MHB Remainder/factor theorem question

[ai93]

Question

A function $$f\left(x\right)$$ is defined by f\left(x\right)=x^{4}+4x^{3}-xx^{2}-16x-12
a) Show that there is no remainder when $$f\left(x\right)$$ is divided by $$(x+1)$$

b)Use the factor theorem to show that $$(x+2)$$ is a factor of $$f\left(x\right)$$

c) Using answers to a) and b) determine the remaining factors by the long division method.

MY SOLUTIONa) $$(x+1)\sqrt{x^{4}+4x^{3}-x^{2}-16x-12}$$ = $$x^{2}+3x^{2}-4x-12$$
Remainder = 0

b) If $$(x+2)$$ is a factor then $$f(-2)=0$$

$$\therefore f(-2)=(-2)^{4}+4(-2)^{3}-(-2)^{2}-16(-2)-12$$ = 0
So (x+2) is a factor.

c)
(x+1)(x+2)=$$x^{2}+3x+2$$

I tried to use long division

$$(x^{2}+3x+2)\sqrt{x^{4}+4x^{3}-xx^{2}-16x-12}$$

But I am having trouble finding the last remaining factors?
How to solve by long division?

 

[MarkFL]

The indicated long division is carried out as follows:

$$\begin{array}{r}x^2+x-6\hspace{102px}\\x^2+3x+2\enclose{longdiv}{x^4+4x^3-x^2-16x-12} \\ -\underline{\left(x^4+3x^3+2x^2\right)} \hspace{62px} \\ x^3-3x^2-16x \hspace{38px} \\ -\underline{\left(x^3+3x^2+2x\right)} \hspace{35px} \\ -6x^2-18x-12 \\ -\underline{\left(-6x^2-18x-12\right)} \hspace{-10px} \\ 0 \end{array}$$

Now you just need to factor the dividend. :D

 

[ai93]

The indicated long division is carried out as follows:

$$\begin{array}{r}x^2+x-6\hspace{102px}\\x^2+3x+2\enclose{longdiv}{x^4+4x^3-x^2-16x-12} \\ -\underline{\left(x^4+3x^3+2x^2\right)} \hspace{62px} \\ x^3-3x^2-16x \hspace{38px} \\ -\underline{\left(x^3+3x^2+2x\right)} \hspace{35px} \\ -6x^2-18x-12 \\ -\underline{\left(-6x^2-18x-12\right)} \hspace{-10px} \\ 0 \end{array}$$

Now you just need to factor the dividend. :D

I had initially got that with my rough working out, but was confused as there are 5 terms in the square root!

So the other factors would be $$(x-2)(x+3)$$ :D

 

[MarkFL]

I had initially got that with my rough working out, but was confused as there are 5 terms in the square root!

So the other factors would be $$(x-2)(x+3)$$ :D

Your factorization of the dividend is correct. That's not a square root though, that is the long division symbol, which denotes a very different operation. :D