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## 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) = x^{3} – 5x^{2} + 4x – 3; g(x) = x – 2

Solution:

p(x) = x^{3} – 5x^{2} + 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) = x^{3} – 2x^{2} – 4x – 1; g(x) = x + 1

Solution:

p(x) = x^{3} – 2x^{2} – 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) = 4x^{3} – 12x^{2} + 14x – 3; g(x) = 2x – 1

Solution:

p(x) = 4x^{3} – 12x^{2} + 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) = x^{3} – 3x^{2} + 4x + 50; g(x) = x – 3

Solution:

p(x) = x^{3} – 3x^{2} + 4x + 50

p(3) = 3^{3} – 3(3)^{2} + 4(3) + 50

= 27 – 27 + 12 + 50

= 62

The remainder is 62.

Question 3.

Find the remainder when 3x^{3} – 4x^{2} + 7x – 5 is divided by (x + 3)

Solution:

p(x) = 3x^{3} – 4x^{2} + 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 x^{2018} + 2018 is divided by x – 1.

Solution:

p(x) = x^{2018} + 2018

When it is divided by x – 1,

p(1) = 1^{2018} + 2018

= 1 + 2018

= 2019

The remainder is 2019.

Question 5.

For what value of k is the polynomial

p(x) = 2x^{3} – kx^{2} + 3x + 10 exactly divisible by x – 2

Solution:

p(x) = 2x^{3} – kx^{2} + 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 2x^{3} + ax^{2} + 4x – 12 and x^{3} + x^{2} – 2x + a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.

Solution:

p(x_{1}) = 2x^{3} + ax^{2} + 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 ……….(R_{1})

p(x_{2}) = x^{3} + x^{2} – 2x + a

When it is divided by x – 3,

p(3) = 3^{3} + 3^{2} – 2(3) + a

= 27 + 9 – 6 + a

= 30 + a ………(R_{2})

The given remainders are same (R_{1} = R_{2})

∴ 54 + 9a = 30 + a

9a – a = 30 – 54

8a = -24

∴ a = -24/8

= -3

Consider R_{2},

Remainder = 30 – 3

= 27

Question 7.

Determine whether (x – 1) is a factor of the following polynomials:

(i) x^{3} + 5x^{2} – 10x + 4

Solution:

p(x) = x^{3} + 5x^{2} – 10x + 4

p(1) = 1^{3} + 5(1) – 10(1) + 4

= 1 + 5 – 10 + 4

= 10 – 10

= 0

∴ x – 1 is a factor of p(x)

(ii) x^{4} + 5x^{2} – 5x + 1

Solution:

p(1) = 1^{4} + 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

2x^{3} – 5x^{2} – 28x + 15

Solution:

p(x) = 2x^{3} – 5x^{2} – 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 x^{3} – 3x^{2} – mx + 24.

Solution:

p(x) = x^{3} – 3x^{2} – 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 ax^{2} + 5x + b, then show that a = b.

Solution:

p(x) = ax^{2} + 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 kx^{3} – 2x^{2} + 25x – 26 without remainder, then find the value of k.

Solution:

p(x) = kx^{3} – 2x^{2} + 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 x^{2} – 2x – 8 by using factor theorem.

Solution:

Let the area of a rectangle be p(x)

p(x) = x^{2} – 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