### Theory:

If you divide $\frac{5}{2}$, you will obtain $2$ as the quotient and $1$ as the remainder.

Also here $5$ is the

**dividend**, $2$ is the**divisor**, $2$ is the**quotient**, and $1$ is the**remainder**.So we can write $5=(2\times 2)+1$.

or Dividend \(=\) (Divisor \(×\) Quotient) \(+\) Remainder

Note that the remainder is always less than the divisor.

In algebra, long division polynomial is an algorithm for dividing a polynomial by a same or lower degree polynomial, a generally used version of the common arithmetic technique known as long division.

Let’s try to divide the two polynomials.

Example:

$\frac{p(x)}{q(x)}$ where $p(x)\; =\phantom{\rule{0.147em}{0ex}}{2x}^{3}+\phantom{\rule{0.147em}{0ex}}{x}^{2}+\phantom{\rule{0.147em}{0ex}}x$ & $q(x)\; =\phantom{\rule{0.147em}{0ex}}x$.

\((2x^3 + x^2 +x) ÷ x\)

\(=\) \(\frac {2x^3}{x}+\) \(\frac{x^2}{x}+\)\(\frac{x}{x}\)

\(=\) \(2x^2\) \(+\) \(x\) \(+\) \(1\)

In here, \((2x^3 + x^2 +x)\) is the dividend, \(x\) is the divisor, \(2x^2\) \(+\) \(x\) \(+\) \(1\) is the quotient and \(0\) is the remainder.

Note that the remainder is \(0\) as the polynomial \(q(x)\) divides the polynomial \(p

(x)\) without any remainder.

(x)\) without any remainder.

Important!

In this case, $x$ is common to each term of $\phantom{\rule{0.147em}{0ex}}{2x}^{3}+\phantom{\rule{0.147em}{0ex}}{x}^{2}+\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}$. So we can write $\phantom{\rule{0.147em}{0ex}}{2x}^{3}+\phantom{\rule{0.147em}{0ex}}{x}^{2}+\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}$ as $x\left(\phantom{\rule{0.147em}{0ex}}{2x}^{2}+\phantom{\rule{0.147em}{0ex}}x+\phantom{\rule{0.147em}{0ex}}1\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\right)$.

We say that $x$ and $\phantom{\rule{0.147em}{0ex}}{2x}^{2}+\phantom{\rule{0.147em}{0ex}}x+1\phantom{\rule{0.147em}{0ex}}$ are factors of $\phantom{\rule{0.147em}{0ex}}{2x}^{3}+\phantom{\rule{0.147em}{0ex}}{x}^{2}+\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}$, and $\phantom{\rule{0.147em}{0ex}}{2x}^{3}+\phantom{\rule{0.147em}{0ex}}{x}^{2}+\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}$ is a multiple of $x$ as well as a multiple of $\phantom{\rule{0.147em}{0ex}}{2x}^{2}+\phantom{\rule{0.147em}{0ex}}x+1\phantom{\rule{0.147em}{0ex}}$.

Example:

Let us try to divide \(p(x) =\) \(3x^2 + x + 1\) with \(q(x)= x\).

\((3x^2 + x + 1) ÷ x\) \(=\) \((3x^2 ÷ x)\) \(+ (x ÷ x)\) \(+ (1 ÷ x)\)

\(= 3x + 1 + \frac{1}{x}\).

Since the last term \((1)\) is not divisible by \(x\), we will write this expression as follows:

\(3x^2 + x + 1\) \(=\) \((3x^2 + x)\) \(+1\)

Separate the \(x\) value from the expression.

\(((3x +1) (x)) +1\)

\(((3x +1) × x) +1\)

\(3x^2 + x + 1 = \) \(((3x +1) × x) +1\)

Now compare the above expression with division algorithm, Dividend \(=\) (Divisor \(×\) Quotient) \(+\) Remainder.

Important!

Note that \(1\) is remainder here. In this case, \(3x+1\) is the quotient and \(1\) is the remainder. Also here, \(x\) is not a factor of \(3x^2 + x + 1\), since the remainder is not zero.

Example:

Let us try to divide \(3x^3 – 5x^2 + 10x – 3\) by \(3x + 1\).

Start with the setup of the long-division:

Just look at the leading terms, dividing \(3x^3\) by \(3x\) will get \(x^2\). Keep this quotient to the top:

Multiplying this \(x^2\) by \(3x + 1\), we will get \(3x^3 + 1x^2\) as below:

Then change the sign, add it down, and remember to carry the "\(+10x – 3\)" from the original dividend, giving a new bottom line of \(– 6x^2 + 10x – 3\):

Dividing the new leading term, \(– 6x^2\), by the leading term divisor, \(3x\), we get \(– 2x\). Keep it to the top:

Then multiply \(– 2x\) by \(3x + 1\) \(=\)\(– 6x^2 – 2x\) and enter under the bottom value. Change the signs, add them down, and remember to carry the dividend \(– 3\):

The last line is then \(12x-3\). Divide the new \(12x\) lead term by \(3x\), the lead term of the divisor, we will get \(+4\). Keep it on top.

To get \(12x + 4\) multiply \(4\) by \(3x + 1\). Change signs, then add them down. Thus, we end with the remainder \(-7\):

**Therefore**,

**Dividend**(\(3x^3 – 5x^2 + 10x – 3\)) \(=\)

**Divisor**(\(3x + 1\)) \(×\)

**Quotient**(\(x^2-2x + 4\)) \(+\)

**Remainder**(\(-7\)).