We carry out the process of division by means of the following steps:
`text (Step 1 : )` We write the dividend `x + 3x^2 – 1` and the divisor `1 + x` in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is `3x^2 + x –1` and divisor is `x + 1`.
`text (Step 2 : )` We divide the first term of the dividend by the first term of the divisor, i.e., we divide `(3x^2)/x =3x` = first term of quotient of the quotient.
`text (Step 3 : )` We multiply the divisor by the first term of the quotient, and subtract this product from
the dividend, i.e., we multiply `x + 1 `by `3x` and subtract the product `3x^2 + 3x` from the dividend `3x^2 + x – 1`. This gives us the remainder as `–2x – 1` .
`text (Step 4 : )` We treat the remainder `–2x – 1` as the new dividend. The divisor remains the same. We repeat Step 2 to get the
next term of the quotient, i.e., we divide `(-2x)/x = -2` the first term `– 2x` of the (new) dividend by the first term `x` of the divisor and obtain = second term of quotient `– 2`. Thus, `– 2` is the second term in the quotient.
`text (Step 5 : )` We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply `x + 1` by `– 2` and subtract the product `– 2x – 2` from the dividend `– 2x – 1`. This gives us` 1` as the remainder.
This process continues till the remainder is 0 or the degree of the new dividend is less
than the degree of the divisor. At this stage, this new dividend becomes the remainder
and the sum of the quotients gives us the whole quotient.
`text (Step 6 : )` Thus, the quotient in full is `3x – 2` and the remainder is `1`.
Let us look at what we have done in the process above as a whole:
Notice that `3x^2 + x – 1 = (x + 1) (3x – 2) + 1`
i.e.,`text ( Dividend ) = ( text (Divisor) × text (Quotient ) ) + text ( Remainder)`
In general, if `p(x)` and `g(x)` are two polynomials such that degree of `p(x) ≥` degree of
`g(x)` and `g(x) ≠ 0`, then we can find polynomials `q(x)` and `r(x)` such that:
`p(x) = g(x)q(x) + r(x)`,
where `r(x) = 0` or degree of `r(x) <` degree of `g(x)`. Here we say that `p(x)` divided by
`g(x)`, gives `q(x)` as quotient and `r(x)` as remainder.
In the example above, the divisor was a linear polynomial. In such a situation, let us
see if there is any link between the remainder and certain values of the dividend.
In `p(x) = 3x^2 + x – 1`, if we replace `x` by `–1`, we have
`p(–1) = 3(–1)^2 + (–1) –1 = 1`
So, the remainder obtained on dividing `p(x) = 3x^2 + x – 1` by `x + 1` is the same as the
value of the polynomial `p(x)` at the zero of the polynomial `x + 1`, i.e., `–1`.
Let us consider some more examples.