Dividing Polynomials

Polynomial Long Division

Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example:

Example 1

Divide $x^{2}-2x-15$ (the dividend or numerator) by $x+3$ (the divisor or denominator)}} Similar to long division of numbers, we set up our problem as follows:

{\begin{array}{rl}\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\end{array}}

First we have to answer the question, how many times does $x+3$ go into $x^{2}$ ? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in $x$ times. We record this above the leading term of the dividend:

{\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\\end{array}}

, and we multiply $x+3$ by $x$ and write this below the dividend as follows:

{\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\\end{array}}

Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend:

{\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\&\!\!\!\!~~~~~~-5x-15~~~\\\end{array}}

Now we repeat, treating the bottom line as our new dividend:

{\begin{array}{rl}&~~\,x-5\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\&\!\!\!\!~~~~~~-5x-15~~~\\&\!\!\!\!~~~-{\underline {(-5x-15)~~~}}\\&\!\!\!\!~~~~~~~~~~~~~~~~~~~0~~~\\\end{array}}

In this case we have no remainder.

Example 2

What about a non-divisible polynomials like this?

(3x^{2}+3x-4)/(x-4)

Long division method
1	We first consider the highest-degree terms from both the dividend and divisor, the result is the first term of our quotient.	$(3x^{2})/(x)=3x$	${\begin{array}{r\|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$
2	Then we multiply this by our divisor.	$(3x)\times (x-4)=3x^{2}-12x$
3	And subtract the result from our dividend.	$(3x^{2}+3x-4)-(3x^{2}-12x)=15x-4$
4	Now once again with the highest-degree terms of the remaining polynomial, and we got the second term of our quotient.	$(15x)/(x)=15$
5	Multiplying...	$(15)\times (x-4)=15x-60$
6	Subtracting...	$(15x-4)-(15x-60)=56$
7	We are left with a constant term - our remainder:	${\begin{array}{lcr}Q(x)=3x+15&&R=56\end{array}}$