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Perform the following divisions. For each, express your answer in both quotient-remainder form, and then as the sum of a polynomial and proper rational expression.$\require{color}$
$(x^3 + 10x^2 + 12x + 4) \div (x+6)$
$\cfrac{x^3 - 5x + 4}{x-2}$
$(x^3+4x^2-6x+3) \div (x^2-5x+7)$
$\cfrac{x^4 - 2x^2 + 1}{3x-1}$
$\displaystyle{
\begin{array}[t]{l}
{\color{white}x+6 \ )\ x^3 + 10}x^2 + \ 4x - 12\\
x+6 \ \overline{\smash{\raise{.6ex}{) \ }}{\vphantom{)}} x^3 + 10x^2 + 12x + \,4}\\
{\color{white}x+6 \ )\ }\underline{x^3 + \ 6x^2}\\
{\color{white}x+6 \ )\ x^3 + {\ }}4x^2\\
{\color{white}x+6 \ )\ x^3 + {\ }}\underline{4x^2 + \,24x}\\
{\color{white}x+6 \ )\ x^3 + {\ }4x^2} \ -12x\\
{\color{white}x+6 \ )\ x^3 + {\ }4x^2}\underline{\ -12x - 72}\\
{\color{white}x+6 \ )\ x^3 + {\ }4x^2 - 12x + {}} \,76
\end{array}
}$
Quotient-remainder form:
$x^2 + 4x - 12$, with remainder $76$
Quotient as a sum of a polynomial and (proper) rational expression:
$\displaystyle{x^2 + 4x - 12 + \cfrac{76}{x+6}}$
Quotient-remainder form:
$x^2+2x-1$, with remainder $2$
Quotient as sum of a polynomial and (proper) rational expression:
$\displaystyle{x^2 + 2x - 1 + \cfrac{2}{x-2}}$
$\displaystyle{
\begin{array}[t]{l}
\hphantom{x^2 - 3x + 2 \ )\ x^3 + 5x^2 + \ 6}\ x \ + \ 9\\
x^2 - 5x + 7 \ \overline{\smash{\raise{.6ex}{)}}{\vphantom{)}} \ x^3 + 4x^2 - \ \ 6x \ + \ 3}\\
\hphantom{x^2 - 5x + 6 \ )\ } \underline{x^3 - 5x^2 + \ \ 7x}\\
\hphantom{x^2 - 5x + 6 \ )\ x^3 - {}} 9x^2 - 13x\\
\hphantom{x^2 - 5x + 6 \ )\ x^3 - {}} \underline{9x^2 - 45x + 63}\\
\hphantom{x^2 - 5x + 6 \ )\ x^3 - 5x^2 - {}} 32x - 60
\end{array}
}$
Quotient-remainder form:
$x+9$, with remainder $32x-60$
Quotient as sum of a polynomial and (proper) rational expression:
$\displaystyle{x+9 + \cfrac{32x-60}{x^2-5x+7}}$
$\displaystyle{
\begin{array}[t]{l}
\hphantom{3x - 1 \ )\ x^4 + {}}\frac{1}{3}x^3 + \frac{1}{9}x^2 - \frac{17}{27}x - \frac{17}{81}\\
3x - 1 \ \overline{\smash{\raise{.6ex}{) \ }}{\vphantom{)}} x^4 + \ 0x^3 - 2x^2 + \ \ 0x + \ 1}\\
\hphantom{3x - 1 \ )\ }\underline{x^4 - \frac{1}{3}x^3}\\
\hphantom{3x - 1 \ )\ x^4 - {}} \frac{1}{3}x^3\\
\hphantom{3x - 1 \ )\ x^4 - {}} \underline{\frac{1}{3}x^3 - \frac{1}{9}x^2}\\
\hphantom{3x - 1 \ )\ x^4 - {} \frac{1}{3}x^3} \,{-\frac{17}{9}x^2}\\
\hphantom{3x - 1 \ )\ x^4 - {} \frac{1}{3}x^3} \,\underline{-\frac{17}{9}x^2 + \frac{17}{27}x}\\
\hphantom{3x - 1 \ )\ x^4 - {} \frac{1}{3}x^3 \,-\frac{17}{9}x^2} {-\frac{17}{27}x}\\
\hphantom{3x - 1 \ )\ x^4 - {} \frac{1}{3}x^3 \,-\frac{17}{9}x^2} \underline{-\frac{17}{27}x + \frac{17}{81}}\\
\hphantom{3x - 1 \ )\ x^4 - {} \frac{1}{3}x^3 \,-\frac{17}{9}x^2 -\frac{17}{27}x + }\frac{64}{81}\\
\end{array}
}$
Quotient-remainder form:
$\frac{1}{3}x^3 + \frac{1}{9}x^2 - \frac{17}{27}x - \frac{17}{81}$, with remainder $\frac{64}{81}$
Quotient as sum of a polynomial (with rational coefficients) and (proper) rational expression:
$\frac{1}{3}x^3 + \frac{1}{9}x^2 - \frac{17}{27}x - \frac{17}{81} + \cfrac{\frac{64}{81}}{3x-1}$
Note, factoring out a $\frac{1}{81}$ will make the answer look a bit "cleaner":
$\cfrac{1}{81} (27x^3 + 9x^2 + 51x - 17) + \cfrac{\frac{64}{81}}{3x-1}$
For each pair of polynomials, express the first as a multiple of the second, plus some remainder as needed.
$(x^4 + 5x^3 - 9x + 8); \ (x - 2)$
$(x^4 - x^2 + 3x -7); \ (x^2 + x - 3)$
$(x^5 - 1); \ (x-1)$
$(x^4 - 2x^3 + 4x^2 - 6); \ (x^2 + 2x + 1)$
$(x^3+7x^2-5x-8); \ (x-3)$
$(x^4-16); \ (x-2)$
$(x^3 + 7x^2 + 14x + 19)(x-2) + 46$
$(x^2 - x + 3)(x^2 + x - 3) + (-3x+2)$
$(x^4 + x^3 + x^2 + x + 1)(x-1)$
$(x^2 - 4x + 11)(x^2 + 2x + 1) + (-18x - 17)$
$(x^2 + 10x + 25)(x-3) + 67$
$(x^3 + 2x^2 + 4x + 8)(x-2)$
For each, decide if $(x-c)$ is a factor of the given polynomial without doing any division.
$x^3 + 2x^2 + 3x - 6; \ c = 1$
$x^4 - x^3 + 2x^2 - 7x - 2; \ c = 2$
$x^5 + 1; \ c = -1$
$x^3 + 2x + 1; \ c = 1$
$(x-1)$ is a factor of $f(x) = x^3 + 2x^2 + 3x - 6$ as $f(1) = 0$
$(x-2)$ is a factor of $f(x) = x^4 - x^3 + 2x^2 - 7x - 2$ as $f(2) = 0$
$(x+1)$ is a factor of $f(x) = x^5 + 1$ as $f(-1) = 0$
$(x-1)$ is not a factor of $f(x) = x^3 + 2x + 1$ as $f(1) \neq 0$
Find $(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + x^{n-4}y^3 + x^{n-5}y^4 + x^{n-6}y^5) \div (x^2 + xy + y^2)$
$x^{n-3} + x^{n-6}y^3$
Do the division: $\cfrac{a^3 + b^3 + c^3 - 3abc}{a+b+c}$
Given that this division is of polynomials of multiple variables, you may find it challenging to set this up as a typical "long division" problem. Instead, focus on the process for doing division that inspired that "long division" algorithm.