If you divide 52, 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=(2×2)+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.
 p(xq(x where p(x) =2x3+x2+xq(x) =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.
In this case, x is common to each term of 2x3+x2+x. So we can write 2x3+x2+x as x2x2+x+1.
We say that x and  2x2+x+1 are factors of 2x3+x2+x, and 2x3+x2+x is a multiple of x as well as a multiple of 2x2+x+1.
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
\(3x^2 + x + 1\) \(=\) \((3x^2 + x)\) \(+1\)
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.
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.
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 (\(-7\)).