CBSE Factorisation Mathematics Class VIII(8TH)
Factorisation
Factors: When an algebraic equation can be
written as the product of two or more expressions then each of these is called
a factor of the give equation.
Factorisation: The process of finding two or more
expressions whose product is the given expression is called factorisation.
Factorisation
is the reverse process of multiplication.
Product
|
Factorisation
|
(i) 3x(4X-5Y)=12Xᶻ-15XY
|
(i) 12xᶻ-15xy=3x(4x-5y)
|
(ii) (a+3)(a-2)=aᶻ+a-6
|
(ii) aᶻ+a-6=(a+3)(a-2)
|
Method to
Find Factorisation When a Common Monomial Factor Occurs is Each Case:
Step1: Find
the HCF of all terms of the give expression.
Step2:
Divide each term of the given expression by this HCF.
Step3: Write
the given expression as the product of this HCF
and the quotient obtained is step2.
Factorise:
1(i) 12x+15
=3(4x+5)
(ii) 14m+21
=7(2m+3)
(iii) 9n+12nᶻ
=3n(3+4n)
2(i) 16aᶻ-24ab
= 8a(2a+3b)
(ii) 15abᶻ-20aᶻb
= 5ab(3b-4a)
(iii) 12xᶻyᶾ-21xᶾyᶻ
= 3xᶻyᶻ(4x-7y)
3(i) 24xᶾ-36xᶻy
= 12xᶻ(x-3y)
(ii) 10xᶾ-15xᶻ
= 5xᶻ(2x-3)
(iii) 36xᶾy-60xᶻyᶾz
= 12xᶻy(3x-5yᶻz)
4(i) 9xᶾ-6xᶻ+12x
= 3x(3xᶻ-2x+4)
(ii) 8xᶻ-72XY+12X
= 4x(2x-9y+3)
(iii) 18aᶾbᶾ-27aᶻbᶾ+36aᶾbᶻ
= 9aᶻbᶻ(2ab-3b+4a)
5(i) 14xᶾ+21x⁴y-28xᶻyᶻ
= 7xᶻ(2x+3xᶻy-4yᶻ)
(ii) -5-10t+20tᶻ
= 20tᶻ-10t-5
= 5(4tᶻ-2t-1)
6(i) x(x+3)+5(x+3)
= (x+3)(x+5)
(ii) 5x(x-4)-7(x-4)
= (x-4)(5x-7)
(7) 6a(a-2b)+5b(a-2b)
=(a-2b)(6a+5b)
(8) xᶾ(2a-b)+xᶻ(2a-b)
=xᶻ(2a-b)(x-1)
(9) 9a(3a-5b)-12aᶻ(3a-5b)
=3a(3a-5b)(3-4a)
(10) (x+5)ᶻ-4(x+5)
=(x+5){(x+5)-4}
= (x+5)(x+5-4)
=(x+5)(x+1)
(11) 3(a-2b)ᶻ-5(a-2b)
=(a-2b){3(a-2b)-5}
=(a-2b)(3a-6b-5)
(12) 2a+6b-3(a+3b)ᶻ
=2(a+3b)-3(a+3b)ᶻ
=(a+3b){2-3(a+3b)}
=(a+3b)(2-3a-9b)
(13) 16(2p-3q)ᶻ-4(2p-3q)
=4(2p-3q){4(2p-3q)-1}
=4(2p-3q)(8p-12q-1)
(14) x(a-3)+4(3-a)
=x(a-3)-4(a-3)
=(a-3)(x-4)
(15) 12(2x-3y)ᶻ-16(3y-2x)
=12(2x-3y)ᶻ+16(2x-3y)
=4(2x-3y){3(2x-3y)+4)
4(2x-3y)(6x-9y+4)
(16) (x+y)(2x+5)-(x+y)(x+3)
=(x+y){(2x+5)-(x+3)}
=(x+y)(2x+5-x-3)
=(x+y)(x+2)
(17) ar+br+at+bt
=ar+at+br+bt
=a(r+t)+b(r+t)
=(r+t)(a+b)
(18) xᶻ-ax-bx+ab
=x(x-a)-b(x-a)
=(x-a)(x-b)
(19) abᶻ-bcᶻ-ab+cᶻ
=abᶻ-ab-bcᶻ+cᶻ
=ab(b-1)-cᶻ(b-1)
=(b-1)(ab-cᶻ)
(20) xᶻ-xz+xy-yz
=x(x-z)+y(x-z)
=(x-z)(x+y)
(21) 6ab-bᶻ+12ac-2bc
=6ab+12ac-bᶻ-2bc
=6a(b+2c)-b(b+2c)
=(b+2c)(6a-b)
(22) (x-2y)ᶻ+4x-8y
=(x-2y)ᶻ+4(x-2y)
=(x-2y){(x-2y)+4}
=(x-2y)(x-2y+4)
(23) yᶻ-xy(1-x)-xᶾ
=yᶻ-xy+xᶻy-xᶾ
=y(y-x)+xᶻ(y-x)
=(y-x)(y+xᶻ)
CBSE Factorisation Mathematics Class VIII(8TH)
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