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 $\displaystyle f(-2)=0$.
Now then, we know the remainder should be 0, so....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
$\displaystyle k^2-8k+15=0$
Solve for the two possible values of k and you're done.
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.
Not so. Your attachment is incorrect. Review Synthetic Division
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.
Hello, mwok
We are expected to know this theorem:Find all values of $\displaystyle k$ such that $\displaystyle f(x) \:= \:kx^3 + x^2 + k^2x + 3k^2 + 11$ is divisible by $\displaystyle x + 2$
. . 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: .$\displaystyle k^2 - 8k + 15 \:=\:0 \quad\hdots\quad See?$