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Curriculum on factorization
Factorization a2-b2, trinomial with
perfect square, trinomial in the form
of ax2  bx c.
PRESENTED BY
MANINATH NEUPANE
CLASS 8

Q 1.Factorise: 2ax+4ay
ax ay
a
x
a
y
a
x
a
x
a
y
a
y
a
y
a
y
2a
(x+2y)

Q 1.Factorise: 2ax+4ay
=2 .
=2.2
2ax + 4aySolution:
2ax
4ay
=
a . x
. a . y
( + )
Rule 1. We have to take common
factors as common and remaining
factors should be kept inside
parenthesis.

Q 2.Factorise: 4a2b-6ab2+10ab
(
4a2b-6ab2+10ab
4a2b
6ab2
10ab
=2. 2. a. a. b
2. 3. a. b. b=
2. 5. a. b=
- + )
Solution:
=
Rule 1. We have to take common factors as common
and remaining factors should be kept inside
parenthesis.

Q 3. Resolve into factors: 2x (a + b)- 3y (a + b)
Solution:
2x (a + b)- 3y (a + b) 2x (a + b)=
3y (a + b)=
2 . x . (a+b)
3 . y . (a+b)
= ( - )
Rule 1. We have to take common factors as common
and remaining factors should be kept inside
parenthesis.

Q 4.Factorise: a2-ax+ab-bx
Solution: a2-ax + ab-bx
= (
a2= a . a
ax= a . x- ) + b(a - x)
=(a-x)(a+b)
Rule 2. We have to arrange the terms in groups such
that each group has common factor.

Q 5.Factorise: a(x2-y2)+x(y2-a2)
Solution: a(x2-y2)+x(y2-a2)
=ax2-ay2+xy2-xa2
=ax2+xy2-ay2-xa2
=x(ax+y2)-a(y2+xa)
=x(ax+y2)-a(xa+y2)
=(ax+y2)(x-a)
We have to
simplify
the
expression
first so that
we can
factorize.

Q 6.Factorise: x2-9
Solution:
x 2 - 9
= x 2 – 3 2
∴ x2 – 32 = a 2 – b 2
Let x=a and 3=b
= (a – b)(a + b)
= ( x – 3 )( x + 3 )
a 2 – b 2 = ( a – b ) ( a + b )
Putting the value of a and b.
Rule 3. We have to
use the suitable
formula.

Q 6.Factorise: x2-9
Solution:
x 2 - 9
= x 2 – 3 2
= (
x
–
3
)(
x
+
3
)
a 2 – b 2 = ( a – b ) ( a + b )
Rule 3. We have to
use the suitable
formula.

Q 7.Factorise: 25x2-9y2
Solution: 25x2-9y2
a2 –b2 =(a–b)(a+b)= (5x)2 –(3y)2
= (
5x
–
3y
)(
5x
+
3y
)
Rule 3. We have to
use the suitable
formula.

Q 8.Factorise: a4-16
Solution: a4-16
=(a2)2-42
=(a2-4)(a2+4)
=(a2-22)(a2+4)
=(a-2)(a+2)(a2+4)
a2 –b2 =(a–b)(a+b)
Rule 3. We have to
use the suitable
formula.

Q 9.Factorise: (x2+y2)2-x2y2
Solution: (x2+y2)2-x2y2
=(x2+y2)2-(xy)2
={(x2+y2)-(xy)}{(x2+y2)+(xy)}
=(x2+y2-xy)(x2+y2+xy)
=(x2-xy+y2)(x2+xy+y2)
a2 –b2 =(a–b)(a+b)
Rule 3. We have to
use the suitable
formula.

Q 10.Factorise: 4-(a-b)2
Solution: 4-(a-b)2
=22-(a-b)2
={2-(a-b)}{2+(a-b)}
=(2-a+b)(2+a-b)
a2 –b2 =(a–b)(a+b)
Rule 3. We have to
use the suitable
formula.

Q 11. Simplify by factorization process: 722-622.
Solution: 722-622
=(72-62)(72+62)
=10×134
=1340
We have to use the
suitable formula.

Q 12. Simplify by factorization process: 101×99
Solution: 101×99
= (100+1)(100-1)
= (10000-1)
= 9999
We have to use the
suitable formula.

Q 13. If a+b=8 and ab =15, find the value of a and b.
numbers(a & b) sum(a+b) = 8 product(a.b) =15
3+5=8
1.7=7
∴Numbers are 3 &5.
When a=3 then b=5
and when a=5 then b=3.
1 and 7
2 and 6
3 and 5
1+7=8
2+6=8 2.6=12
3.5=15
Solution:
We use hit and trial
method.

Q 14. If a+b=-10 and ab=24,find the value of a and b.
numbers(a & b) a.b=24 a+b=-10
-3.-8=24
-1-24=-25
∴Numbers are -4 & -6.
When a=-4 then b=-6
and when a=-6 then b=-4.
-1 and -24
-2 and -12
-3 and -8
-1.-24=24
-2.-12=24 -2-12=-14
-3-8=-11
Solution:
-4 and -6 -4.-6=24 -4-6=-10
We use hit and trial
method.

Q 15. Factorize: x2+5x+6
x2 1x 1x 1x 1x 1x
1
1
1
1
1
1
x+3
x+2
∴ x2+5x+6=(x+3)(x+2)
What did we do here?

Solution: x2+5x+6
Rule 4. Method
1. Multiply coefficient of x2
and constant term.
1×6=6(product)
2. Find the possible factors of
the product 6.
6=1×6
6=2×3
3. Remember the sign of
constant term.
= x2+(3+2)x+6
= x2+3x+2x+6
= x(x+3)+2(x+3)
= (x+3)(x+2)
4. If sign of constant term is +
then choose the pair of factors
whose sum is 5 (coefficient of x)

Q 16. Resolve into factors:
x2-9x+20
Rule 4. Method
1. Multiply coefficient of x2 and
constant term.
1×20=20(product)
2. Find the possible factors of the
product 20.
20=1×20
20=2×10
4. If sign of constant term is + then
choose the pair of factors whose sum is
9 (coefficient of x)
20=4×5
Solution: x2-9x+20
= x2-(5+4)x+20
= x2-5x-4x+20
= x(x-5)-4(x-5)
= (x-5)(x-4)
3. Remember the sign of constant term.

Q 17. Factorize: x2+3x-18 Rule 4. Method
constant term.
1×18=18(product)
product 18.
18=1×18
18=2×9
18=3×6
Solution: x2+3x-18
=x2+(6-3)x-18
=x2+6x-3x-18
=x(x+6)-3(x+6)
=(x+6)(x-3) 4. If sign of constant term is - then
choose the pair of factors whose
difference is 3 (coefficient of x)

Q 18. Factorize: x2-5x-14 Rule 4. Method
constant term.
1×14=14(product)
product 14.
14=1×14
14=2×7
4. If sign of constant term is - then
difference is 3 (coefficient of x)
Solution: x2-5x-14
=x2-(7-2)x-14
=x2-7x+2x-14
=x(x-7)+2(x-7)
=(x-7)(x+2)

Q 19. Factorize: 2x2+7x+3 Rule 4. Method
constant term.
2×3=6(product)
product 6.
6=1×6
6=2×3
4. If sign of constant term is + then
choose the pair of factors whose sum is
7 (coefficient of x)
Solution: 2x2+7x+3
=2x2+7x+3
=2x2+(6+1)x+3
=2x2+6x+1x+3
=2x(x+3)+1(x+3)
=(x+3)(2x+1)

Q 20. Factorize: x2+x-30
Solution: x2+x-30
Rule 4. Method
constant term.
1×30=30(product)
product 30.
30=1×30
30=2×15
4. If sign of constant term is - then
difference is 1 (coefficient of x).
30=3×10
30=5×6
=x2+x-30
=x2+(6-5)x-30
=x2+6x-5x-30
=x(x+6)-5(x+6)
=(x+6)(x-5)

Solution:
x2+4x+4
=x2+2.x.2+22
=(x+2)2
Rule 3. We have to
use the suitable
formula.
Rule 4.
Or,

Q 22. Factorize: 4x2-12x+9
Solution:
4x2-12x+9
= (2x)2-2.2x.3+32
= (2x-3)2
Rule 3. We have to
use the suitable
formula.
Rule 4.
Or,

Q 23. Factorize: 4x – 8
Q 24. Factorize: 4x3 - 6x2 + 8x
Q 25. Factorize: xm + ym + xa + ya
Q 26. Factorize: a( a + 4 ) + 2 ( a + 4 )
Q 27. Factorize: a( a – 3 ) - 4( 3 – a )
Q 28. Factorize: – a – b + 1 + ab
Q 29. Factorize: x2y + xy2z + zx + yz2

Q 30. Factorize: x2 – 1
Q 31. Factorize: 4x2 – 9b2
Q 32. Factorize: 1-36x2
Q 33. Factorize: x2-36
Q 34. Factorize: 16a2 – 25b2
Q 35. Factorize: 72x3 – 50x
Q 36. Find the area of shaded part.
y
y
2
2

Q 37. Factorize: x2 + 10x +25
Q 38. Factorize: p2 – 24p+144
Q 39. Factorize: 9a2 – 66a + 121
Q 40. Factorize: x2 + 9x +18
Q 41. Factorize: x2 - 9x +18
Q 42. Factorize: x2 + 3x – 18
Q 43. Factorize: x2 - 3x – 18
Q 44. Factorize: 15x2 – x – 2
Q 45. Factorize: 125x3 + 8a3
Q 46. Factorize: (x – y )3 + 8(x + y)3

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