Synthetic Division

Related Calculators: Synthetic Division Calculator , Polynomial Calculator

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`.

Advantages of this method:

  • less space on a paper
  • fewer calculations
  • calculations are made without variables
  • fewer sign errors, because additions are made instead of subtractions (as in long division)

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

Example 1. Divide `x^2-7x+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^2color(cyan)(-7)xcolor(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*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`: `(-5)*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 `1x-5` or simply `x-5`.

Answer: `(x^2-7x+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 `(-5x+3x^3-4)/(x+1)`.

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

There is missing term, involving `x^2`, so we add it with zero coefficient: `3x^3+color(red)(0x^2)-5x-4`.

Since, `x+1=x-(-1)`, 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*(-1)=-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`: `(-3)*(-1)=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`: `(-2)*(-1)=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 `3x^2-3x-2` and remainder is `-2`.

Therefore, we can write, that `(3x^3-5x-4)=(3x^2-3x-2)(x+1)-2`.

Or `(3x^3-5x-4)/(x+1)=3x^2-3x-2-2/(x+1)`.

Answer: `(3x^3-5x-4)/(x+1)=3x^2-3x-2-2/(x+1)`.

Last example will be solved faster.

Example 3. Find quotient and remainder of `(-10+4x+x^4-3x^3)/(x-3)`.

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

There is missing term, involving `x^2`, so we add it with zero coefficient: `x^4-3x^3+color(red)(0x^2)+4x-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 `1x^3+0x^2+0x+4=x^3+4` and remainder is `2`.

Answer: `(x^4-3x^2+4x-10)/(x-3)=x^3+4+2/(x-3)`.

Now, it is time to exercise.

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

Answer: `(x^2-2x+5)/(x-3)=x+1+8/(x-3)`.

Exercise 2. Divide the following: `(x^3+8)/(x+2)`.

Answer: `x^2-2x+4`.

Exercise 3. Find quotient and remainder of `(2x^4+7x^3-15x^2+x+9)/(x+5)`.

Answer: Quotient is `2x^3-3x^2+1` and remainder is `4`.