Factorising quadratic expressions by inspsection with quadratic coefficient equal to 1

1. Factorise the following expressions:
6 y2
­ 5y =
9 a3
­ 3 a2
=
x2
+ xy + 3 x z =
x2
+ 5 x + 6 =

2. x2
+ 5 x + 6
This expression is called a quadratic expression
because the highest power of any of its terms is 2.

3. Factorising quadratic expressions
By the end of the lesson you will be able to:
• Factorise quadratic expressions of the form
x2
+ bx + c .

4. Expand and simplify the following:
(x+2) ( x+3) =
(x+4) ( x ­5) =
(x ­ 6) ( x+3) =
(x+5) ( x ­ 2) =
(x ­ 5) ( x ­ 3) =
(x+4) ( x+5) =
Do you observe any pattern?

5. Expand and simplify the following:
(x+2) ( x+3) = x2
+ 5 x + 6
(x+4) ( x ­5) = x2
­ x ­ 20
(x ­ 6) ( x+3) = x2
­ 3 x ­ 18
(x+5) ( x ­ 2) = x2
+ 3 x ­ 10
(x ­ 5) ( x ­ 3) = x2
­ 8 x + 15
(x+4) ( x+5) = x2
+ 9 x + 20
Do you observe any pattern?

6. (x+2) ( x+3) = x2
+ 5 x +6
(x+4) ( x+5) = x2
+ 9 x + 20
(x ­ 6) ( x+3) = x2
­ 3 x ­ 18

7. (x+2) ( x+3) = x2
+ 5 x +6
2 x 3
2 + 3
4 x 5
(x+4) ( x+5) = x2
+ 9 x + 20
4+ 5
(x ­ 6) ( x+3) = x2
­ 3 x ­ 18
­6 x 3­6+ 3

8. (x + 4) ( x ­ 5) = x2
­ x ­ 20
(x+5) ( x ­2) = x2
+ 3 x ­ 10
(x ­ 5) ( x ­ 3) = x2
­ 8 x + 15

9. ( x + 3 ) ( x + 2 ) = x2
+ 5 x + 6
Expand
Factorise

10.
Therefore,
(x +2 ) ( x + 3) x2
+ 5 x + 6 =
product : 6
sum: 5
x2
+ 7 x + 12 = ( x ) ( x )
product : 12
sum: 7
x2
+ 4x ­ 12 = ( x ) ( x )
product : ­12
sum: 4
two numbers such that:
two numbers such that:

11. x2
­ 6x + 8 = ( ) ( )
product :
sum:
x2
+ 8 x + 15 = ( ) ( )
product :
sum:
x2
­ 12 x + 20 = ( ) ( )
product :
sum:
x2
­ 9 x ­ 36 = ( ) ( )
product :
sum:

12. Solve worksheet "Factorisation of quadratics "
Extra­practice: Book Ex. 6.10 : 1 to 5

13. http://www.maths­bingo.com/FactoringQuadraticTrinomials.html

14. worksheets to print

15. Expand and simplify the following:
(x+2) ( x+3) =
(x+4) ( x ­5) =
(x ­ 6) ( x+3) =
(x+5) ( x ­ 2) =
(x ­ 5) ( x ­ 3) =
(x+4) ( x+5) =
Do you observe any pattern?

16. (x+2) ( x+3) = x2
+ 5 x +6
(x+4) ( x+5) = x2
+ 9 x + 20
(x ­ 6) ( x+3) = x2
­ 3 x ­ 18

17. (x + 4) ( x ­ 5) = x2
­ x ­ 20
(x+5) ( x ­2) = x2
+ 3 x ­ 10
(x ­ 5) ( x ­ 3) = x2
­ 8 x + 15

18. Therefore,
(x +2 ) ( x + 3) x2
+ 5 x + 6 =
product : 6
sum: 5
x2
+ 7 x + 12 = ( x ) ( x )
product : 12
sum: 7
x2
+ 4x ­ 12 = ( x ) ( x )
product : ­12
sum: 4
two numbers such that:
two numbers such that:

19. x2
­ 6x + 8 = ( ) ( )
product :
sum:
x2
+ 8 x + 15 = ( ) ( )
product :
sum:
x2
­ 12 x + 20 = ( ) ( )
product :
sum:
x2
­ 9 x ­ 36 = ( ) ( x )
product :
sum: