# Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Students can download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Check whether p(x) is a multiple of g(x) or not.
(i) p(x) = x3 – 5x2 + 4x – 3; g(x) = x – 2
Solution:
p(x) = x3 – 5x2 + 4x – 3
P(2) = (2)3 – 5(2)2 + 4(2) – 3
= 8 – 5(4) + 8 – 3
= 8 – 20 + 8 – 3
= 16 – 23
= -7
p{2) ≠ 0
∴ p(x) is not a multiple of g(x)

Question 2.
By remainder theorem, find the remainder when p(x) is divided by g(x) where,
(i) p(x) = x3 – 2x2 – 4x – 1; g(x) = x + 1
Solution:
p(x) = x3 – 2x2 – 4x – 1
p(-1) = (-1)3 – 2(-1)2 – 4(-1) – 1
= 1 – 2 + 4 – 1
= 4 – 4 = 0
∴ The remainder = 0

(ii) p(x) = 4x3 – 12x2 + 14x – 3; g(x) = 2x – 1
Solution:
p(x) = 4x3 – 12x2 + 14x – 3

= 4 × $$\frac{1}{8}$$ – 12 × $$\frac{1}{4}$$ + 14 × $$\frac{1}{2}$$ – 3
= $$\frac{1}{2}$$ – 3 + 7 – 3
= $$\frac{1}{2}$$ – 6 + 7
= $$\frac{1}{2}$$ + 1
= $$\frac{3}{2}$$
∴ The reminder is $$\frac{3}{2}$$

(iii) p(x) = x3 – 3x2 + 4x + 50; g(x) = x – 3
Solution:
p(x) = x3 – 3x2 + 4x + 50
p(3) = 33 – 3(3)2 + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
The remainder is 62.

Question 3.
Find the remainder when 3x3 – 4x2 + 7x – 5 is divided by (x + 3)
Solution:
p(x) = 3x3 – 4x2 + 7x – 5
When it is divided by x +3,
p(-3) = 3(-3)3 – 4(-3)2 + 7(-3) – 5
= 3(-27) – 4(9) – 21 – 5
= -81 – 36 – 21 – 5
= -143
The remainder is -143.

Question 4.
What is the remainder when x2018 + 2018 is divided by x – 1.
Solution:
p(x) = x2018 + 2018
When it is divided by x – 1,
p(1) = 12018 + 2018
= 1 + 2018
= 2019
The remainder is 2019.

Question 5.
For what value of k is the polynomial
p(x) = 2x3 – kx2 + 3x + 10 exactly divisible by x – 2
Solution:
p(x) = 2x3 – kx2 + 3x + 10
When it is exactly divided by x – 2,
P(2) = 0
2(2)3 – k(2)2 + 3(2) + 10 = 0
2(8) – k(4) + 6 + 10 = 0
16 – k(4) + 6 + 10 = 0
16 – 4k + 6 + 10 = 0
32 – 4k = 0
32 = 4k
∴ k = $$\frac{32}{4}$$
= 8
The value of k = 8

Question 6.
If two polynomials 2x3 + ax2 + 4x – 12 and x3 + x2 – 2x + a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.
Solution:
p(x1) = 2x3 + ax2 + 4x – 12
When it is divided by x – 3,
p(3) = 2(3)3 + a(3)2 + 4(3) – 12
= 54 + 9a + 12 – 12
= 54 + 9a ……….(R1)
p(x2) = x3 + x2 – 2x + a
When it is divided by x – 3,
p(3) = 33 + 32 – 2(3) + a
= 27 + 9 – 6 + a
= 30 + a ………(R2)
The given remainders are same (R1 = R2)
∴ 54 + 9a = 30 + a
9a – a = 30 – 54
8a = -24
∴ a = -24/8
= -3
Consider R2,
Remainder = 30 – 3
= 27

Question 7.
Determine whether (x – 1) is a factor of the following polynomials:
(i) x3 + 5x2 – 10x + 4
Solution:
p(x) = x3 + 5x2 – 10x + 4
p(1) = 13 + 5(1) – 10(1) + 4
= 1 + 5 – 10 + 4
= 10 – 10
= 0
∴ x – 1 is a factor of p(x)

(ii) x4 + 5x2 – 5x + 1
Solution:
p(1) = 14 + 5(1)2 – 5(1) + 1
= 1 + 5 – 5 + 1
= 7 – 5
= 2
= 0
∴ x – 1 is not a factor of p(x)

Question 8.
Using factor theorem, show that (x – 5) is a factor of the polynomial
2x3 – 5x2 – 28x + 15
Solution:
p(x) = 2x3 – 5x2 – 28x + 15
x – 5 is a factor
p(5) = 2(5)3 – 5(5)2 – 28(5) + 15
= 250 – 125 – 140 + 15
= 265 – 265
= 0
∴ x – 5 is a factor of p(x)

Question 9.
Determine the value of m, if (x + 3) is a factor of x3 – 3x2 – mx + 24.
Solution:
p(x) = x3 – 3x2 – mx + 24
when x + 3 is a factor
P(-3) = 0
(-3)3 – 3(-3)2 – m(-3) + 24 = 0
-27 – 27 + 3m + 24 = 0
-54 + 24 + 3m = 0
-30 + 3m = 0
3m = 30
m = $$\frac{30}{3}$$
= 10
The value of m = 10

Question 10.
If both (x-2) and (x – $$\frac{1}{2}$$) are the factors of ax2 + 5x + b, then show that a = b.
Solution:
p(x) = ax2 + 5x + b
when (x-2) is a factor
P(2) = 0
a(2)2 + 5(2) + b = 0
4a + 10 + b = 0
4a + b = -10 …….(1)
when (x – $$\frac{1}{2}$$) is a factor
p($$\frac{1}{2}$$) = 0
a$$(\frac{1}{2})^2$$ + 5($$\frac{1}{2}$$) + b = 0
Multiply by 4
a + 10 + 4b = 0
a + 46 = -10 …….(2)
From (1) and (2) we get
4a + b = a + 4b
4a – a = 4b – b
3a = 3b
a = b
Hence it is proved.

Question 11.
If (x – 1) divides the polynomial kx3 – 2x2 + 25x – 26 without remainder, then find the value of k.
Solution:
p(x) = kx3 – 2x2 + 25x – 26
When it is divided by x – 1
P(1) = 0
k(1)3 – 2(1)2 + 25(1) – 26 = 0
k – 2 + 25 – 26 = 0
k + 25 – 28 = 0
k – 3 = 0
k = 3
The value of k = 3

Question 12.
Check if (x + 2) and (x – 4) are the sides of a rectangle whose area is x2 – 2x – 8 by using factor theorem.
Solution:
Let the area of a rectangle be p(x)
p(x) = x2 – 2x – 8
When x + 2 is the side of the rectangle
p(-2) = (-2)2 – 2(-2) – 8
= 4 + 4 – 8
= 8 – 8
= 0
When x – 4 is the side of the rectangle.
P(4) = (4)2 – 2(4) – 8
= 16 – 8 – 8
= 16 – 16
= 0
(x + 2) and (x – 4) are the sides of a rectangle

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