### Theory:

In the previous section, we dealt with quadratic equations and their zeroes. However, in this section, we are going to deal with the division algorithm for polynomials.

Here, the polynomials have a degree of $$3$$ and more.
Division algorithm for polynomials
Let us divide the expression $$x^3 - 3x^2 - 3 + 5x$$ by $$x^2 - 2$$, try to apply division algorithm on the same, and find the zeroes of the expression step-by-step.

Step 1: Arrange the terms in the ascending order of their degrees.

The term $$x^3 - 3x^2 - 3 + 5x$$ becomes $$x^3 - 3x^2 + 5x - 3$$.

$$x^2 - 2$$ is already in the proper format, and hence it does not require a change.

Step 2: Divide the highest degree of the dividend using the highest degree of the divisor.

Here, the first term of the dividend is $$x^3$$, and the first term of the divisor is $$x^2$$.

$$\frac{\text{The first term of the dividend}}{\text{The first term of the divisor}} = \frac{x^3}{x^2}$$ $$=$$ $$x$$.

Thus, $$x$$ becomes the first term of the quotient. Step 3: Multiply the now-formed quotient and divisor, and subtract from the dividend.

$$\text{Quotient} \times \text{Divisor} = x \times (x^2 - 2)$$

$$= x^3 - 2x$$

Let us now subtract $$x^3 - 2x$$ from $$x^3 - 3x^2 + 5x - 3$$. Step 4: Divide the highest degree of the newly-formed dividend using the highest degree of the divisor.

Here, the first term of the dividend is $$-3x^2$$, and the first term of the divisor is $$x^2$$.

$$\frac{\text{The first term of the dividend}}{\text{The first term of the divisor}} = \frac{-3x^2}{x^2}$$ $$=$$ $$-3$$.

Thus, $$-3$$ becomes the second term of the quotient. Step 5: Multiply the second term of the quotient with the divisor, and subtract from the dividend.

$$\text{Quotient} \times \text{Divisor} = (-3) \times (x^2 - 2)$$

$$= -3x^2 + 6$$

Let us now subtract $$-3x^2 + 6$$ from $$-3x^2 + 7x - 3$$. Hence, on dividing $$x^3 - 3x^2 - 3 + 5x$$ by $$x^2 - 2$$, we get $$x - 3$$ as the quotient and $$7x - 9$$ as the remainder.

A dividend is formed by multiplying the quotient with the divisor and then adding the remainder to the result.

Irrational pairs of zeroes:

In a quadratic equation with rational coefficients has a irrational or surd root $$a + √b$$, where $$a$$ and $$b$$ are rational and $$b$$ is not a perfect square, then it has also a conjugate root $$a - √b$$.

Example:
Consider the quadratic equation $$x^2-2x-1$$ with one of its roots $$x = 1+\sqrt2$$.

Let us find another root by long division method.

$\begin{array}{l}\phantom{\rule{2.646em}{0ex}}x-\left(1-\sqrt{2}\right)\\ x-\left(1+\sqrt{2}\right)\overline{){x}^{2}\phantom{\rule{0.735em}{0ex}}-2x\phantom{\rule{1.029em}{0ex}}-1}\\ \phantom{\rule{5.439em}{0ex}}\underset{¯}{\begin{array}{l}{x}^{2}-\left(1+\sqrt{2}\right)x\\ \left(-\right)\phantom{\rule{0.588em}{0ex}}\left(+\right)\phantom{\rule{2.646em}{0ex}}\end{array}}\phantom{\rule{0.147em}{0ex}}\\ \phantom{\rule{6.909em}{0ex}}\underset{¯}{\begin{array}{l}-\left(1-\sqrt{2}\right)x\phantom{\rule{0.147em}{0ex}}-1\\ -\left(1-\sqrt{2}\right)x-\phantom{\rule{0.147em}{0ex}}1\\ \left(-\right)\phantom{\rule{0.588em}{0ex}}\left(+\right)\phantom{\rule{2.646em}{0ex}}\end{array}}\\ \phantom{\rule{9.114em}{0ex}}0\end{array}$

Here the other root of the quadratic equation is $$x-(1-\sqrt2) = 0$$. That is, $$x = 1-\sqrt2$$.

Therefore, the other root is $$x = 1-\sqrt2$$.

It can be noticed that $$(1+\sqrt2)$$ and $$(1-\sqrt2)$$ are conjuage pairs.
Important!
Let $$q(x)$$, $$d(x)$$, and $$r(x)$$ be the quotient, divisor and the remainder of the dividend respectively.

$$\text{The dividend} = (\text{The quotient} \times \text{The divisor}) + \text{The remainder}$$

$$=$$ $$(q(x)$$ $$\times$$ $$d(x))$$ $$+$$ $$r(x)$$