Factorising Quadratics
Factorising Quadratics Revision
Factorising Quadratics
Quadratics are expressions that, in their most general form, look like
ax^{2}+bx+c
where a, b and c are all numbers, called coefficients.
Some quadratics factorise into double brackets, which take the form as follows:
ax^{2}+bx+c=(px+q)(rx+s)
where p, q, r and s are all numbers.
There are different rules on how to factorise quadratics, depending on if a=1 or a\neq1
Note: Positives and Negatives
Assuming a>0, you can easily determine whether each bracket will contain a + or a - by finding which of the three cases below your quadratic falls into.
Case 1: b>0 and c>0
In this case, both brackets will contain a + sign.
Example: x^{2}+10x+24=(x+6)(x+4)
In this example, b=10>0 and c=24>0, so both brackets contain a + sign.
Case 2: b<0 and c>0
In this case, both brackets will contain a - sign.
Example: x^{2}-6x+8=(x-4)(x-2)
In this example, b=-6<0 and c=8>0, so both brackets contain a - sign.
Case 3: c<0
In this case, regardless of the value of b, one bracket will contain a + sign and the other bracket will contain a - sign.
Example: x^{2}-x-6=(x+2)(x-3)
In this example, c=-6<0 so one bracket contains a + sign and the other bracket contains a - sign.
Factorising Quadratics (a=1)
Quadratics where a=1 are quadratics of the form:
x^{2}+bx+c
The brackets will be of the form (x+p)(x+q) where p and q add to give b and multiply to give c. In other words:
p+q=b
pq=c
Example: Factorise x^{2}+4x-12
We need two numbers that add to give 4 and multiply to give -12.
We know that one of the numbers must be positive and one of the numbers must be negative, because c=-12 so c<0.
\textcolor{blue}{-2}+\textcolor{red}{6}=4
\textcolor{blue}{-2}\times\textcolor{red}{6}=12
So the numbers must be \textcolor{blue}{-2} and \textcolor{red}{6}.
Putting it all together, we have:
x^{2}+4x-12=(x-2)(x+6)
Factorising Quadratics (a\neq1)
When a\neq1, there will be coefficients in front of the x terms inside the brackets, i.e. the factorisation will take the form:
ax^{2}+bx+c=(px+q)(rx+s)
From this, we know that a=pr and c=qs, so we can generate many possible pairs for these numbers. However, the only way to find the right set is to try them all and see which gives the correct value for b.
Example: Factorise 3x^{2}+13x+12
Firstly, since all terms are positive, we know all of p, q, r and s are going to be positive.
Next, the only pair of numbers that multiply to give 3 are 3 and 1, so these must be p and r
Next, to get 12, we have lots of options:
1\times12=12
2\times6=12
3\times4=12
One of these pairs will be q and s.
Next, start expanding every possible factorisation combination, until the correct b value is found.
(3x+12)(x+1)=3x^{2}+15x+12 has b=15\neq13
(x+12)(3x+1)=3x^{2}+37x+12 has b=37\neq13
(3x+6)(x+2)=3x^{2}+12x+12 has b=12\neq13
(x+6)(3x+2)=3x^{2}+20x+12 has b=20\neq13
(3x+4)(x+3)=3x^{2}+13x+12 has b=13 is the correct factorisation.
At this point, even though we have one combination left to try, we can stop because we have found the correct set of brackets.
Hence, we can conclude that 3x^{2}+13x+12=(3x+4)(x+3)
Example 1: Factorising Quadratics (a=1)
Factorise x^{2}+3x+2
[2 marks]
Since b=3>0 and c=2>0, both brackets will contain a + sign.
We need to find two numbers that add to give 3 and multiply to give 2.
These numbers are 1 and 2, so 1 and 2 are the numbers in the brackets.
Hence, our factorised quadratic is:
x^{2}+3x+2=(x+1)(x+2)
Example 2: Factorising Quadratics (a\neq1)
Factorise 10x^{2}-3x-1
[3 marks]
Since a\neq1, we know that this factorisation must take the form (px+q)(rx+s).
We also know that pr=10 and qs=-1.
There are multiple pairs for p abnd r:
10\times1=10
5\times2=10
For q and s, there is only one pair: 1 and -1.
Try every combination:
(10x+1)(x-1)=10x^{2}-9x-1
(10x-1)(x+1)=10x^{2}+9x-1
(5x+1)(2x-1)=10x^{2}-3x-1
We can stop here because we have found the correct value for b.
Hence, we can conclude that:
10x^{2}-3x-1=(5x+1)(2x-1)
Factorising Quadratics Example Questions
Question 1: Factorise y^{2}-2y+1
[2 marks]
We need two numbers that add to give -2 and multiply to give 1.
Since b=-2<0 and c=1>0, we know that both numbers will be negative.
The numbers are -1 and -1.
Hence,
y^{2}-2y+1=(y-1)(y-1)=(y-1)^{2}Question 2: Factorise x^{2}+4x-21
[2 marks]
We need two numbers that add to give 4 and multiply to give -21.
Since c=-21<0, we know that one number will be positive and the other will be negative.
The possible pairs to multiply to -21 are:
1 and -21
-1 and 21
3 and -7
-3 and 7
The correct pair is -3 and 7, because these add to give 4
Hence,
x^{2}+4x-21=(x-3)(x+7)Question 3: Factorise x^{2}+52x+147
[2 marks]
We need two numbers that add to give 52 and multiply to give 147.
Since b=52>0 and c=147>0, we know that both numbers will be positive.
The possible pairs to multiply to 147 are:
1 and 147
3 and 49
7 and 21
The correct pair is 3 and 49, because these add to give 52
Hence,
x^{2}+52x+147=(x+3)(x+49)Question 4: Factorise 2x^{2}+13x+11
[3 marks]
We know that this will take the form (px+q)(rx+s).
We also know that, since b=13>0 and c=11>0, all of p, q, r and s must be positive.
pr=2 – the only pair that makes 2 is 1 and 2.
qs=11 – the only pair that makes 11 is 1 and 11.
This gives only two possibilites: (2x+1)(x+11) and (2x+11)(x+1)
Multiply both out to find correct answer:
(2x+1)(x+11)=2x^{2}+23x+11 has b=23\neq13
(2x+11)(x+1)=2x^{2}+13x+11 has b=13 so must be the correct answer.
Hence,
2x^{2}+13x+11=(2x+11)(x+1)Question 5: Factorise 12x^{2}-29x-221
[3 marks]
We know this will take the form (px+q)(rx+s)
Since c=-221<0, we know that p>0, r>0 and one of q and s is positive while the other is negative.
pr=12, the possible pairs for which are:
1 and 12
2 and 6
3 and 4
qs=-221, the possible pairs for which are:
1 and -221
-1 and 221
13 and -17
-13 and 17
Now start trying combinations:
(x+1)(12x-221)=12x^{2}-209x-221
(x-1)(12x+221)=12x^{2}+209x-221
(x+13)(12x-17)=12x^{2}+139x-221
(x-13)(12x+17)=12x^{2}-139x-221
(2x+1)(6x-221)=12x^{2}-436x-221
(2x-1)(6x+221)=12x^{2}+436x-221
(2x+13)(6x-17)=12x^{2}+42x-221
(2x-13)(6x+17)=12x^{2}-42x-221
(4x+1)(3x-221)=12x^{2}-881x-221
(4x-1)(3x+221)=12x^{2}+881x-221
(4x+13)(3x-17)=12x^{2}-29x-221
Although there are several other combinations to try, we can stop here because we have found the correct b value, so we know this must be the correct factorisation.
Hence,
12x^{2}-29x-221=(4x+13)(3x-17)You May Also Like...
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