Class maths students are Interesting.all the best guy for you homework

1. Find the following products
Question 1.
2a3
(3a + 5b)
Solution:
2a3
(3a + 5b) = 2a3
x 3a + 2a3
x 5b
= 6a3 +1
+ 10a3
b
= 6a4
+ 10a3
b
Question 2.
-11a (3a + 2b)
Solution:
-11a (3a + 2b) = -11a x 3a – 11a x 2b
= -33a2
– 22ab
Question 3.
-5a (7a – 2b)
Solution:
-5a (7a – 2b) = -5a x 7a- 5a x (-2b)
= -35a2
+ 10ab
Question 4.
-11y2
(3y + 7)
Solution:
-11y2
(3y + 7) = -11y2
x 3y – 11y2
x 7
= -33y2+1
-77y2
= 33y3
-77y2
Question 5.
(frac { 6x }{ 5 }) (x3
+y3
)
Solution:
Question 6.
xy (x3
-y3
)
Solution:
xy (x3
– y3
) =xy x x3
– xy x y3
= x1 + 3
x y – x x y1+3
= x4
y – xy4

2. Question 7.
0.1y (0.1x5
+ 0.1y)
Solution:
0.1y (0.1x5
+ 0.1y) = 0.1y x 0.1x5
+ 0.1y x 0.1y
= 0.01x5
y + 0.01y2
Question 8.
Solution:
Question 9.
Solution:

3. Question 10.
Solution:
Question 11.
5x (10x2
y – 100xy2
)
Solution:
5x (10x2
y – 100xy2
)
= 1.5x x 10x2
y – 1.5x x 100xy2
= 15x1 + 2
y- 150x1+1
x y2
15 x3
y- 150x2
y2
Question 12.
4.1xy (1.1x-y)
Solution:
4.1xy (1.1x-y) = 4.1xy x 1.1x – 4.1xy x y
= 4.51x2
y-4.1xy2
Question 13.
Solution:

4. Question 14.
Solution:
Question 15.
(frac { 4 }{ 3 }) a (a2
+ 62
– 3c2
)
Solution:
Question 16.
Find the product 24x2
(1 – 2x) and evaluate its value for x = 3.
Solution:
24x2
(1 – 2x) = 24x2
x 1 + 24x2
x (-2x)
= 24x2
+ (-48x2+1
)
= 24x2
– 48x3
If x = 3, then
= 24 (3)2
– 48 (3)3
= 24 x 9-48 x 27 = 216- 1296
= -1080

5. Question 17.
Find the product of -3y (xy +y2
) and find its value for x = 4, and y = 5.
Solution:
-3y (xy + y2
) = -3y x xy – 3y x y2
= -3xy2
-3y2 +1
= -3xy2
– 3y3
If x = 4, y = 5, then
= -3 x 4 (5)2
– 3 (5)3
= -12 x 25 – 3 x 125
= -300 – 375 = – 675
Question 18.
Multiply – (frac { 3 }{ 2 }) x2
y3
by (2x-y) and verify the answer for x = 1 and y =
2.
Solution:

6. Question 19.
Multiply the monomial by the binomial and find the value of each for x = -1, y =
25 and z =05 :
(i) 15y2
(2 – 3x)
(ii) -3x (y2
+ z2
)
(iii) z2
(x – y)
(iv) xz (x2
+ y2
)
Solution:

7. Question 20.
Simplify :

8. (i) 2x2
(at1
– x) – 3x (x4
+ 2x) -2 (x4
– 3x2
)
(ii) x3
y (x2
– 2x) + 2xy (x3
– x4
)
(iii) 3a2
+ 2 (a + 2) – 3a (2a + 1)
(iv) x (x + 4) + 3x (2x2
– 1) + 4x2
+ 4
(v) a (b-c) – b (c – a) – c (a – b)
(vi) a (b – c) + b (c – a) + c (a – b)
(vii) 4ab (a – b) – 6a2
(b – b2
) -3b2
(2a2
– a) + 2ab (b-a)
(viii) x2
(x2
+ 1) – x3
(x + 1) – x (x3
– x)
(ix) 2a2
+ 3a (1 – 2a3
) + a (a + 1)
(x) a2
(2a – 1) + 3a + a3
– 8
(xi) (frac { 3 }{ 2 })-x2
(x2
– 1) + (frac { 1 }{4 })-x2
(x2
+ x) – (frac { 3 }{ 4 })x
(x3
– 1)
(xii) a2
b (a – b2
) + ab2
(4ab – 2a2
) – a3
b (1 – 2b)
(xiii )a2
b (a3
– a + 1) – ab (a4
– 2a2
+ 2a) – b (a3
– a2
-1)
Solution:
(i) 2x2
(x3
-x) – 3x (x4
+ 2x) -2 (x4
– 3x2
)
= 2x2
x x3
-2x2
x x-3x x x4
-3x x 2x-2x4
+ 6x2
= 2x2 + 3
– 2x2 +1
– 3x,1+ 4
-6x,1+1
-2x4
+ 6x2
= 2x5
– 2x3
– 3x5
— 6x2
– 2x4
+ 6x2
= 2x5
– 3x5
– 2a4
– 2x3
+ 6x2
– 6x2
= -x5
– 2x4
– 2x3
+ 0
= -x5
-2x4
-2x3
(ii) x3
y (x2
– 2x) + 2xy (x3
– x4
)
= x3
y x x2
– x3
y x 2x + 2ay x ac3
– 2xy x x4
= x3 + 2
y-2x3 + 1
y + 2x1 + 3
y – 2yx4+1
= x5
y – 2x4
y + 2x4
y – 2yx5
= -x5
y
(iii) 3a2
+ 2 (a + 2) – 3a (2a + 1)
= 3a2
+ 2a + 4 – 6a2
– 3a
= 3a2
– 6a2
+ 2a – 3a + 4
= -3a2
– a + 4
(iv) x (x + 4) + 3x (2x2
– 1) + 4x2
+ 4
= x2
+ 4x + 3x x 2x2
– 3x x 1 + 4x2
+ 4
= x2
+ 4x + 6x2 +1
– 3x + 4x2
+ 4
= x2
+ 4x + 6x3
– 3x + 4x2
+ 4
= 6a3
+ 4x2
+ x2
+ 4x – 3x + 4
= 6x3
+ 5x2
+ x + 4
(v) a (b – c)-b (c – a) – c (a – b)
= ab – ac – be + ab – ac + bc
= 2ab – 2ac
(vi) a (b – c) + b (c – a) + c (a – b)
= ab – ac + bc – ab + ac – bc
= ab – ab + bc – be + ac – ac
= 0
(vii) 4ab (a – b) – 6a2
(b – b2
) -3b2
(2a2
– a) + 2ab (b – a)
= 4a2
b – 4ab2
– 6a2
b + 6a2
b2
– 6a2
b2
+ 3ab2
+ 2ab2
– 2a2
b
= 4a2
b- 6a2
b – 2 a2
b – 4ab2
+ 3 ab2
+ 2ab2
+ 6a2
b2
– 6a2
b2
= 4a2
b – 8a2
b – 4ab2
+ 5 ab2
+ 0
= – 4a2
b + ab2
(viii) x2
(x2
+ 1) – x3
(x + 1) – x (x3
– x)

9. = x2 + 2
+ x2
– x3 + 1
– x3
– x1 + 3
+ x1 + 1
= x4
+ x2
-x4
-x3
-x4
+ x2
= x4
-x4
-x4
-x3
+ x2
+ x2
= -x4
– x3
+ 2x2
(ix) 2a2
+ 3a (1 – 2a3
) + a (a + 1)
= 2a2
+ 3 a – 3 a x 2a3
+ a2
+ a
= 2a2
+ 3a – 6a1 + 3
+ a2
+ a
= 2a2
+ 3a – 6a4
+ a2
+ a
= -6a4
+ 3a2
+ 4a
(x) a2
(2a – 1) + 3a + a3
– 8
= 2 a2
x a – a2
x 1+3a + a3
-8
= 2a3
– a2
+ 3a + a3
– 8
= 2a3
+ a3
– a2
+ 3a – 8
= 3a3
– a2
+ 3a – 8
(xii) a2
b (a – b2
) + ab2
(4ab – 2a2
) – a3
b (1 – 2b)
= a2
b x a – a2
b x b1
+ ab2
x 4ab – ab1
x2a2
-a3
b x 1 + a3
b x 2b
= a2+1
b-a2
b2 +1
+ 4a1 +1
b2 +1
-2a2+1
b2
-a3
b + 2a3
b1 +1
= a’b – a2
b3
+ 4a2
b3
– 2a3
b2
– a3
b + 2a3
b2
= a3
b – a3
b – a2
b3
+ 4a2
b3
– 2a3
b2
+ 2a3
b2
= 0 + 3a2
b3
+ 0 = 3 a2
b3
(xiii) a2
b (a3
– a + 1) – ab (a4
– 2a2
+ 2a) – b (a3
-a2
– 1)
= a2
b x a
3
– a2
b x a + a2
b – ab x a2
+ ab x 2a2
– ab x 2a- ba3
+ ba2
+ b
= a2+ 3
b – a2+1
b + a2
b -a1 + 4
b + 2a1 + 2
b- 2a1+1
b- a3
b + a2
b + b

10. = a5
b – a3
b + a2
6 – a5
b + 2a3
b – 2a2
b – a3
b + a2
b + b
= a5
b – a3
b + 2a3
b – a3
6 – a3
b + a2
b – 2a2
b + a2
b + b
= a3
b – a5
b + 2a3
b – 2a3
b + 2a2
b-2a2
b + b
= 0 + 0 + 0 + b = b