# Find k so that polynomial division has remainder 0

Find $k$ so that $x^3-kx^2+3x+7k$ has remainder $0$ when divided by $x+2$.

How do I approach this problem? I know how to do polynomial long division and synthetic division, but I don't know how to apply it in this equation.

• You know the following: $$x^3 - kx^2 + 3x + 7k = (x + 2)(ax^2 + bx + c)$$ Just do the multiplication and find $a$, $b$, and $c$. – Jared Oct 6 '16 at 4:01
• Can you give more detail about "I don't know how to apply it"? e.g. are you able to perform polynomial long division here to obtain a remainder? – Hurkyl Oct 6 '16 at 9:21
• Yes I know long and synthetic division. – thunderbolt Oct 7 '16 at 2:09

### Hint:

Let $F(x)=x^3-kx^2+3x+7k$. Then $F(-2)=0$

• Best answer!!!! – Jared Oct 6 '16 at 4:11
• I chose this answer because this question is short and simple, and also gives a reason why substituting -2 for x would be the correct thing to do. – thunderbolt Oct 6 '16 at 4:21

Hint: Question is same as asking when will $x=-2$ be a root of the polynomial. Evaluate your polynomial at $x=-2$ to see for what value of $k$ the expression will vanish.

• Good answer, +1. – thunderbolt Oct 6 '16 at 4:21

Divide your polynomial $x^3-kx^2+3x+7k$ by $x+2$ like usual.

When you do, you find that the numerator of the remainder term becomes $7k+2(2k-1)$ (by my quick calculation). Then, we desire that the remainder is $0$, so really the problem is asking us to solve $7k+2(2k-1)=0$ for a suitable k.

Hope this helps!

You know the following:

$$x^3 - kx^2 + 3x + 7k = (x + 2)(ax^2 + bx + c)$$

Just do the multiplication and find $a$, $b$, and $c$.

$$(x + 2)(ax^2 + bx + c) = ax^3 + (2a + b)x^2 + (c + 2b)x + 2c$$

Now we set each coefficient equal:

$$a = 1 \\ 2a + b = -k \rightarrow b = -k - 2\\ c + 2b = 3 \rightarrow c = 3 - 2(-k - 2) \\ 2c = 7k \rightarrow 2(3 - 2(-k - 2) = 7k$$

Solving the last equation gives:

$$6 + 4k + 8 = 7k \rightarrow 14 = 3k \rightarrow k = \frac{14}{3}$$

Let's call your polynomial $p(x)$. By the division theorem, you know you $p(x)=(x+2)q(x)+r(x)$, where $q(x)$ is the quotient and $r(x)$ is the remainder term (with degree two or less, why?).

Then what happens to $p(x)=(x+2)q(x)+r(x)$ when you let $x=-2$?

• Good answer, +1. – thunderbolt Oct 6 '16 at 4:21

Since the remainder is $0$, it implies $(x+2)$ is a factor of $x^3-kx^2+3x+7k$

i.e., $x^3-kx^2+3x+7k = g(x)(x+2)$

This is valid for all real values of x. Therefore, substituting $x=-2$ in the above equation, we get:

$(-2)^3-k(-2)^2+3(-2)+7k = 0$

$\implies 3k -14 = 0$

Hence, $k = \frac{14}{3}$