Remainder/factor theorem question

[ai93]

Question

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

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

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

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

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

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

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

I tried to use long division

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

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:

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

Now you just need to factor the dividend. :D

 

[ai93]

The indicated long division is carried out as follows:

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

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 \(\displaystyle (x-2)(x+3)\) :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 \(\displaystyle (x-2)(x+3)\) :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

 

[CellCoree]

To solve this problem using long division, we need to divide the given polynomial by the known factor (x+2). The steps are as follows:

1. Write the polynomial in descending order of powers of x. We have f(x) = x^4 + 4x^3 - x^2 - 16x - 12.

2. Write the divisor (x+2) on the left side and the dividend (f(x)) on the right side, leaving spaces for each term. We have:

x^2 + 3x + 2) x^4 + 4x^3 - x^2 - 16x - 12

3. Determine the first term of the quotient by dividing the first term of the dividend (x^4) by the first term of the divisor (x^2). This gives us x^2 as the first term of the quotient.

4. Multiply the first term of the quotient by the divisor and write the result below the dividend, lining up the like terms. We have:

x^2 + 3x + 2) x^4 + 4x^3 - x^2 - 16x - 12
x^4 + 2x^3

5. Subtract the result from the dividend, bringing down the next term (4x^3) to the right of the difference. We have:

x^2 + 3x + 2) x^4 + 4x^3 - x^2 - 16x - 12
x^4 + 2x^3
--------------
2x^3 - x^2

6. Repeat the process by determining the next term of the quotient and multiplying it by the divisor. We have:

x^2 + 3x + 2) x^4 + 4x^3 - x^2 - 16x - 12
x^4 + 2x^3
--------------
2x^3 - x^2
2x^3 + 4x^2

7. Subtract the result from the previous difference, bringing down the next term (-16x). We have:

x^2 + 3x + 2) x^4 + 4x^3 - x^2 - 16x - 12
x^