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 | |
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Algebraic Fractions | Direct & Inverse Proportion |
Congruence And Similarity | Factorising Quadratic Expressions |
Further Expansion And Factorisation | Quadratic Equations And Graphs |
Simultaneous Equation |