Class 8, Exercise 20B, Factorization

Class 8: Factorization – Exercise 20B

Date: July 16, 2016Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

Factorize

Question 1: $5xy+20x-9y-36$

Answer:

$5xy+20x-9y-36 = 5x(y+4)-9(y+4) = (y+4)(5x-9)$

$\\$

Question 2: $5ab-10b-7a+14$

Answer:

$5ab-10b-7a+14 = 5b(a-2)-7(a-2) = (a-2)(5b-7)$

$\\$

Question 3: $9xy+7y-9y^2-7x$

Answer:

$9xy+7y-9y^2-7x = 9y(x-y)-7(x-y) = (x-y)(9y-7)$

$\\$

Question 4: $3ax-6ay-8by+4bx$

Answer:

$3ax-6ay-8by+4bx = 3a(x-y)+4b(x-y)= (x-y)(3a+4b)$

$\\$

Question 5: $4x^2-10xy-6xz+15yz$

Answer:

$4x^2-10xy-6xz+15yz = 2x(2x-5y)-3z(2x-5y) = (2x-5y)(2x-3z)$

$\\$

Question 6: $3ab-6bc-3ax+6cx$

Answer:

$3ab-6bc-3ax+6cx = 3b(a-2c)-3x(a-2c) = (a-2c)(3b-3x)$

$\\$

Question 7: $a^2+ab(1+b)+b^3$

Answer:

$a^2+ab(1+b)+b^3$

$= a^2+ab+ab^2+b^3$

$= a(a+b)+b^2 (a+b)$

$= (a+b)(a+b^2)$

$\\$

Question 8: $xy^2+(x-1)y-1$

Answer:

$xy^2+(x-1)y-1$

$= xy^2+xy-y-1$

$= xy(y+1)-1(y+1)$

$= (y+1)(xy-1)$

$\\$

Question 9: $b^2-ab(1-a)-a^3$

Answer:

$b^2-ab(1-a)-a^3$

$= b^2-ab+a^2 b-a^3$

$= b(b-a)+a^2 (b-a)$

$= (b-a)(b+a^2)$

$\\$

Question 10: $y^2-y(2b+a)+2ab$

Answer:

$y^2-y(2b+a)+2ab$

$= y^2-2yb-ya+2ab$

$= y(y-2b)-a(y-2b)$

$= (y-2b)(y-a)$

$\\$

Question 11: $x-2-(x-2)^2+ax-2a$

Answer:

$x-2-(x-2)^2+ax-2a$

$= (x-2)[1-(x-2)]+a(x-2)$

$= (x-2)(1-x+2-a)$

$= (x-2)(3-x-a)$

$\\$

Question 12: $x-1-(x-1)^2+4-4x$

Answer:

$x-1-(x-1)^2+4-4x$

$= (x-1)(1-(x-1))-4(x-1)$

$= (x-1)(1-x+1-4)$

$= (x-1)(-x-2)$

$= (1-x)(x+2)$

$\\$

Question 13: $b(c-d)^2-a(d-c)+5c-5d$

Answer:

$b(c-d)^2-a(d-c)+5c-5d$

$= b(c-d)^2-a(d-c)+5(c-d)$

$= (c-d)[b(c-d)-a+5]$

$= (c-d)(bc-bd-a+5)$

$\\$

Question 14: $ab(c^2+b^2 )-a^2 cd-b^2 cd$

Answer:

$ab(c^2+b^2 )-a^2 cd-b^2 cd$

$= abc^2+abd^2-a^2 cd-b^2 cd$

$= ac(bc-ad)+bd(ad-bc)$

$= (bc-ad)(ac-bd)$

$\\$

Question 15: $ab(x^2+y^2 )+xy(a^2+b^2 )$

Answer:

$ab(x^2+y^2 )+xy(a^2+b^2 )$

$= abx^2+aby^2+xya^2+xyb^2$

$= ax(bx+ay)+by(ay+bx)$

$= (bx+ay)(ax+by)$

$\\$

Question 16: $ab(x^2+y^2 )-xy(a^2+b^2 )$

Answer:

$ab(x^2+y^2 )-xy(a^2+b^2 )$

$= abx^2+aby^2-xya^2-xyb^2$

$= ax(bx-ay)+by(ay-bx)$

$= (bx-ay)(ax-by)$

$\\$

Question 17: $(ax+by)^2+(bx-ay)^2$

Answer:

$(ax+by)^2+(bx-ay)^2$

$= a^2 x^2+b^2 y^2+2abxy+b^2 x^2+a^2 y^2-2abxy$

$= a^2 x^2+b^2 y^2+b^2 x^2+a^2 y^2$

$= a^2 (x^2+y^2 )+b^2 (x^2+y^2 )$

$= (x^2+y)(a^2+b^2)$

$\\$

Question 18: $a(a+b+c)-bc$

Answer:

$a(a+b+c)-bc$

$= a^2+ab-ac-bc$

$= a(a-c)+b(a-c)$

$= (a-c)(a+b)$

$\\$

Question 19: $a(a-2b-c)+2bc$

Answer:

$a(a-2b-c)+2bc$

$= a^2-2ab-ac+2bc$

$= a(a-c)-2b(a-c)$

$= (a-c)(a-2b)$

$\\$

Question 20: $3x^5-6x^4-2x^3+4x^2+x-2$

Answer:

$3x^5-6x^4-2x^3+4x^2+x-2$

$= 3x^4 (x-2)-2x^2 (x-2)+(x-2)$

$= (x-2)(3x^4-2x^2+1)$

$\\$

Question 21: $2a+2b-3cx-3bx+2c-3ax$

Answer:

$2a+2b-3cx-3bx+2c-3ax$

$= a(2-3x)+b(2-3x)+c(2-3x)$

$= (2-3x)(a+b+c)$

$\\$

Question 22: $a^2 b^2+b^2+a^2+1$

Answer:

$a^2 b^2+b^2+a^2+1$

$= b^2 (a^2+1)+(a^2+1)$

$= (a^2+)(b^2+1)$

Related

Leave a Reply Cancel reply

%d bloggers like this: