1. Factorising Harder Quadratics
Slideshow 16
Mathematics
Mr Richard Sasaki

2. Objectives
Factorise quadratics in the form 𝑎 𝑥 2 + 𝑏𝑥+𝑐 where 𝑏 and 𝑐 are divisible by 𝑎.
Factorise other quadratics in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐

3. Easy Quadratics
We can use methods we already know to solve some quadratics in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐.
Example
Solve 2 𝑥 2 −2𝑥−12.
2 𝑥 2 −2𝑥−12=
2 (𝑥 2 −𝑥−6)
=2(𝑥−3)(𝑥+2)
If we can divide all terms by the 𝑥 2 coefficient, we can remove it and factorise the quadratic in the form 𝑥 2 +𝑎𝑥+𝑏.

4. 2(𝑥+1)(𝑥+2)
2(𝑥+3)(𝑥+5)
3(𝑥+2)(𝑥+1)
2(𝑥−1)(𝑥+3)
2(𝑥−3)(𝑥+4)
4(𝑥+1)(𝑥+2)
3(𝑥+5)(𝑥−7)
4(𝑥+7)(𝑥−2)
3(𝑥−5)(𝑥−3)
5(𝑥+3)(𝑥−2)
7(𝑥−3)(𝑥−1)
2(𝑥+5)(𝑥−7)
3(𝑥−11)(𝑥+2)
4(𝑥+12)(𝑥−4)
2(𝑥−7)(𝑥+16)
6(𝑥−13)(𝑥+31)

5. Typical Quadratics
As you should know, we were lucky with the questions on the last worksheet. Usually it’s more complicated.
Let’s think back to expressions in the form 𝑥 2 + 𝑎𝑥+𝑏≡(𝑥+𝑐)(𝑥+𝑑).
What are our two simultaneous equations for the numbers 𝑎, 𝑏, 𝑐 and 𝑑?
𝑎=𝑐+𝑑
𝑏=𝑐∙𝑑
We also use this idea for quadratics in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐.

6. Typical Quadratics
For a quadratic in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐, we need to consider two numbers.
Example
𝑎=−3, 𝑏=4
(Or vice-versa.)
Factorise 2 𝑥 2 +𝑥−6. ×
𝑎∙𝑏=
−12
𝑎+𝑏= 1
We can write the coefficient of 𝑥 (1) as the sum of −3+4.
2 𝑥 2 +𝑥−6=
2 𝑥 2 −3𝑥+4𝑥−6
Finally, we group the first two and last two terms.
(2 𝑥 2 −3𝑥)+(4𝑥−6) =
𝑥(2𝑥−3) +
2(2𝑥−3) =
(𝑥+2)(2𝑥−3)

7. Typical Quadratics Let’s try one more example. Example
Factorise 6 𝑥 2 +8𝑥−8.
𝑎=−4, 𝑏=12 (Or vice-versa.) ×
𝑎∙𝑏=
−48
𝑎+𝑏= 8
6 𝑥 2 +8𝑥−8=
6 𝑥 2 −4𝑥+12𝑥−8
=(6 𝑥 2 −4𝑥)+(12𝑥−8)
=2𝑥(3𝑥−2) +
4(3𝑥−2)
=(2𝑥+4)(3𝑥−2)
=2(𝑥+2)(3𝑥−2)

8. (2𝑥+1)(𝑥+3)
(3𝑥+2)(𝑥−1)
(4𝑥−1)(𝑥+2)
(2𝑥−3)(𝑥+1)
(2𝑥+1)(2𝑥+3)
(4𝑥+3)(2𝑥+3)
(2𝑥−1)(3𝑥+2)
(4𝑥+5)(2𝑥+7)
(𝑥+2)(2𝑥−7)
2(𝑥−2)(2𝑥+1)
(5𝑥−2)(2𝑥+5)
(2𝑥−1)(2𝑥+1)

9. 2(𝑥+2)(2𝑥−1)
2(3𝑥+2)(𝑥−4)
6(4𝑥−1)(𝑥+1)
3(2𝑥+7)(2𝑥−7)
5𝑥+4 2
8𝑥+1 2
3(2𝑥−5)(𝑥−9)
2(2𝑥−9)(3𝑥+5)
(3𝑥−2)(7𝑥+1)
(5𝑥+8)(3𝑥−11)
−2(3𝑥+2)(2𝑥+5)
2𝑥(3𝑥+4)(2𝑥+3)

10. 16 49
144 3 7 9
196
289
256 8 11 1 1
400
529 12 14 18
729
576
1296 21 30 27
2304
2809
3844 43 48 52

11. Evil Quadratics
Some of those quadratics were tough to factorise because some big numbers have a lot of factors…or factors that are hard to find.
For a difficult quadratic, it may be more sensible to pretend its equal to 0, solve it and then factorise the expression.
If a quadratic equation has solutions 𝑥=1 or −2, how does it factorise?
(𝑥−1)(𝑥+2), we learned this before! This is correct…right?

12. Evil Quadratics
Do you remember the formula for solving a quadratic equation?
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
For some 𝑎 𝑥 2 +𝑏𝑥+𝑐=0…
Let’s try factorising using this with an easy example.
Example
Factorise 𝑥 2 +3𝑥+2.
First, let 𝑥 2 +3𝑥+2=0. Now…𝑎=1, 𝑏=3, 𝑐=2.
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 =
−3± −4∙1∙2 2∙1 =
−3± 1 2
We get 𝑥=−1 𝑜𝑟 −2. So how does our expression factorise?
(𝑥+1)(𝑥+2)

13. Evil Quadratics Let’s try 15 𝑥 2 −31𝑥−88 again. Example
Factorise 15 𝑥 2 −31𝑥−88.
Let 15 𝑥 2 −31𝑥−88=0. (𝑎=15, 𝑏=−31, 𝑐=−88.)
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 =
31± (−31) 2 −4∙15∙(−88) 2∙15
= 31±
= 31±
= 31±79 30
𝑥− 𝑥+ 8 5
We get 𝑥= or − So we have…
This is 15 𝑥 2 −31𝑥− factorised.
Is this correct?
No!
We need to multiply it by 15.

14. Evil Quadratics
In fact, when we solve and get two solutions 𝑥 1 and 𝑥 2 …we substitute these into…
𝑎(𝑥− 𝑥 1 )(𝑥− 𝑥 2 )
So, as 𝑥 1 = 11 3 , 𝑥 2 =− 8 5 , and 𝑎=15, we get…
15 𝑥− 𝑥+ 8 5
=3 𝑥− ∙5 𝑥+ 8 5
= 3𝑥−11 5𝑥+8
And that’s it! It is a long process but it’s easier with big numbers, especially when 𝑎∙𝑐 is a big product! Let’s try some questions using this method.

15. 3𝑥−1 𝑚𝑚 (𝑎+𝑏)(𝑎−𝑏) 12 𝑎𝑛𝑑 13 2𝑚 1 2 and − 2 3 (any order)
𝑎=− 1 4 , 𝑏= 1 3
3𝑥−1 𝑚𝑚
(𝑎+𝑏)(𝑎−𝑏)
12 𝑎𝑛𝑑 13 2𝑚