Exercise 3.4
1.
Find the remainder using remainder theorem, when
(i)
3x3
+ 4x2 - 5x + 8 is divided by x – 1
Solution:
Let p(x) = 3x3 + 4x2 - 5x + 8
The zero of (x – 1) is 1.
When p(x) is divided by (x – 1) the remainder is p(1). Now,
p(1) = 3(1)3 + 4(1)2 - 5(1) + 8
= 3 + 4 – 5 + 8
= 10
؞ The remainder is 10.
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(ii)
5x3
+ 2x2 - 6x + 12 is divided by x + 2
Solution:
Let p(x) = 5x3 + 2x2 - 6x + 12
The zero of (x + 2) is -2.
When p(x) is divided by (x + 2) the remainder is p(-2). Now,
p(-2) = 5(-2)3 + 2(-2)2 - 6(-2) + 12
= -40 + 8 + 12 + 12
= -8
؞ The remainder is -8.
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(iii)
2x3
- 4x2 + 7x + 6 is divided by x – 2
Solution:
Let p(x) = 2x3 - 4x2 + 7x + 6
The zero of (x - 2) is 2.
When p(x) is divided by (x - 2) the remainder is p(2). Now,
p(2) =2(2)3 - 4(2)2 + 7(2) + 6
= 16 - 16 + 14 + 6
= 20
؞ The remainder is 20.
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(iv)
4x3
- 3x2+ 2x - 4 is divided by x + 3
Solution:
Let p(x) = 4x3 - 3x2+ 2x - 4
The zero of (x + 3) is -3.
When p(x) is divided by (x+ 3) the remainder is p(-3). Now,
p(-3) = 4(-3)3 – 3(-3)2+ 2(-3) - 4
= -108 – 27 – 6 – 4
= -145
؞ The remainder is -145.
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(v)
4x3
- 12x2 + 11x - 5 is divided by 2x – 1
Solution:
Let p(x) = 4x3 - 12x2 + 11x - 5
The zero of (2x - 1) is ½ .
When p(x) is divided by (2x- 1) the remainder is p(1/2). Now,
p(1/2) = 4(1/2)3 – 12(1/2)2 + 11(1/2) - 5
= 4/8 – 12/4 + 11/2 - 5
=
½ - 3 + 11/2 - 5
= 12/2 – 8 = 6 – 8
= -2
؞ The remainder is -2.
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(vi)
8x4+
12x3- 2x2- 18x+14 is divided by x + 1
Solution:
Let p(x) = 8x4+ 12x3- 2x2- 18x+14
The zero of (x + 1) is -1.
When p(x) is divided by (x+1) the remainder is p(-1). Now,
p(-1) = 8(-1)4+ 12(-1)3- 2(-1)2- 18(-1)+14
= 8 – 12 – 2 + 18 + 14
= 26
؞ The remainder is 26.
(vii)
x3-
ax2- 5x+ 2a is divided by x – a
Solution:
Let p(x) = x3- ax2- 5x+ 2a
The zero of (x - a) is a.
When p(x) is divided by (x- a) the remainder is p(a). Now,
p(a) = (a)3- a(a)2- 5(a)+ 2a
= a3 – a3 – 5a + 2a
= -3a
؞ The remainder is -3a.
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2. When the polynomial 2x3 – ax2
+ 9x - 8 is divided by x - 3 the remainder is
28.
Find the value of a.
Solution:
Let p(x) = 2x3 – ax2 + 9x - 8
When p(x) is divided by(x – 3) the remainder is p(3).
Given that p(3) = 28
ð 2(3)3 – a(3)2 + 9(3) – 8 = 28
54 – 9a + 27 – 8 = 28
73 – 9a = 28
-9a = 28 – 73
a = -45/-9
a = 5
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3. Find the value
of m if x3
- 6x2+
mx+ 60 leaves the remainder 2 when divided by
(x + 2).
Solution:
Let p(x) = x3 - 6x2+ mx+ 60
When
p(x) is divided by (x +2) the remainder is p(-2).
Given that p(-2) = 2
ð (-2)3 – 6(-2)2+ m(-2)+ 60 = 2
-8 – 24 – 2m + 60 = 2
28 – 2m = 2
-2m = 2 – 28
m = -26/-2
m
= 13
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Solution:
Let p(x) = mx3 - 2x2 + 25x - 26
When p(x) is divided by (x -1) the remainder is p(1).
Given that p(1) = 0
ð m(1)3 - 2(1)2 + 25(1) – 26 = 0
m – 2 + 25 – 26 = 0
m – 3 = 0
m = 3
|
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5. If the
polynomials x3 + 3x2-
m and 2x3-
mx + 9 leave the same remainder when
they are divided by(x – 2), find the value of m. Also find the remainder.
Solution:
Let p(x) = x3 + 3x2- m
When p(x) is divided by (x -2) the remainder is p(2).
Let q(x) = 2x3- mx
+ 9
When q(x) is divided by (x -2) the remainder is q(2).
Given that p(2) = q(2)
ð (2)3 + 3(2)2- m = 2(2)3- m(2) + 9
8 + 12 – m =
16 – 2m +9
20 –
m = 25 – 2m
2m –
m = 25 – 20
m =
5
|
Now,
p(2) = (2)3 + 3(2)2- m
= 8 + 12 – 5
Remainder
= 15---------------------------------------------------------------------------------------------------------