# Synthetic Division

Synthetic Division is a method, similar to polynomial long division, but it requires less writing and fewer calculations. However, it can be used only for dividing polynomial in one variable by linear polynomial ${x}-{a}$.

• less space on a paper
• fewer calculations
• calculations are made without variables

Disadvantage: it is only used for dividing polynomial by LINEAR polynomial. In other cases it won't work.

Example 1. Divide ${{x}}^{{2}}-{7}{x}+{10}$ by ${x}-{2}$, using synthetic division.

First, we need to order degrees of polynomial from greatest to lowest.

It is already done: ${\color{green}{{{1}}}}{{x}}^{{2}}{\color{cyan}{{-{7}}}}{x}{\color{purple}{{+{10}}}}$.

Also, ${\color{brown}{{{a}={2}}}}$.

Next, write coefficients in the special form:

$\begin{array}{r|lll}\phantom{2}&x^2&x^1&x^0\\\color{brown}{2}&\color{green}{1}&\color{cyan}{-7}&\color{purple}{10}\\\phantom{1}&\phantom{-7}&\phantom{10}\\\hline \end{array}$

Now, perform the following steps:

 Take coefficient of the leading term and carry it down unchanged. $\begin{array}{r|lll}2&1&-7&10\\\phantom{2}&\color{red}{\downarrow}&\phantom{-7}&\phantom{10}\\\hline\phantom{2}&\color{red}{1}&\phantom{-7}&\phantom{10}\end{array}$ Multiply carry-down value by ${a}$: ${1}\cdot{2}={2}$, and write result into the next column. $\begin{array}{r|lll}\color{red}{2}&1&-7&10\\\phantom{2}&\downarrow&\color{red}{+2}&\phantom{10}\\\hline\phantom{2}&\color{red}{1^{\nearrow}}&\phantom{-7}&\phantom{10}\end{array}$ Add down the column. $\begin{array}{r|lll}2&1&\color{red}{-7}&10\\\phantom{2}&\downarrow&\color{red}{+2}&\phantom{10}\\\hline \phantom{2}&1^{\nearrow}&\color{red}{-5}&\phantom{10}\end{array}$ Multiply calculated result by ${a}$: ${\left(-{5}\right)}\cdot{2}=-{10}$, and write the result into the next column. $\begin{array}{r|lll}\color{red}{2}&1&-7&\phantom{-}10\\\phantom{2}&\downarrow&+2&\color{red}{-10}\\\hline \phantom{2}&1^{\nearrow}&\color{red}{-5^{\nearrow}}&\phantom{10}\end{array}$ Add down the column. $\begin{array}{r|lll}2&1&-7&\phantom{-}\color{red}{10}\\\phantom{2}&\downarrow&+2&\color{red}{-10}\\\hline \phantom{2}&1^{\nearrow}&-5^{\nearrow}&\phantom{-1}\color{red}{0}\end{array}$

We are done.

Coefficients under the horizontal line (except last) are coefficients of the quotient. Last coefficient is a remainder (it is zero).

Thus, quotient is ${1}{x}-{5}$ or simply ${x}-{5}$.

Answer: $\frac{{{{x}}^{{2}}-{7}{x}+{10}}}{{{x}-{2}}}={x}-{5}$.

Sometimes, there will be missing terms.

For example, in ${{x}}^{{3}}-{1}$ missing terms are the terms, that involve ${{x}}^{{2}}$ and ${x}$.

To fix that, just add missing terms with zero coefficient. This doesn't change anything.

Example 2. Divide $\frac{{-{5}{x}+{3}{{x}}^{{3}}-{4}}}{{{x}+{1}}}$.

Order degrees of dividend from greatest to lowest: ${3}{{x}}^{{3}}-{5}{x}-{4}$.

There is missing term, involving ${{x}}^{{2}}$, so we add it with zero coefficient: ${3}{{x}}^{{3}}+{\color{red}{{{0}{{x}}^{{2}}}}}-{5}{x}-{4}$.

Since, ${x}+{1}={x}-{\left(-{1}\right)}$, then ${a}=-{1}$.

Next, write coefficients in the special form:

$\begin{array}{r|llll}\phantom{-1}&x^3&x^2&x^1&x^0\\-1&3&0&-5&-4\\\phantom{1}&\phantom{-7}&\phantom{10}\\\hline \end{array}$

Now, perform the following steps:

 Take coefficient of the leading term and carry it down unchanged. $\begin{array}{r|llll}-1&3&0&-5&-4\\\phantom{-1}&\color{red}{\downarrow}&\phantom{0}&\phantom{-5}&\phantom{-4}\\\hline\phantom{-1}&\color{red}{3}&\phantom{0}&\phantom{-5}&\phantom{-4}\end{array}$ Multiply carry-down value by ${a}$: ${3}\cdot{\left(-{1}\right)}=-{3}$, and write result into the next column. $\begin{array}{r|llll}\color{red}{-1}&3&\phantom{-}0&-5&-4\\\phantom{-1}&\downarrow&\color{red}{-3}&\phantom{-5}&\phantom{-4}\\\hline\phantom{-1}&\color{red}{3^{\nearrow}}&\phantom{-3}&\phantom{-5}&\phantom{-4}\end{array}$ Add down the column. $\begin{array}{r|llll}-1&3&\phantom{-}\color{red}{0}&-5&-4\\\phantom{-1}&\downarrow&\color{red}{-3}&\phantom{-5}&\phantom{-4}\\\hline\phantom{-1}&3^{\nearrow}&\color{red}{-3}&\phantom{-5}&\phantom{-4}\end{array}$ Multiply calculated result by ${a}$: ${\left(-{3}\right)}\cdot{\left(-{1}\right)}={3}$, and write the result into the next column. $\begin{array}{r|llll}\color{red}{-1}&3&\phantom{-}0&-5&-4\\\phantom{-1}&\downarrow&-3&\color{red}{+3}&\phantom{-4}\\\hline\phantom{-1}&3^{\nearrow}&\color{red}{-3^{\nearrow}}&\phantom{-5}&\phantom{-4}\end{array}$ Add down the column. $\begin{array}{r|llll}-1&3&\phantom{-}0&\color{red}{-5}&-4\\\phantom{-1}&\downarrow&-3&\color{red}{+3}&\phantom{-4}\\\hline\phantom{-1}&3^{\nearrow}&-3^{\nearrow}&\color{red}{-2}&\phantom{-4}\end{array}$ Multiply calculated result by ${a}$: ${\left(-{2}\right)}\cdot{\left(-{1}\right)}={2}$, and write the result into the next column. $\begin{array}{r|llll}\color{red}{-1}&3&\phantom{-}0&-5&-4\\\phantom{-1}&\downarrow&-3&+3&\color{red}{+2}\\\hline\phantom{-1}&3^{\nearrow}&-3^{\nearrow}&\color{red}{-2^{\nearrow}}&\phantom{-4}\end{array}$ Add down the column. $\begin{array}{r|llll}-1&3&\phantom{-}0&-5&\color{red}{-4}\\\phantom{-1}&\downarrow&-3&+3&\color{red}{+2}\\\hline\phantom{-1}&3^{\nearrow}&-3^{\nearrow}&-2^{\nearrow}&\color{red}{-2}\end{array}$

We are done.

Coefficients under the horizontal line (except last) are coefficients of the quotient. Last coefficient is a remainder (it is non-zero).

Thus, quotient is ${3}{{x}}^{{2}}-{3}{x}-{2}$ and remainder is $-{2}$.

Therefore, we can write, that ${\left({3}{{x}}^{{3}}-{5}{x}-{4}\right)}={\left({3}{{x}}^{{2}}-{3}{x}-{2}\right)}{\left({x}+{1}\right)}-{2}$.

Or $\frac{{{3}{{x}}^{{3}}-{5}{x}-{4}}}{{{x}+{1}}}={3}{{x}}^{{2}}-{3}{x}-{2}-\frac{{2}}{{{x}+{1}}}$.

Answer: $\frac{{{3}{{x}}^{{3}}-{5}{x}-{4}}}{{{x}+{1}}}={3}{{x}}^{{2}}-{3}{x}-{2}-\frac{{2}}{{{x}+{1}}}$.

Last example will be solved faster.

Example 3. Find quotient and remainder of $\frac{{-{10}+{4}{x}+{{x}}^{{4}}-{3}{{x}}^{{3}}}}{{{x}-{3}}}$.

Order degrees of polynomial from greatest to lowest: ${{x}}^{{4}}-{3}{{x}}^{{3}}+{4}{x}-{10}$.

There is missing term, involving ${{x}}^{{2}}$, so we add it with zero coefficient: ${{x}}^{{4}}-{3}{{x}}^{{3}}+{\color{red}{{{0}{{x}}^{{2}}}}}+{4}{x}-{10}$.

Also, ${a}={3}$.

Table for synthetic division is the following:

$\begin{array}{r|lllll}3&1&-3&\phantom{-}0&\phantom{-}4&-10\\\phantom{-1}&\downarrow&+3&+0&+0&+12\\\hline\phantom{3}&1^{\nearrow}&\phantom{-}0^{\nearrow}&\phantom{-}0^{\nearrow}&\phantom{-}4^{\nearrow}&\phantom{+1}2\end{array}$

So, quotient is ${1}{{x}}^{{3}}+{0}{{x}}^{{2}}+{0}{x}+{4}={{x}}^{{3}}+{4}$ and remainder is ${2}$.

Answer: $\frac{{{{x}}^{{4}}-{3}{{x}}^{{2}}+{4}{x}-{10}}}{{{x}-{3}}}={{x}}^{{3}}+{4}+\frac{{2}}{{{x}-{3}}}$.

Now, it is time to exercise.

Exercise 1. Use synthetic division to divide ${{x}}^{{2}}-{2}{x}+{5}$ by ${x}-{3}$.

Answer: $\frac{{{{x}}^{{2}}-{2}{x}+{5}}}{{{x}-{3}}}={x}+{1}+\frac{{8}}{{{x}-{3}}}$.

Exercise 2. Divide the following: $\frac{{{{x}}^{{3}}+{8}}}{{{x}+{2}}}$.

Answer: ${{x}}^{{2}}-{2}{x}+{4}$.

Exercise 3. Find quotient and remainder of $\frac{{{2}{{x}}^{{4}}+{7}{{x}}^{{3}}-{15}{{x}}^{{2}}+{x}+{9}}}{{{x}+{5}}}$.

Answer: Quotient is ${2}{{x}}^{{3}}-{3}{{x}}^{{2}}+{1}$ and remainder is ${4}$.