Factorisation of Polynomials

ICSE Class 10 Mathematics • Chapter 06

1. Remainder Theorem

Statement: When a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$.

Remainder = p(a)

Example 1: Find remainder when $p(x) = x^3 - 2x^2 + 3x - 1$ is divided by $(x - 2)$.

Remainder = $p(2) = 8 - 8 + 6 - 1 = 5$

2. Factor Theorem

Statement: $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$

In other words, if $p(a) = 0$, then $a$ is a zero/root of $p(x)$.

To check if (x - a) is a factor: Simply calculate p(a). If p(a) = 0, it's a factor!

Example 2: Show that $(x - 1)$ is a factor of $x^3 - 6x^2 + 11x - 6$.

$p(1) = 1 - 6 + 11 - 6 = 0$ → Yes, $(x-1)$ is a factor ✓

3. Factorising Cubic Polynomials

Steps to Factorise $ax^3 + bx^2 + cx + d$:
  1. Find one root: Try $\pm 1, \pm 2, \pm 3, ...$ (factors of d)
  2. If $p(a) = 0$, then $(x - a)$ is a factor
  3. Divide: Use long division or synthetic division to get the quadratic quotient
  4. Factorise the quadratic by splitting middle term

Example 3: Factorise $x^3 - 6x^2 + 11x - 6$

Step 1: Try x = 1: $p(1) = 1 - 6 + 11 - 6 = 0$ ✓

So $(x - 1)$ is a factor

Step 2: Divide by $(x - 1)$:

$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$

Step 3: Factorise quadratic:

$x^2 - 5x + 6 = (x - 2)(x - 3)$

Answer: $(x - 1)(x - 2)(x - 3)$

4. Finding Unknown Constants

Given: $(x - a)$ is a factor of polynomial $p(x)$

Use: $p(a) = 0$ to find unknown constants

Example 4: If $(x + 1)$ is a factor of $x^3 + ax^2 - x + 2$, find $a$.

Since $(x + 1)$ is a factor, $p(-1) = 0$

$(-1)^3 + a(-1)^2 - (-1) + 2 = 0$

$-1 + a + 1 + 2 = 0$

$a + 2 = 0$

$a = -2$

5. Special Algebraic Identities

Identity Factorised Form
$a^3 + b^3$ $(a + b)(a^2 - ab + b^2)$
$a^3 - b^3$ $(a - b)(a^2 + ab + b^2)$
$a^3 + b^3 + c^3 - 3abc$ $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Special Case: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$

Example 5: Factorise $8x^3 + 27$

$= (2x)^3 + (3)^3$

$= (2x + 3)((2x)^2 - (2x)(3) + 3^2)$

$= (2x + 3)(4x^2 - 6x + 9)$

6. Quick Reference

Theorem Application
Remainder Theorem $p(x) ÷ (x-a)$ → Remainder = $p(a)$
Factor Theorem If $p(a) = 0$, then $(x-a)$ is a factor
Sum of Cubes $a^3 + b^3 = (a+b)(a^2-ab+b^2)$
Diff of Cubes $a^3 - b^3 = (a-b)(a^2+ab+b^2)$

Exam Practice (PYQ Trends)

PYQ: 2023

BOARD Using Factor Theorem, factorise: $x^3 + x^2 - 4x - 4$

PYQ: 2022

BOARD Find values of a and b if $(x - 1)$ and $(x - 2)$ are factors of $x^3 + ax^2 + bx + 6$.