Factorization Class 8 ICSE
Factorization :
The process of writing an algebraic expression as the product of two or more expressions is called factorization or the resolution of the given expression into its factors.
Factorisation Of An Expression By Taking Out The Common Factor :
Factorize :
1 . 6xy - 4xz = 2x(3y - 2z)
2. 9bx² + 15b²x = 3bx(3x + 5b)
3. 4a²b - 6ab² + 8ab = 2ab(2a - 3b + 4)
4. 5x² + 15x3 - 20x = 5x (x + 3x² - 4)
5. 2x3 + x² - 4x = x (2x² + x - 4)
6. a3b - a²b² - b3 = b(a3- a²b - b²)
7. 28a²b²c - 42ab²c² = 14ab²c(2a - 3c)
8. 15x²y - 6xy² + 9y3 = 3y(5x² - 2xy + 3y²)
9. 10x²y + 15xy² - 25x²y² = 5xy(2x + 3y - 5xy)
10. 18x²y - 24xyz = 6xy(3x - 4z)
11. 27a3b3 - 45a4b2 = 9a3b2(3b -5a)
12. 4(a + b) - 6(a + b)2 = 2(a + b) (2 - 3(a + b)) = 2(a + b) ( 2 - 3a - 3b)
13. 2a(3x + 5y) - 5b(3x + 5y) = (2a - 5b) (3x + 5y)
14. 2x (p² + q²) + 4y (p² + q²) = (2x + 4y) (p² + q²)
= 2 (x + 2y) (p² + q²)
15. 8 (3a - 2b)² - 10 (3a - 2b)
let (3a - 2b) be x
so, 8x² - 10x = 2x (4x - 5)
put value of x
= 2(3a - 2b) (4(3a - 2b) -5)
= 2(3a - 2b) (12a - 8b -5)
16. x(x +y)3 - 3x²y (x + y)
let (x + y) be a
so,
xa3 - 3x²ya = xa (a² - 3xy)
put value of a
= x(x + y) ((x + y)² -3xy)
= x(x + y) (x² + 2xy +y² - 3xy)
= x(x + y) (x² +y² - xy)
17. a(3x - 2y) + b (2y - 3x)
= a(3x - 2y) - b (3x - 2y)
= (a - b) (3x - 2y)
18. a(b - c)² - d(c - d) = a(b -c)² + d(b - c)
let (b - c) be x
so, ax² + dx = x(ax + d)
put value of x
= (b - c) (a(b - c) + d)
= (b - c) (ab -ac + d)
19. a(4a -b) +2b(4a - b) - c(4a - b) = (a + 2b -c) (4a -b)
20. m(mx + ny)² + mn (mx + ny) + n (mx +ny)
let (mx + ny) be a
so, ma² + mna + na = a(ma + mn + n)
put value of a
= (mx + ny) (m(mx + ny) + mn + n)
= (mx + ny) (m²x + mny + mn + n)
Factorization of an expression by grouping the terms :
Certain algebraic expressions can be factorized by grouping the terms in pairs in such a way that in each pair, the terms have a common factor and when this common factor is taken out, the same expression is left in each pair.
Exercise
1. 5xy + 20x - 9y -36 = 5xy + 20x - 9y -36 (make groups)
= 5x(y + 4) - 9(y + 4)
= (y + 4)(5x - 9)
2. 5ab - 10b - 7a + 14 = 5b (a - 2) - 7 (a - 2)
= (a - 2) (5b - 7)
3. 9xy + 7y - 9y2 - 7x = 9xy - 9y2 + 7y - 7x
= 9y(x-y) - 7 (x - y)
= (x - y) (9y - 7)
4. 3ax - 6ay - 8by + 4bx = 3a(x - 2y) - 4b(2y - x)
= 3a(x - 2y) + 4b(x - 2y)
= (x - 2y) (3a + 4b)
5. 4x2- 10xy - 6xz + 15yz = 2x(2x - 5y) - 3z(2x - 5y)
= (2x - 5y) (2x - 3z)
6. 3ab - 6bc - 3ax + 6cx = 3b(a - 2c) - 3x(a - 2c)
= (a - 2c) (3b - 3x)
= 3(b - x)(a - 2c)
7. a2+ ab (1 + b) + b3 = a2+ ab + ab2+ b3
= a (a + b) + b2(a + b)
= (a + b) ( a + b2)
8. xy2 + (x - 1)y - 1 = xy2 + xy - y - 1
= xy (y + 1) - 1 (y + 1)
= (xy - 1) (y + 1)
9 . b2- ab( 1 - a) - a3 = b2- ab + a2b - a3
= b(b - a) + a2(b - a)
= (b - a) (b + a2)
10. y2- y( 2b + a) + 2ab = y2- 2yb - ya + 2ab
= y (y - 2b) - a (y - 2b)
= (y - 2b) (y - a )
11. x - 2 - ( x - 2)2 + ax - 2a = x - 2 - ( x - 2)(x - 2) + a(x - 2)
= (x - 2) [1 -(x - 2) + a]
= (x - 2) ( 1 - x + 2 + a)
= (x - 2) (3 - x + a)
12. x - 1 - ( x - 1)2 + 4 - 4x = x - 1 - (x - 1) (x - 1) + 4(1 - x)
= x - 1 - (x - 1) (x - 1) - 4 (x - 1)
= (x - 1) [1 - (x - 1) -4]
= (x - 1) [1 - x + 1 -4]
= (x - 1) ( -x -2)
= - (x - 1) (x + 2)
13. b(c - d)2 - a(d - c) + 5c - 5d = b(c - d)2 + a(c - d) + 5(c - d )
= (c - d) [ b(c - d) + a + 5]
= ( c - d) (bc - bd + a + 5)
14. ab (c2 +d2) - a2cd - b2cd = abc2 + abd2 - a2cd - b2cd
= abc2 - a2cd - b2cd + abd2
= ac ( bc - ad ) - bd ( bc - ad )
= ( bc - ad ) (ac - bd)
15. ab (x2 + y2) + xy (a2 + b2) = abx2+ aby2+ xya2 + xyb2
= abx2+ xya2 + aby2+ xyb2
= ax (bx + ay) + by(ay + bx)
= (bx + ay) (ax + by)
16. ab (x2 + y2 ) - xy (a2 + b2) = abx2+ aby2 - xya2- xyb2
= abx2 - xya2 + aby2- xyb2
= ax(bx - ay) + by( ay - bx)
= ax(bx - ay) - by(bx - ay)
= (bx - ay) (ax - by)
17. a(a + b - c) - bc = a2 + ab - ac - bc
= a(a + b) - c(a + b)
= (a + b) ( a - c)
Factorizing the difference of two Squares
Exercise 14C :
Post a Comment