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Factorising with a Common Factor
Factorising with a Common Factor
Exercise 8.06
(Page 102-103)
2nd Edition
Pages 18-19
Factorising Quadratics (Double Brackets)
Factorising Quadratics (Double Brackets)
Notes:
Notes:
The starting point whenever we are asked to factorise is to look for a common factor.
Following this if we are left with a quadratic (an equation in the form (ax2 + bx + c) then we will need to factorise it into two brackets (x-g)(x-h).
There is a little method on how this is done.
The two numbers that go in the brackets
Multiply together to the final term (the number)
Add together to the middle term (the 'x' term)
Examples
Examples
1)x2 + 3x + 2
No common factor. Therefore think of factors of 2 that add to 3.
2 x 1 = 2
2+1=3
=(x+1)(x+2)
2) x2 + x – 72
No common factor. Therefore think of factors of -72 that add to 1.
9x-8 = -72
9+-8 = 1
=(x+9)(x-8)
3) 4x2 + 6x
Common factor 0f 2x
=2x(2x + 3) No further factorising
4) x2 – 25
Think about this being x2 +0x– 25
Factors of -25 that add to 0
=(x+5)(x-5)
5) 2x2 + 6x + 4
Step 1 – Common factor is 2
2(x2 + 3x + 2)
Step 2 – Factorise further eg x2 + 3x + 2
2(x+2)(x+1)
Exercise 8.07-8.09
(Pages 104-106)
2nd Edition
Pages 41-43
Factorising Quadratics (Double Brackets) with coefficients and no common factor
Factorising Quadratics (Double Brackets) with coefficients and no common factor
Example
Example
Factorise
2x2 + 7x + 3
STEP 1 multiply these numbers (ac x bd)
2 x 3 = 6
STEP 2 Find the factors of 6 that add to give the coefficient of the middle term (ad + bc)
1 and 6, 2 and 3
1 + 6 = 7
STEP 3 Rewrite the equation breaking up the middle term
2x2 + 1x + 6x + 3
STEP 4 Factorise in little parts the goal is to have 2 similar brackets!!!!!
x(2x + 1) + 3(2x +1)
STEP 5 Fully factorise
(x+3)(2x+1) Look above x and 3 are multiplied by (2x+1)
Exercise 8.10
(Page 107)
2nd Edition
Pages 44-45
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