrightarrow 3-2k=0 \Longleftrightarrow k=\frac{3}{2}$
Thanks so much for the help, but I've actually never done/been taught synthetic division. Is there another method I could use?

4. Originally Posted by starswept
Thanks so much for the help, but I've actually never done/been taught synthetic division. Is there another method I could use?
you're original method was fine. you simply made an error in calculation somewhere

$(-2)^3 + 4(-2)^2 - 2k - 5 = 0$

$\Rightarrow -8 + 16 - 2k - 5 = 0$

$\Rightarrow 3 - 2k = 0$

$\Rightarrow \boxed { k = \frac {3}{2} }$

Using synthetic division is fine, but it does not really rely on the factor theorem as this method does, so i believe this is the method you should use

5. That's simpler and less work than through synthetic division.

6. Oh gosh, I'm sorry, that was a dumb mistake on my part. After 12 straight hours of math, apparently I've forgotten how to add and subtract. Following through on the original method is perfect. Thank you both for the help!

7. Originally Posted by starswept
After 12 straight hours of math, apparently I've forgotten how to add and subtract.
I know how that feels, so let's all put that little mistake behind us