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Factorising Quadratics
Factorising Quadratics Revision
Factorising Quadratics
Quadratics are algebraic expressions that include the term, x^2, in the general form,
ax^2 + bx + c
Where a, b, and c are all numbers. We’ve seen already seen factorising into single brackets, but this time we will be factorising quadratics into double brackets.
(nx+m)(px+q)
There are 2 main types of quadratics you will need to be able to factorise; one where a=1 and the other where a\neq1.
Make sure you are happy with the following topics before continuing.
Take Note: The Factorising Trick
There is a quick trick to determine whether you should add a + or - sign to your brackets. There are three sub-types which we will go over here.
Sub-type (a): contains all positives
These quadratics contain all positive terms, e.g. x^2 +5x + 6. When factorised, both brackets will contain \bf{\bf{\large{+}}}.
x^2 + 5x + 6 = (x+3)(x+2)
Sub-type (b): b is negative and c is positive
These quadratics contain a negative b value and a positive c value. When factorised, both brackets will contain \bf{\large{-}}.
x^2 -10x +21 = (x-7)(x-3)
Sub-type (c): c is negative.
If c is negative, when factorised, one bracket will contain a \bf{+} the other will contains a \bf{\large{-}}. The order or these will need to be determined. These are the hardest type and require the most thought.
x^2 + 3x -18 = (x+6)(x-3)
Type 1: Factorising quadratics (a=1)
When we say a=1, we mean the number before x^2 in x^2+bx+c will be 1 (typically we don’t write the 1). Any number that appears before an x term is called a coefficient, so in this case, a is the coefficient of x^2 which has a value of 1.
Example: Factorise the following quadratic into two brackets, x^2\textcolor{blue}{-3}x+\textcolor{red}{2}
Step 1: First we can write two brackets with an x placed in each bracket.
(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)
Step 2: We can identify that this is a sub-type (b) quadratic, meaning both brackets will contain \large{-}
(x\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,)
Step 3: We have to find two numbers which multiply to make \textcolor{red}{2} and when added together make \textcolor{blue}{-3}.
We know both numbers will be negative.
-2 \times -1 = \textcolor{red}{2}
-2 + -1 = \textcolor{blue}{-3}
Finally add these numbers to the brackets.
(x-2) (x-1)
Type 2: Factorising quadratics (a> 1)
In this instance the general form of the equation is ax^2+bx+c where a>1.
Example: Factorise the following quadratic 4x^2+\textcolor{blue}{3x}\textcolor{red}{-1}
Step 1: When a>1 it makes things more complicated. It is not immediately obvious what the coefficient of each x term should be. There are two possible options,
(4x \kern{1 cm} ) (x \kern{1 cm} ) or (2x \kern{1 cm} ) (2x \kern{1 cm} )
Step 2: We can identify that this quadratic is part of sub-type (c) meaning it can contain + and -
This is most important for quadratic pairs which are non-symmetrically creating a third option, all three are shown below.
\begin{aligned}(4x \kern{0.4 cm} +\kern{0.4 cm} )&(x \kern{0.4 cm}-\kern{0.4 cm} )\\ (4x \kern{0.4 cm} -\kern{0.4 cm} )&(x \kern{0.4 cm}+\kern{0.4 cm} )\\(2x \kern{0.4 cm}+\kern{0.4 cm} )&(2x \kern{0.4 cm}-\kern{0.4 cm} )\end{aligned}
Step 3: We need to find two numbers which when multiplied make \textcolor{red}{-1}
\textcolor{red}{-1} has only one factor.
-1 \times 1 = -1
Step 4: We need to find a combination which gives \textcolor{blue}{3x}
We can test our 3 possibilities,
\begin{aligned}(4x+1)(x -1) &= 4x^2 -3x -1 \\(4x-1)(x+1) &= 4x^2 + 3x -1 \\(2x+1)(2x-1) &= 4x^2 -1\end{aligned}
As we can see (4x-1) (x+1) gives the correct expansion and is therefore the answer.
Example 1: Factorising Simple Quadratics
Factorise x^2 - x - 12.
[2 marks]
Step 1: Draw empty brackets
(x \kern{1 cm} ) (x \kern{1 cm} )
Step 2: Identify sub-type (b)
(x \kern{0.4cm} + \kern{0.4cm} ) (x \kern{0.4 cm} - \kern{0.4cm} )
Step 3: We are looking for two numbers which multiply to make \textcolor{red}{-12} and add to make \textcolor{blue}{-1}. Let’s consider some factor pairs of -12.
\begin{aligned}(-1)\times12&=-12 \,\,\text{ and } -1 + 12 = 11\\(-2)\times6&=-12 \,\,\text{ and } -2 + 6 = 4\\ (-6)\times2&=-12 \,\,\text{ and } -6 + 2 = -4\\ (-3)\times4&=-12 \,\,\text{ and } -3 +4 = 1\\ \textcolor{red}{(-4)\times3}&\textcolor{red}{=-12 }\,\,\text{ and } \textcolor{blue}{-4 + 3 = -1}\end{aligned}
We could keep going, but there’s no need because the last pair, -4 and 3, add to make -1. This pair fills both criteria, (as highlighted above) so the factorisation of x^2 - x - 12 is
(x - 4)(x + 3)
Note: You can try expanding the double brackets to check your answer is correct. You should always get your original quadratic equation if you do this correctly.
Example 2: Factorising Harder Quadratics
Factorise 2x^2 + 7x + 3.
[3 marks]
Step 1: Draw the empty brackets. Even though a>1 there is only one possible option this time.
(2x \kern{1 cm} ) (x \kern{1 cm} )
Step 2: Identify sub-type (a), meaning both brackets contain +.
(2x \kern{0.4cm} + \kern{0.4cm} ) (x \kern{0.4 cm} + \kern{0.4cm} )
Step 3: Find two numbers which multiply to give \textcolor{red}{3}
3 only has one factor.
3 \times 1 = 3
Step 4: Find the combination which gives 7x
\begin{aligned}\textcolor{blue}{(2x \,\, +\,\, 1) (x\,\, + \,\,3) }&\textcolor{blue}{= 2x^2 + 7x +3} \\ (2x \,\, +\,\, 3) (x\,\, + \,\,1) &= 2x^2 +5x + 3\end{aligned}
As we can see, (2x + 1)(x + 3), gives the correct expansion and is therefore the correct answer.
Factorising Quadratics Example Questions
Question 1: Factorise a^2 + a - 30
[2 marks]
We are looking for two numbers which add to make 1 and multiply to make -30.
The factors of 30 that satisfy theses two requirements are 5 and 6.
Therefore, the full factorisation of a^2 + a - 30 is
(a - 5)(a + 6)
We are looking for two numbers which add to make 1 and multiply to make -30.
The factors of 30 that satisfy theses two requirements are 5 and 6.
Therefore, the full factorisation of a^2 + a - 30 is
(a - 5)(a + 6)
Question 2: Factorise k^2 - 5k + 6
[2 marks]
We are looking for two numbers which add to make -5 and multiply to make 6.
The factors of 6 that satisfy theses two requirements are -2 and -3.
Therefore, the full factorisation of k^2 - 5k + 6 is
(k - 2)(k - 3)
Question 3: Factorise x^2 + 7x + 12
[2 marks]
We are looking for two numbers which add to make 7 and multiply to make 12.
The factors of 12 that satisfy theses two requirements are 3 and 4.
Therefore, the full factorisation of x^2 + 7x + 12 is,
(x + 3)(x + 4)
Question 4: Factorise 3x^2 + 11x + 6
[4 marks]
In the quadratic, a = 3, 11 is positive and c is positive. We can set up the brackets as follows: (3x \,\,\, + \,\,\,)(x \,\,\, + \,\,\,).
We are looking for two positive numbers which multiply to make 6. The possible factors of 6 are
\begin{aligned}(6)\times(1)&=6 \\ (3)\times(2)&=6 \end{aligned}We now test all the combinations:
\begin{aligned}(3x+6)(x+1)&=3x^2+6x+3x+6 \\ (3x+1)(x+6)&= 3x^2+x+18x+6 \\ (3x+3)(x+2)&= 3x^2+6x+3x+6 \\ (3x+2)(x+3) &= 3x^2+9x+2x+6\end{aligned}
Hence the correct factorisation is (3x+2)(x+3)
Question 5: Factorise 4m^2 - 5m - 6
[3 marks]
We can see this is a sub-type (c) meaning it will contain both + and -
Factors of -6
(-1) \times 6 = -6
(-6) \times 1 = -6
(-2) \times 3 = -6
(-3) \times 2 = -6
Lets find the options which give - 5m
(2m+2)(2m-3) = 4m^2 - 2m - 6
(4m + 1)(m-6) = 4m^2 -23m-6
(4m-1)(m+6) = 4m^2 +23m -6
…
…
(4m+3)(m-2) = 4m^2 -5m -6
We can see that last option with +3 and -2 is the correct combination.
This gives the final answer to be:
(4m+3)(m-2)
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