Factorisation using synthetic division — lesson. Mathematics State Board, Class 9.

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Synthetic division helps in factorising a cubic polynomial into its linear factors.

Procedure:

To perform the synthetic division, first, we need to find one of its linear factor using the factor theorem by trial and error method.

$p(x)$ is a polynomial of degree

$n \geq 1$ and ‘

$a$ ’ is any real number then:
(i)

$p(a) = 0$ implies

$(x - a)$ is a factor of

$p(x)$ .
(ii)

$(x-a)$ is a factor of

$p(x)$ implies

$p(a) = 0$ .

Perform the synthetic division with the given polynomial as the dividend and the obtained linear factor as the divisor to get a quadratic factor.
Simplify the quadratic factor further into linear factors either by performing synthetic division or by resolving the quadratic expression into factors by one the methods of factorisation.

Important!

$(x-1)$ is the factor of any polynomial

$p(x)$ if and only if sum of the coefficients of

$p(x)$ is equal to zero.

$(x+1)$ is the factor of any polynomial

$p(x)$ if and only if sum of the coefficients of even power of

$x$ including the constant is equal to sum of the coefficients of odd power of

$x$ .

Let us learn how to factorise a cubic polynomial by synthetic division with an example.

Example:

Consider the polynomial

$p(x)$

$=$

$x^3 - 9x^2 + 26x - 24$ .

First find one of the linear factors of

$p(x)$ as follows:

Sum of the coefficients of even power of

$x$ including the constant

$=$

$-9-24$

$=$

$-33$

Sum of the coefficients of odd power of

$x$

$=$

$1+26$

$=$

$27$

Here

$-33 \neq 15$ .

Hence,

$(x+1)$ is not a factor.

Sum of all the coefficients of

$p(x)$

$=$

$1 - 9 + 26 - 24$

$=$

$-6$

$\neq 0$

Hence,

$(x-1)$ is not a factor.

Using factor theorem, verify if

$(x-2)$ is a factor of

$p(x)$ .

This implies

$x = 2$ .

$p(2)$

$=$

$(2)^3 - 9(2)^2 + 26(2) - 24$

$=$

$8 - 36 + 52 - 24$

$=$

$0$

Since

$p(x)$

$=$

$0$ , this implies

$(x-2)$ is a factor of

$p(x)$ .

Now perform the synthetic division with

$p(x)$

$=$

$x^3 - 9x^2 + 26x - 24$ as the dividend and

$x=2$ as the zero of the divisor.

$\begin{array}{r|rrrrr}{2} & 1 & -9 & 26 & -24 \\& 0 & 2 & -14 & 24 \\\hline & 1 & -7 & 12 &\underline{\begin{array}{|r} {0} \end {array}} \end{array}$

The quadratic factor obtained by synthetic division is

$x^2 - 7x + 12$ .

If possible this quadratic factor is further resolved into linear factors by either of the following two methods.

Method 1: Resolve further by the method of synthetic division.

$\begin{array}{r|rrrrr}{2} & 1 & -9 & 26 & -24 \\& 0 & 2 & -14 & 24 \\\hline 3 & 1 & -7 & 12 &\underline{\begin{array}{|r} {0}\end {array}} & \text{(Remainder)}\\ & 0 & 3 & -12 \\\hline & 1 & -4 &\underline{\begin{array}{|r} {0}\end {array}} & & \text{(Remainder)}\end{array}$

Thus, the linear factors of

$p(x)$ is

$(x-2)(x-3)(x-4)$ .

Method 2: Factorise

$x^2 - 7x + 12$ .

Split the coefficient of

$x$ into two numbers in such a way that the sum of the numbers is

$-7$ and the product of the numbers is

$12$ .

$x^2 - 7x + 12$

$=$

$x^2 - 3x -4x + 12$