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Download now on Google PlaySynthetic division helps in factorising a cubic polynomial into its linear factors.

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*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 \((x-2)(x-3)(x-4)\).