Study S2 Mathematics Maths Factorising Quadratic Expressions - Geniebook

Factorising Quadratic Expressions

In this chapter, we will be discussing the below-mentioned topics in detail:

  • Factorisation as the reverse process of Expansion
  • Factorisation using a Multiplication Frame
  • Factorisation using a Factorisation Table
  • Factorisation involving extracting common factor and quadratic expressions

 

Factorisation Of Algebraic Expressions

Types Of Factorisation \(\begin{align} \xrightarrow [] {\quad \text{ Expansion }\quad } \end{align}\)
Extract Common Factor \(\begin{align*} 2x(x-1) &= \end{align*}\) \(\begin{align*} 2x^2-2x \end{align*}\)
Quadratic Factorisation \(\begin{align} (x-2)(x-1) &= \end{align}\) \(\begin{align} x^2-3x+2 \end{align}\)
  \(\xleftarrow{\quad \text{ Factorisation }\quad}\)

Factorisation is the opposite of expansion. 

Let’s understand this with the help of some examples:

 

Question 1: 

Factorise \(3x^2-27x\) completely.

 

Solution: 

The common factor is \(3x\).

\(3x^2-27=3x(x-9)\)

 

 

Question 2: 

Factorise \(x^2+5x+6\) completely.

 

Solution: 

Method (a): Factorisation using Multiplication Frame

\(x^2+5x+6\)

As we know, \(x\) times \(x\) will give us \(x^2\), and plot both the \(x\) in Row \(I\) Column \(II\) and Row \(II\) Column \(I\), respectively. 

Also, either 3 times 2 is 6 or 1 times 6 is 6, so using trial and error, put 3 in Row \(III\) Column \(I\) and put 2 in Row \(I\) Column \(III\).

Multiply Row \(I\) Column \(III\), i.e. 2, by Row \(II\) Column \(I\), i.e. \(x\) and put \(2x\) in Row \(II\) Column \(III\).

Multiply Row \(III\) Column \(I\), i.e. 3, by Row \(I\) Column \(II\), i.e. \(x\) and plot \(3x\) in Row \(III\) Column \(II\).

Now, \(+2x\) and \(+3x\) give us \(+5x\)

 

However, if we use 1 times 6 instead, see below.

Now, \(+6x\) and \(+x\) do not give us \(+5x\), and it gives us \(+7x\).

Hence, the answer would be \((x+2)(x+3)\)

 

Method (b): Factorisation using Factorisation Table

\(x^2+5x+6\)

\(x^2+5x+6=(x+3)(x+2)\)

 

 

Question 3: 

Factorise \(x^2+6x+5\) completely.

 

Solution: 

Using a factorisation table,

\(\therefore x^2+6x+5=(x+1)(x+5)\)

 

 

Question 4:

Factorise \(3x^2-18x+15\) completely.

 

Solution:

Firstly, we extract out a common factor 3.

\(3x^2-18x+15=3(x^2-6x+5)\)

\(\begin{align} x^2 +5 -6x \end{align}\)

\(\begin{align*} \therefore \quad 3x^2-18x+15&=3(x^2-6x+5)\\ &=3(x-1)(x-5) \end{align*}\)

 

 

Continue Learning
Algebraic Fractions Direct & Inverse Proportion
Congruence And Similarity Factorising Quadratic Expressions
Further Expansion And Factorisation Quadratic Equations And Graphs
Simultaneous Equation
Resources - Academic Topics
Primary
Primary 1
Primary 2
Primary 3
Primary 4
Primary 5
Primary 6
Secondary
Secondary 1
Secondary 2
English
+ More
Maths
Algebraic Fractions
Direct & Inverse Proportion
Congruence And Similarity
Factorising Quadratic Expressions
Further Expansion And Factorisation
Quadratic Equations And Graphs
Simultaneous Equation
+ More
Science
+ More
Secondary 3
Secondary 4
+ More
Sign up for a free demo
(P1 to S4 levels)
Our Education Consultants will get in touch to offer a complimentary product demo and Strength Analysis to your child.
Ready to power up your
child's academic success?
Let our Education Consultants show you how.
*By submitting your phone number, we have your permission to contact
you regarding Geniebook. See our Privacy Policy.