- 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 +4b4 (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 ( 7/2x2 – 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
(vi) x6 – 1 by x3 – 2x2 + 2x – 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
- 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.
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