Page 50 - SM inner class 8.cdr

E xercise 6.2

1. Multi ply the first expres sion by the second expres sion:
2
1. 2a , 3ab 2. 6ab,- 9b 2
2
2
3
2
5
3. 5a x , 2ax y 4. 2ax ,- 7a x
3
2
2
2
2
5 4
5. -3bx , 10b x yz 6. -3p q , -5p q
4
4
3
4
7. -4abxy, -8a x by 2 8. -14xy , -5x yz
3
5
3 2
5 2
,
9. 7a xy z ,- 2z x a y 4 10. 2x - 3 4xy
y
2
2
3
2
2
2
11. 2a - 3b - c abc 12. -2c d + 3d c -5cd , -6c 2 -d 4
,
3
3
4
,
13. a - b + 3 abc a b 14. x + 5, x + 4
15. a b a b+ , - 16. a b c b c+ - , +
2
2
+
17. 5x - 8 2x - 7y 18. x + 2 ab b a b
-
,
,
y
2
2
2
2
+
,
+
19. a - ab b 3 a 4 b 20. a - 2 ax + a x + a
,
4
2
2
4
4
21. x + y x - y 4 22. x + 2 ax + a x + a
,
,
2
2
2
-
x
+
23. 4p - 4pq q , 2p q 24. x - x 1 1 + + x 2
+
,
2
2
2
y
+
25. a + ab b b - ab a 2 26. x + + z x + + z
,
+
,
y
5
3
(
Division of Algebraic Expressions \ 18x ¸ - 3x ) = -6x 2
-12y 3
The process of division of a polynomial by a (ii) -12y 3 ¸ 4y =
2
polynomial is similar to that of division in 4y 2
numbers (i.e. integers). æ -12 ö y 3 3 -2
3
ç ÷ ´ = - ´ (y ) = -3y
In both the cases we get two types of results. è 4 ø y 2
1. When dividend is exactly divisible by -12y 3 ¸ 4y = -3y
2
divisor.
In this case, we get the remainder zero. It 2. Division of a polynomial by a
is termed ‘Zero remainder’ e.g. 10 ¸ 5 = 2, monomial:
remainder = 0 Example 15. Divide (6x - 8x + 10x + 12x )
3
2
4
2. When dividend is not exactly divisible by by (2x )
divisor. 4 3 2
¸
Solution: (6x - 8x + 10x + 12x ) 2x
In this case, we get the remainder other 6x 4 - 8x 3 + 10x 2 + 12x
than zero. =
2x
It is termed as ‘non-zero remainder’ e.g. 6x 4 8x 3 10x 2 12x
17 ¸ 6 = 2, remainder = 5 = - + +
2x 2x 2x 2x
Division of a Polynomial by a Polynomial = 3x 4 -1 - 4x 3 -1 + 5x 2 -1 + 6x 1 -1
= 3x 3 - 4x 2 + 5x + 6
(Zero Remainder)
3
2
Example 16. Divide 4x + 8y - 6y by 4y
1. Division of monomial by a monomial 3 2
¸
Solution: (4y + 8y - 6y ) 4y
Example 14. Divide: 2 6
= 4y ´ y + 4y ´ 2y - 4y ´ ¸ 4y
5
(i)18x by (-3x 3 ) 4
3
æ
ö
3
(ii) -12y by 4y 2 = 4y y 2 + 2y - 6 ö æ ç 2 + 2y - ÷ ¸ 4y
÷ ¸ 4y= 4y y
ç
18x 5 è 4 ø è 2 ø
5
3
Solution: (i) 18x ¸ -( 3x ) = 3
ö
-3x 3 æ 2 + 2y - ÷
ç y
18 x 5 è 2 ø 2 3
= ´ = 4y = y + 2y -
-3 x 3 4y 2
50
Mathematics In Focus - 8