1. ## polynomial division problem

Find all values of k such that f(x) = kx^3 + x^2 + k^2x + 3k2 + 11 is completely divisible by polynomial p(x) = x + 2

I have no idea on how to solve this.

2. Originally Posted by mwok
Find all values of k such that f(x) = kx^3 + x^2 + k^2x + 3k2 + 11 is completely divisible by polynomial p(x) = x + 2

I have no idea on how to solve this.
Well, mwok. Let's try synthetic division on this one.

$\displaystyle f(x)=kx^3+x^2+k^2x+3k^2+11$. Since $\displaystyle (x + 2)$ is a factor, then $\displaystyle f(-2)=0$.

Code:

-2 | k      1         k^2      3k^2  +11
-2k        4k-2     -2k^2-8k+4
------------------------------------
k    -2k+1    k^2+4k-2     k^2-8k+15
Now then, we know the remainder should be 0, so....

$\displaystyle k^2-8k+15=0$

Solve for the two possible values of k and you're done.

3. Originally Posted by masters
Well, mwok. Let's try synthetic division on this one.

$\displaystyle kx^3+x^2+k^2x+3k^2+11$. Since $\displaystyle (x + 2)$ is a factor, then $\displaystyle f(-2)=0$.

Code:

-2 | k      1         k^2      3k^2  +11
-2k        4k-2     -2k^2-8k+4
------------------------------------
k    -2k+1    k^2+4k-2     k^2-8k+15
Now then, we know the remainder should be 0, so....

$\displaystyle k^2-8k+15=0$

Solve for the two possible values of k and you're done.

I'm not sure I understand, with synthetic divisions...don't you just use the coefficients?

4. Originally Posted by mwok
I'm not sure I understand, with synthetic divisions...don't you just use the coefficients?
I did use the coefficients.

$\displaystyle f(x)={\color{red}k}x^3+{\color{red}1}x^2+{\color{r ed}k^2}x+{\color{blue}3k^2+11}$.

The coefficients are in red and the constant is in blue.

5. x is the variable in the polynomial, not k. k is an undetermined coefficient, but it is still a coefficient, so you can use synthetic division and get an equation for k, and figure out what value it has.

6. Originally Posted by masters
I did use the coefficients.

$\displaystyle f(x)={\color{red}k}x^3+{\color{red}1}x^2+{\color{r ed}k^2}x+{\color{blue}3k^2+11}$.

The coefficients are in red and the constant is in blue.

Sorry, I'm a bit confused.

Why do you use a constant in synthetic division? I thought synthetic division only used the coefficients.

Attached is what I got.

7. Originally Posted by mwok
Sorry, I'm a bit confused.

Why do you use a constant in synthetic division? I thought synthetic division only used the coefficients.

Attached is what I got.
There are only 3 terms with an x in them. In those 3 terms, whatever has been multiplied by x is a coefficient. Any other terms (without x) would constitute the constant.

$\displaystyle f(x)=ax^2+bx+c$

a is the coefficient of the quadratic term.
b is the coefficient of the linear term.
c is the constant.

It just so happens that the c in your function is $\displaystyle 3k^2+11$, but a constant none the less.

Originally Posted by mwok
I thought synthetic division only used the coefficients.
Not so. Your attachment is incorrect. Review Synthetic Division

8. As was demonstrated in my post #2, k can be found by solving the following equation:

$\displaystyle k^2-8k+15=0$

$\displaystyle (k-5)(k-3)=0$

$\displaystyle k=5 \ \ or \ \ k=3$

If you substitute each of these values for k back into your original function, you can verify that f(-2)=0.

9. Hello, mwok

Find all values of $\displaystyle k$ such that $\displaystyle f(x) \:= \:kx^3 + x^2 + k^2x + 3k^2 + 11$ is divisible by $\displaystyle x + 2$
We are expected to know this theorem:

. . If $\displaystyle f(a) \:=\:0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle f(x)$.

For this problem, if $\displaystyle f(\text{-}2) = 0$, then $\displaystyle (x + 2)$ is a factor of $\displaystyle f(x).$

We have: .$\displaystyle f(\text{-}2) \;=\;k(\text{-}2)^4 + (\text{-}2)^2 + k^2(\text{-}2) + 3k^2 + 11 \;=\;0$

. . which simplifies to: .$\displaystyle k^2-8k+15 \;=\;0$

. . which factors: .$\displaystyle (k-3)(k-5) \:=\:0$

. . and has roots: .$\displaystyle \boxed{k \;=\;3,\:5}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you dare, you can try Long Division . . .

$\displaystyle \begin{array}{ccccccc} & & & kx^2 & +(1-2k)x & +(k^2+4k -2) \\ & & ---&-----&-------&-------&--- \\ x+2 & ) & kx^3 & +x^2 & +k^2x & + 3k^2 & +11 \\ & & kx^3 & +2kx^2 \\ & & --- & ----- \\ & & & (1-2k)x^2 & +k^2x \\ & & & (1-2k)x^2 & + 2(1-2k)x \\ & & & ----- & ------- \\ & & & & (k^2+4k-1)x & + 3k^2 & +11\\ & & & & (k^2+4k-2)x & +2(k^2+4k-2) \\ & & & & -------& ------- \\\end{array}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle k^2 - 8k + 4 \quad\;\;+ 11$

If the division "comes out even", then the remainder is zero.

So we have: .$\displaystyle k^2 - 8k + 15 \:=\:0 \quad\hdots\quad See?$

10. Thanks for the help! Will read on synthetic division further.