Factorising the cubic polynomial — lesson. Maths CBSE Mentoring, Class 9 Mentoring product.

Steps to factorize the cubic polynomial \(p(x)\).

Step 1: Find \(x = a\) where \(p(a) = 0\). That is, we have to find one of the factors.

Step 2: Then \(x-a\) is a factor of \(p(x)\).

Step 3: Now divide \(p(x)\) by \(x-a\). That is \(\frac{p(x)}{(x-a)}\).

Step 4: Then factorize the quotient(quadratic equation) by splitting its middle term.

Step 1: Let us find one of factors by trial method.

Consider the polynomial

e^{3} - e^{2} + 13 e \underline{- 13}

The product of coefficient of \(e^3\) and constant \(=\) \(1 \times -13\) \(=\) \(-13\).

We shall now look for the factors of \(-13\).

Factors of \(-13\): \(\pm1, \pm13\)

Let us start with the first factor \(e = 1\).

\begin{array}{l} p (1) = {(1)}^{3} - {(1)}^{1} + 13 (1) - 13 \\ p (1) = (1) - (1) + 13 - 13 \\ p (1) = 1 - (1) + 13 - 13 \\ P (1) = 0 \\ P (1) = 0 + 0 = 0 \end{array}

So at \(e =1\),

p (y) = 0

Thus, the factor \(e =1\) satisfies step 1.

Step 2:

We can conclude that \(e -1\) is a factor of

p (k)

Step 3: Now divide

p (k)

by \(e -1\).

Let the quotient be \(g(e)\).

\begin{array}{l} g (e) = \frac{p (e)}{(e - 1)} \\ e^{2} + 13 \\ e - 1 e^{3} - e^{2} + 13 e - 13 \\ e^{3} - e^{2} \\ \underline{(-) (+)} \\ 0 + 13 e - 13 \\ 13 e - 13 \\ \underline{(-) (+)} \\ 0 \end{array}

So, \(g(e) =\)

e^{2} + 13

\begin{array}{l} p (e) = (e - 1) g (e) \\ = (e - 1) (e^{2} + 13) \end{array}

Thus, the factorisation of the cubic polynomial

e^{3} - e^{2} + 13 e - 13

(e - 1) (e^{2} + 13)

Did you find an error?

2. Factorising the cubic polynomial