5.4 Division of Polynomials

Let's first look at normal division. Note that this can be used for dividing any $2$ polynomials. Let's take a simple example of $5x^{3} + 6x^{2} + 3x - 9$ divided by $(x + 2)$. It involves the following steps.

1) First term in the quotient will be the term which when multiplied with the highest-degree term in the divisor (i.e. $x$) will produce the highest-degree term in the dividend $(5x^{3})$.

2) Write down the remainder after subtracting the product of the first term and the divisor from the dividend.

3) Second term in the quotient will be the term which when multiplied with the highest-degree term in the divisor (i.e. $x$) will produce the highest-degree term in the dividend and this process continues.

4) This process continues till the degree of the remainder is lower than that of the divisor.

(When the divisor has a degree of $1$, then the remainder will have a degree of $1$ less than the divisor, which is $0$. Therefore, the remainder will be a constant.)

In the case to the right, when $5x^{3} + 6x^{2} + 3x - 9$ is divided by $(x + 2)$, we get a quotient of $5x^{2} - 4x + 11$ and a remainder of $-31$.

5.5 Remainder Theorem

For any real number $a$, the remainder when a polynomial $f(x)$ is divided by $(x - a)$ is $f(a)$.

Taking the above example, when f(x) $=$ $5x^{3} + 6x^{2} + 3x - 9$ is divided by $(x + 2)$,

Remainder $= f(-2) = 5(-2)^{3} + 6(-2)^{2} + 3(-2) - 9$ $= -40 + 24 - 6 - 9 = -31$

This is definitely a quick way to find the remainder. However, we will not be able to find the quotient using this approach. The following process of synthetic division will help save time in these questions.

5.6 Synthetic Division

Synthetic division can be applied only when the divisor is of the form $(x - a)$. Using the same example of $5x^{3} + 6x^{2} + 3x - 9$ divided by $(x + 2)$, note the following steps.

1) $a$ will be the divisor. Write the down the coefficients of each of the terms in decreasing order of their powers. 2) Fill $0$ under the first coefficient and add the two values and write it below.
3) Multiply the first sum ($5$) with the divisor ($-2$) and fill the product of ($-10$) under the next coefficient ($6$). Once again, add the two values (-4) and write it below.
4) Repeat the process outlined in step $3$ till the end. The right-most number ($-31$) in the bottom is the remainder. The other numbers to the left of it are the coefficients of the quotient whose degree will be $1$ less than the dividend. Note the the remainder in this case is $-31$. The other numbers in the bottom (i.e., $5, -4$ and $11$) form the coefficients of the quotient, i.e. $5x^{2} - 4x + 11$.

Example 24

If $3x^{3} - 9x^{2} + kx - 12$ is divisible by $(x - 3)$, then it is also divisible by: [FMS 2010]

(1) $3x^{2} - 4$ (2) $3x^{2} + 4$ (3) $3x - 4$ (4) $3x + 4$

Solution

Let $f(x) = 3x^{3} - 9x^{2} + kx -12$

As $(x - 3)$ is a factor, $f(3) = 3 \times 3^{3} - 9 \times 3^{2} + 3k - 12$
⇒ $3k = 12$
⇒ $k = 4$

∴ $f(x) = 3x^{3} - 9x^{2} + 4x - 12$

To find other factors, let's apply synthetic division to divide $f(x)$ by $(x - 3)$ and find the quotient.

The coefficients of the quotient are $3, 0$ and $4$. Therefore, the quotient is $3x^{2} + 0 \times x + 4 = 3x^{2} + 4$

When the remainder is 0, a dividend will also be divisible by its quotient. So,$f(x)$ is divisible by $3x^{2} + 4$.

Answer: (2) $3x^{2} + 4$

5.7 Factorising polynomials

A factor is a number that leaves a remainder of $0$. So, to factorise a higher-order polynomial please apply the following

1) Questions with higher order polynomial will typically have at least $1$ small integral root. So, start by applying remainder theorem and substitute values such as f($1$), f($-1$), f($2$), f($-2$), ..etc. and find the root where the remainder is $0$

2) Once you've found a root, apply synthetic division and find the quotient.

3) Once again, apply remainder theorem on the quotient to find another root and then apply synthetic division to get the next quotient.

4) Continue this process till you're left with a quadratic equation (factorising of which has been explained earlier in this lesson).

Example 25

What is the sum of the squares of the roots of the equation $4x^{3} + 9x^{2} - x - 6$ ?

Solution

Let $f(x) = 4x^{3} + 9x^{2} - x - 6$. We begin with checking if $1, -1, 2, -2,$ etc. are roots.

$f(1) = 6$
$f(-1) = -4 + 9 + 1 - 6$

∴ $x = -1$ is a root. We not apply synthetic division to find the quotient.

As the coefficients of the quotient are $4, 5$ and $-6$,

$4x^{3} + 9x^{2} - x - 6 = (x + 1)(4x^{2} + 5x - 6)$
$= (x + 1)(4x^{2} + 8x - 3x - 6)$ $= (x + 1)(4x(x + 2) - 3(x + 2))$
$= (x + 1)(4x - 3)(x + 2)$

Roots $= -1, \dfrac{3}{4}, -2$

Sum of Squares of roots $= 1 + \dfrac{9}{16} + 4 = \dfrac{89}{16}$

Answer: $\dfrac{89}{16}$