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

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(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

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(iii) a⁴ + 4b⁴ by a² + 2ab + b²

a² + 2ab + b² ) a⁴ +0.a³.b + 0.a².b² + 0.ab³ +4b⁴ (a² – 2ab + 3b²

a⁴ + 2a³b + a²b²

(-)    (-)     (-)

————————
-2a³b – a²b² + 0.ab³

-2a³b – 4a²b² – 2ab³

+       +       +

————————–

+3a²b² + 2ab³ + 4b⁴

+3a²b² + 6ab³ + 3b⁴

–          –          –

———————————

-4ab³ + b⁴

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(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

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(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

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

2x² + 3x – 2) 7x⁴ + 0.x³  + 4x² + 3x – 5 ( ⁷/₂x² – ²¹/₄ x + ¹⁰⁷/₈

7x⁴ + ²¹/₂ x³ – 7x²

(-)      (-)      (+)

—————————-
–²¹/₂x³ + 11x² + 3x

–²¹/₂ x³ + ⁶³/₄ x² – ²¹/₂ x

(+)     (-)         (+)

———————————–
+¹⁰⁷/₄ x² – ¹⁵/₂x  – 5

+¹⁰⁷/₄x² + ³²¹/₈x – ¹⁰⁷/₄

(-)      (-)           (+)

———————————-
–³⁸¹/₈x + ⁸⁷/₄

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(vi) x⁶ – 1 by x³ – 2x² + 2x – 1

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2. 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

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3. 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

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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.

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