# Division – Exercise 3.4.4 – Class IX

1. Divide

(i) 4x5 + 7x4 + 14x3 + 3x2 – 8x + 6 by x2 + 3x + 2

x2 + 3x + 2 ) 4x5 + 7x4 + 14x3 + 3x2 – 8x + 6 ( 4x3 – 5x2 + 21x – 50

4x5 +12x4 +8x3

(-)      (-)      (-)

————————–

-5x4 + 6x3 + 3x2

-5x2 – 15x3 – 10x2

+       +       +

————————–

+21x3 + 13x2 – 8x

+21x3 + 63x2 + 42x

(-)      (-)      (-)

————————-

-50x2 – 50x + 6

-50x2 – 150x – 100

+       +       +

————————–

+100x + 106

(ii) x3 + 5x2 + 4x – 4 by x2 + 3x – 2

x2 + 3x – 2 ) x3 + 5x2 + 4x – 4 (x + 2

x3 + 3x2 – 2x

(-)   (-)    +

———————–
2x2 + 6x – 4

2x2 + 6x – 4

–      –     +

———————-
0

(iii) a4 + 4b4 by a2 + 2ab + b2

a2 + 2ab + b2 ) a4 +0.a3.b + 0.a2.b2 + 0.ab3 +4b(a2 – 2ab + 3b2

a4 + 2a3b + a2b2

(-)    (-)     (-)

————————
-2a3b – a2b2 + 0.ab3

-2a3b – 4a2b2 – 2ab3

+       +       +

————————–

+3a2b2 + 2ab3 + 4b4

+3a2b2 + 6ab3 + 3b4

–          –          –

———————————

-4ab3 + b4

(iv) x4 – 4x2 + 12x + 9 by x2 + 2x – 3

x2 + 2x – 3) x4 + 0.x3 – 4x2 + 12x + 9 (x2 – 2x + 3

x4 + 2x3 – 3x2

(-)   (-)    (+)

———————

-2x3 – x2 + 12x

-2x3 – 4x2 + 6x
+     +     –

———————–

+3x2 + 6x + 9

+3x2 + 6x – 9

–       –                +

———————-

18

(iv) 2x5 – 7x4 – 2x3 + 18x2 – 3x – 8 by x3 – 2x2 + 1

x3 – 2x2 + 0.x + 1 ) 2x5 – 7x4 – 2x3 + 18x2 – 3x – 8( 2x2 – 3x – 8

2x5 – 4x4 + 0 + 2x2

(-)   (+)    (-)    (-)

——————-

-3x4 – 2x3 + 16x2 – 3x

-3x4 + 6x3 + 0 + 3x

(+)    (-)   (-)    (-)

————————–

-8x3 + 16x2 – 8

-8x3 + 16x2 – 8

(+)  (-)   (+)

———————

0

(v) 7x4 + 4x2 + 3x – 5 by  2x2 + 3x – 2

2x2 + 3x – 2) 7x4 + 0.x3  + 4x2 + 3x – 5 ( 7/2x221/4 x + 107/8

7x4 + 21/2 x3 – 7x2

(-)      (-)      (+)

—————————-
21/2x3 + 11x2 + 3x

21/2 x3 + 63/4 x221/2 x

(+)     (-)         (+)

———————————–
+107/4 x215/2x  – 5

+107/4x2 + 321/8x – 107/4

(-)      (-)           (+)

———————————-
381/8x + 87/4

(vi) x6 – 1 by x3 – 2x2 + 2x – 1 1. Find the remainder when x3+px2+qx+1 is divided by x2+px+q

Solution:

x2+px+q) x3+px2+qx+r ( x

x3 + px2 + qx

(-)    (-)     (-)

————————–

0+0+0+ r

1. What should be added to x5 -1 to be completely divisible by x2 +3x -1?

Solution: Thus the remainder is (109x – 34) when x5 -1 is devided by x2 + 3x -1

Therefore, [-(109x-34)]=[-109x+34] should be added to x5 -1 to be completely divisible by x2 +3x -1

4. What should be added to x6-64 to be completely divisible by a4-16

Solution: The reminder is 16a2 – 64 when a6 – 64 is divided by a4 – 16.Therefore,

-16a2 + 64 should be added to x6-64 to be completely divisible by a4-16.