### Theory:

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.
If $$p(x)$$ 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!
If $$(x-1)$$ is the factor of any polynomial $$p(x)$$ if and only if sum of the coefficients of $$p(x)$$ is equal to zero.

If $$(x+1)$$ 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 $$12$$.

$$x^2 - 7x + 12$$ $$=$$ $$x^2 - 3x -4x + 12$$

Now factor out the common terms and group them.

$$=$$ $$x(x-3) -4(x-3)$$

$$=$$ $$(x-3)(x-4)$$

Thus, the linear factors of $$p(x)$$ is $$(x-2)(x-3)(x-4)$$.