Polynomial Long Division

Example 1. Divide $$${{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}$$$ by $$${x}+{3}$$$.

Write division in special form:

$$$\begin{array}{r}\\x+3\hspace{1pt})\overline{\hspace{1pt}x^3-2x^2-8x+21}\\\end{array}$$$

Let's carefully think what to do next.

The goal is to get rid of the leading term of dividend $$${{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}$$$.

In other words, we need to get rid of $$${{x}}^{{3}}$$$.

To do it, we divide $$${{x}}^{{3}}$$$ by the leading term of divisor $$${x}+{3}$$$, i.e. $$${x}$$$ (divide monomials): $$$\frac{{{x}}^{{3}}}{{x}}={\color{red}{{{{x}}^{{2}}}}}$$$.

Now, we subtract $$${\color{red}{{{{x}}^{{2}}}}}{\left({x}+{3}\right)}$$$ from $$${{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}$$$:

$$${{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}-{{x}}^{{2}}{\left({x}+{1}\right)}={{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}-{\left({{x}}^{{3}}+{3}{{x}}^{{2}}\right)}=$$$ (multiply monomial by polynomial)

$$$={{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}-{{x}}^{{3}}-{3}{{x}}^{{2}}=-{5}{{x}}^{{2}}-{8}{x}+{21}$$$.

Nice, we got rid of $$${{x}}^{{3}}$$$ and ended up with a polynomial of lower degree.

Let's write it in vertical division:

$$$\begin{array}{r}\color{green}{x^2}\phantom{-2x^2-8x+21}\\\color{green}{x+3}\hspace{1pt})\overline{\hspace{1pt}x^3-2x^2-8x+21}\\-\underline{\color{green}{(x^3+3x^2)}}\phantom{-8x+21}\\\end{array}$$$

You need to be careful with minus sign (we subtract polynomials). The best way is to change signs of terms, and discard parenthesis and minus sign in front of parenthesis, in other words add polynomials.

$$$\begin{array}{r}\color{green}{x^2}\phantom{-2x^2-8x+21}\\\color{green}{x+3}\hspace{1pt})\overline{\hspace{1pt}x^3-2x^2-8x+21}\\\underline{\color{green}{\color{red}{-}x^3\color{red}{-}3x^2}}\phantom{-8x+21}\\\end{array}$$$

Next, perform addition:

$$$\begin{array}{r}\color{green}{x^2}\phantom{-2x^2-8x+21}\\\color{green}{x+3}\hspace{1pt})\overline{\hspace{1pt}x^3-2x^2-8x+21}\\\underline{\color{green}{-x^3-3x^2}}\phantom{-8x+21}\\-5x^2-8x+21\end{array}$$$

Now, repeat above steps.

We need to get rid of $$$-{5}{{x}}^{{2}}$$$.

To do it, we divide $$$-{5}{{x}}^{{2}}$$$ by the leading term of divisor $$${x}+{3}$$$, i.e. $$${x}$$$: $$$\frac{{-{5}{{x}}^{{2}}}}{{x}}={\color{red}{{-{5}{x}}}}$$$.

Next, subtract $$${\color{red}{{-{5}{x}}}}{\left({x}+{3}\right)}$$$ from $$$-{5}{{x}}^{{2}}-{8}{x}+{21}$$$:

$$$-{5}{{x}}^{{2}}-{8}{x}+{21}-{\left(-{5}{x}{\left({x}+{3}\right)}\right)}=-{5}{{x}}^{{2}}-{8}{x}+{21}-{\left(-{5}{{x}}^{{2}}-{15}{x}\right)}=$$$ (multiply monomial by polynomial)

$$$=-{5}{{x}}^{{2}}-{8}{x}+{21}+{5}{{x}}^{{2}}+{15}{x}={7}{x}+{21}$$$.

We got rid of $$$-{5}{{x}}^{{2}}$$$ and ended up with a polynomial of lower degree:

$$$\begin{array}{r}x^2\color{purple}{-5x}\phantom{-8x^2+21}\\\color{green}{x+3}\hspace{1pt})\overline{\hspace{1pt}x^3-2x^2-8x+21}\\\underline{-x^3-3x^2}\phantom{-8x+21}\\-5x^2-8x+21\\\underline{\color{purple}{\color{red}{+}5x^2\color{red}{+}15x}\phantom{+21}}\\7x+21\end{array}$$$

Finally, get rid of $$${7}{x}$$$ by multiplying $$${x}+{3}$$$ by $$${7}$$$:

$$$\begin{array}{r}x^2-5x+\color{brown}{7}\phantom{x+21}\\\color{brown}{x+3}\hspace{1pt})\overline{\hspace{1pt}x^3-2x^2-8x+21}\\\underline{-x^3-3x^2}\phantom{-8x+21}\\-5x^2-8x+21\\\underline{+5x^2+15x}\phantom{+21}\\7x+21\\\underline{\color{brown}{-7x-21}}\\\color{cyan}{0}\end{array}$$$

We are done, because remainder is 0.

Thus, we can write that $$$\frac{{{{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}}}{{{x}+{3}}}={{x}}^{{2}}-{5}{x}+{7}$$$.

From another side it means, that $$${{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}={\left({x}+{3}\right)}{\left({{x}}^{{2}}-{5}{x}+{7}\right)}$$$.

To better understand the process of long division, let's write, that

$$${\color{purple}{{{{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}={{x}}^{{2}}{\left({x}+{3}\right)}+{\left(-{5}{x}\right)}{\left({x}+{3}\right)}+{7}{\left({x}+{3}\right)}}}}$$$ (you can simplify right hand side to make sure, that this is equality).

Each summand in the right hand side corresponds to the step in the division process.

Now, above equality can be divided by $$${\left({x}+{3}\right)}$$$: $$$\frac{{{{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}}}{{{x}+{3}}}=\frac{{{{x}}^{{2}}{\left({x}+{3}\right)}}}{{{x}+{3}}}+\frac{{-{5}{x}{\left({x}+{3}\right)}}}{{{x}+{3}}}+\frac{{{7}{\left({x}+{3}\right)}}}{{{x}+{3}}}={{x}}^{{2}}-{5}{x}+{7}$$$.

Note, how $$${x}+{3}$$$'s in the denominators and numerators were cancelled.

Answer: $$$\frac{{{{x}}^{{3}}-{2}{{x}}^{{2}}-{8}{x}+{21}}}{{{x}+{3}}}={{x}}^{{2}}-{5}{x}+{7}$$$.

Example 2. Perform the long division: $$$\frac{{{6}{{x}}^{{3}}-{17}{{x}}^{{2}}-{8}{x}+{2}}}{{{3}{{x}}^{{2}}+{2}{x}-{1}}}$$$.

I will do this example a bit faster.

$$$\begin{array}{r}\color{green}{2x}\phantom{-^3}\color{purple}{-7}\phantom{-8x^2+21}\\3x^2+2x-1\hspace{1pt})\overline{\hspace{1pt}6x^3-17x^2-8x+2}\\\underline{\color{green}{\color{red}{-}6x^3\color{red}{-}4x^2\color{red}{+}2x}}\phantom{+21}\\-21x^2-6x+2\\\underline{\color{purple}{\color{red}{+}21x^2\color{red}{+}14x\color{red}{-}7}}\\\color{cyan}{8x-5}\end{array}$$$

We are done, because degree of the remainder $$${8}{x}-{5}$$$ is less than degree of the divisor $$${3}{{x}}^{{2}}+{2}{x}-{1}$$$, however, remainder is not zero.

We can write now (according to the steps), that

$$${6}{{x}}^{{3}}-{17}{{x}}^{{2}}-{8}{x}+{2}={2}{x}{\left({\underline{{{3}{{x}}^{{2}}+{2}{x}-{1}}}}\right)}-{7}{\left({\underline{{{3}{{x}}^{{2}}+{2}{x}-{1}}}}\right)}+{\left({8}{x}-{5}\right)}$$$.

Or $$${\color{purple}{{{6}{{x}}^{{3}}-{17}{{x}}^{{2}}-{8}{x}+{2}={\left({2}{x}-{7}\right)}{\left({3}{{x}}^{{2}}+{2}{x}-{1}\right)}+{8}{x}-{5}}}}$$$ (distributive property of multiplication: $$${a}{c}-{b}{c}={\left({a}-{b}\right)}{c}$$$).

Finally, $$$\frac{{{6}{{x}}^{{3}}-{17}{{x}}^{{2}}-{8}{x}+{2}}}{{{3}{{x}}^{{2}}+{2}{x}-{1}}}={2}{x}-{7}+\frac{{{8}{x}-{5}}}{{{3}{{x}}^{{2}}+{2}{x}-{1}}}$$$.

Answer: $$$\frac{{{6}{{x}}^{{3}}-{17}{{x}}^{{2}}-{8}{x}+{2}}}{{{3}{{x}}^{{2}}+{2}{x}-{1}}}={2}{x}-{7}+\frac{{{8}{x}-{5}}}{{{3}{{x}}^{{2}}+{2}{x}-{1}}}$$$, quotient is $$${2}{x}-{7}$$$, remainder is $$${8}{x}-{5}$$$.

Example 3. Find quotient and remainder of $$$\frac{{{{x}}^{{3}}-{1}}}{{{2}{x}+{8}}}$$$.

When we perform long division, there will appear terms involving $$${x}$$$ and $$${{x}}^{{2}}$$$. However, there are no such terms in $$${{x}}^{{3}}-{1}$$$.

To fix situation, we add mentioned terms with zero coefficient: $$${{x}}^{{3}}-{1}={{x}}^{{3}}+{0}{{x}}^{{2}}+{0}{x}-{1}$$$. We haven't changed anything, because adding zero doesn't change anything.

Now, perform long division:

$$$\begin{array}{r}\color{green}{\frac{1}{2}x^2}\color{purple}{-2x}\color{brown}{+8}\phantom{x-1}\\2x+8\hspace{1pt})\overline{\hspace{1pt}x^3\color{magenta}{+0x^2+0x}-1}\\\underline{\color{green}{\color{red}{-}x^3\color{red}{-}4x^2}}\phantom{+0x-1}\\-4x^2+0x-1\\\underline{\color{purple}{\color{red}{+}4x^2\color{red}{+}16x}}\phantom{-1}\\16x-1\\\underline{\color{brown}{-16x-64}}\\\color{cyan}{-65}\end{array}$$$

Therefore, $$${{x}}^{{3}}-{1}={\left(\frac{{1}}{{2}}{{x}}^{{2}}-{2}{x}+{8}\right)}{\left({2}{x}+{8}\right)}-{65}$$$.

Answer: Quotient is $$$\frac{{1}}{{2}}{{x}}^{{2}}-{2}{x}+{8}$$$, remainder is $$$-{65}$$$.

Example 4. Divide $$${3}{{x}}^{{3}}{{y}}^{{3}}+{{x}}^{{2}}{{y}}^{{3}}+{{x}}^{{2}}{{y}}^{{2}}+{x}{{y}}^{{2}}-{2}{x}{y}$$$ by $$${x}{y}+{1}$$$.

$$$\begin{array}{r}3x^2y^2-2xy+xy^2\phantom{+x^2y^3+x^2y^2}\\xy+1\hspace{1pt})\overline{\hspace{1pt}3x^3y^3+x^2y^2-2xy+x^2y^3+xy^2}\\\underline{-3x^3y^3-3x^2y^2}\phantom{+x^2y^3+xy^2-2xy}\\-2x^2y^2-2xy+x^2y^3+xy^2\\\underline{+2x^2y^2+2xy}\phantom{+x^2y^3+xy^2}\\0+x^2y^3+xy^2\\\underline{-x^2y^3-xy^2}\\\color{cyan}{0}\end{array}$$$

Note, that we swapped some terms to perform division.

Answer: $$$\frac{{{3}{{x}}^{{3}}{{y}}^{{3}}+{{x}}^{{2}}{{y}}^{{3}}+{{x}}^{{2}}{{y}}^{{2}}+{x}{{y}}^{{2}}-{2}{x}{y}}}{{{x}+{1}}}={3}{{x}}^{{2}}{{y}}^{{2}}-{2}{x}{y}+{x}{{y}}^{{2}}$$$.