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Mr Barton’s Maths Notes
Algebra
6. Factorising Quadratics
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6. More Factorising – Quadratics
Again, you knew it was coming…
Just like we had to expand double brackets, it should come as no surprise that we have to
factorise expressions back into double brackets as well!
There is a bit of a trick to this, and to discover it, let’s look back at our answers from
5. Expanding Brackets…
(a  6) (a  4)
a2  10a  24
( p  10) ( p  8)
p  2 p  80
(t  9) (t  2)
t 2  7t  18
(m  7) (m  9)
2
Note: These answers
are called Quadratic
Expressions because
they have a squared
term in them
m2  16m  63
Focus your attention on the numbers…
if you can see how to get from the numbers in the questions to the numbers in the answers…
then you should be able to see how to get from the answer back to the question…
and if you can do that, then you can already factorise quadratic expressions!
How to Factorise Quadratic Expressions
Factorising quadratics means you want to get from:
x
2
 ?x  ?
to
( x  ? ) ( x  ? )
To be able to do this you need to be able to solve a little puzzle
If you look back at the examples, you will see that…
( x  ? ) ( x  ? )
+
x2  ? x  ?
x
In other words, the two numbers in the
bracket (including their sign) must…
ADD TOGETHER to give you the
number (and sign) in front of the x…
And…
MULTIPLY TOGETHER to give you the
number (and sign) at the end
( x  ? ) ( x  ? )
So… if you can discover what two numbers solve that little puzzle, then you can factorise
quadratics… and practice makes perfect!
Example 1
Example 2
p2  2 p  15
x 2  11x  24
Okay, here the question we must ask ourselves…
Okay, here the question we must ask ourselves…
p2  2 p  15
x 2  11 x  24
Which two numbers multiply together to give
24 and add together to give 11?
Which two numbers multiply together to give
-15 and add together to give 2?
Now, if you find it helps, you can write down all
the pairs of numbers which multiply together to
give 24, and see which one also adds up to 11…
Again, nothing wrong with writing down pairs
that multiply together to give -15, but be
careful of your negatives!
1 x 24
1 + 24 = 25
x
1 x -15
1 + -15 = -14 x
2 x 12
2 + 12 = 14
x
-1 x 15
-1 + 15 = 14
3 x 8
3 + 8
3 x -5
3 + -5
-3 x 5
-3 + 5 = 2
= 11
√
Once we have our pair, we just write them in
the brackets, remembering that no sign is just
a disguised plus!
( x  3) (x  8)
or
( x  8) (x  3)
Why not expand the brackets to make doubly
sure you are correct!
= -2
x
x
√
Once we have our pair, we just write the
numbers in the brackets, making sure we get
our signs in the correct place!
( p  3) (p  5)
or
( p  5) (p  3)
Again, expanding the brackets is a good check!
Example 3
Example 4
k 2  13k  14
v2  9v  18
Okay, here the question we must ask ourselves…
k 2  13 k  14
Okay, here the question we must ask ourselves…
v 2  9 v  18
Which two numbers multiply together to give
-14 and add together to give -13?
Which two numbers multiply together to give
18 and add together to give -9?
Unless you can do it in your head, just write
down pairs of numbers that multiply together
to give -14:
Now, switch on your brain here… we CAN’T be
talking two positive numbers, as how will they
add up to give -9?... so we need two negatives!
-1 x 14
-1 + 14 = 13
x
-1 x -18
-1 + -18 = -19 x
1 x -14
1 + -14 = -13
√
-2 x -9
-2 + -9 = -11 x
-3 x -6
-3 + -6
Once we have our pair, we just write the
numbers in the brackets, making sure we get
our signs in the correct place!
(k  1) (k  14)
or (k
√
People tend to mess these up, but we won’t,
because we know that two negatives multiplied
together gives a positive!
 14) (k  1) (v  3) (v  6)
If you expand the brackets you will definitely
know that you are correct!
= -9
or
(v  6) (v  3)
Again, expanding the brackets is a good check!
What abut this funny looking one?…
x 2  16
Okay, looks a bit strange, but let’s ask ourselves the same question as we always do…
Which two numbers multiply together to give -16 and add together to give…
erm…well… erm… 0?
Remember, it is the number in front of the x which tells us what the numbers must add
together to make, but we don’t have any x’s, so the sum of our two numbers must be… 0!
Think of the expression like this is it helps…
x 2  0 x  16
So, isn’t it true that for two numbers to add together to give zero, they must be the same
number, but of opposite sign, so they cancel each other out!
So, which two numbers do we need?... 4 and -4! Expand it to check!
x 2  16
( x  4) ( x  4)
Expressions like this are called “the difference of two squares”, and are always factorised in a
similar way. Look at these three examples and see if you can see how I got the answers…
a 2  25
(a  5) (a  5)
p 2  100
( p  10) ( p  10)
4t 2  49
(2t  7) (2t  7)
When things get a little tricky…
Okay, whilst it was not so tricky to spot how to factorise those types of quadratics,
what about when there is a number in front of the squared term?...
Let’s look back at two examples we did in 5. Expanding Double Brackets, to see if we
can spot where the numbers come from…
(5g  9) ( g  3)
5g 2  6 g  27
(3c  4) (2c  5)
6c 2  23c  20
Any ideas?... It’s not easy to spot, is it?
I think the best thing I can do is to try and take you through two examples as
carefully as I can…
Are you ready?....
2 x2  x  3
Example 1
Okay, let’s start by thinking what the first terms in the two brackets must be?...
Do you agree that they would have to be 2x and x, otherwise we would not get out 2x2!
How about the numbers at the end of each bracket?...
Well, they would have to be a pair of numbers which multiply together to give -3!
I set out this information in a table like this:
The first
terms in the
brackets
2x
-1
3
1
-3
x
3
-1
-3
1
All the pairs of
numbers which
multiply to give -3
Next I multiply DIAGONALLY, and I am looking for a pair of numbers which will add up to
the amount of x’s I need… which from the question is -1x!
2x
-1
3
-3
1
x
3
-1
1
-3
(2x x 3) + (x x -1) = 6x + -1x = 5x x
(2x x -1) + (x x 3) = -2x + 3x = 1x x
(2x x 1) + (x x -3) = 2x + -3x = -1x √
The pair I want is the 3rd column of numbers along
Now, to get my answer, I just put the two terms from the top row in the first bracket, and
the two terms from the second row in my second bracket…
(2 x  3) (x  1)
You have done so much work here, that it
is definitely worth checking you are
correct by expanding the brackets!
Example 2
8 x 2  2x  15
Okay, again let’s start by thinking what the first terms in the two brackets must be?...
Problem: They could be either: 8x and x, or the could be 4x and 2x. Both, when multiplied
together, would give us the 8x2 that we need!
So this time we need two tables!
All the pairs of
numbers which
multiply to give -15
The first
terms in the
brackets
8x
-1
15
1
-15
-3
5
3
-5
x
15
-1
-15
1
5
-3
-5
3
4x
-1
15
1
-15
-3
5
3
-5
2x
15
-1
-15
1
5
-3
-5
3
And once again I must just keep I multiplying DIAGONALLY (in my head if I can), looking for a
pair of numbers which will add up to the amount of x’s I need… which from the question is -2x!
After a long search, I reckon the pair with the rings around them is what I need!
(4x x -3) + (2x x 5) = -12x + 10x = -2x
As before, to get my answer I just put the two terms from the top row in the first bracket,
and the two terms from the second row in my second bracket…
(4 x  5) (2x  3)
Good luck with
your revision!