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

**9th Maths Algebra Exercise 3.3 Question 1.**

Check whether p(x) is a multiple of g(x) or not.

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

Solution:

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

Let g(x) = 0

x – 2 = 0

x = 2

p(2) = 2^{3} – 5(2^{2}) + 4(2) – 3

= 8 – 5 × 4 + 8 – 3 = 8 – 20 + 5 = -7 ≠ 0

⇒ p(x) is not a multiple of g(x)

**9th Maths Exercise 3.3 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

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

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

Solution:

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

Let g(x) = x + 1

x + 1 = 0

x = -1

P(-1) = (-1)^{3} – 2(-1)^{2} – 4(-1) – 1

= -1 -2 × 1 + 4 – 1

= -4 + 4 = 0

∴ Remainder = 0.

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

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

Let g(x) = x – 3

x – 3 = 0

x = 3

p(3) = 3^{3} – 3(3^{2}) + 4(3) + 50

= 27 – 27 + 12 + 50

= 62

∴ Remainder = 62.

**9th Maths 3.3 Question 3.**

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

Solution:

(3x^{3} – 4x^{2} + 7x – 5) + (x + 3)

The remainder is -143.

**9th Class Maths Exercise 3.3 Solution Question 4.**

What is the remainder when x^{2018} + 2018 is divided by x – 1.

Solution:

x^{2018} + 2018 is divided by x – 1

Let g(x) = x – 1 = 0

x = 1

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

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

= 1 + 2018 = 2019

**Samacheer Kalvi 9th Maths Chapter 3 Question 5.**

For what value of k is the polynomial p(x) = 2x^{3} – kx^{2} + 3x + 10 exactly divisible by (x – 2).

Solution:

Let g(x) = x – 2 = 0

x = 2

Since p(x) is exactly divisible by (x – 2)

p(2) = 2(2^{3}) – k(2^{2}) + 3(2)+ 10

= 16 – 4k + 6 + 10

= 32 – 4k = 0

= -k = -32

k= \(\frac{32}{4}\) = 8.

**Ex 3.3 Class 9 Samacheer 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:

Let f(x) = 2x^{3} + ax^{2} + 4x – 12 and g(x) = x^{3} + x^{2} – 2x + a

When f(x) is divided by x – 3, the remainder is f(3).

Now f(3) = 2(3)^{3} + a(3)^{2} + 4(3) – 12 = 54 + 9a + 12 – 12

f(3) = 9a + 54 …………. (1)

When g(x) is divided by x – 3, the remainder is g(3).

Now g(3) = 3^{3} + 3^{2} – 2(3) + a = 27 + 9 – 6 + a

g(3) = a + 30 ……….. (2)

Since, the remainder’s are same (1) = (2)

Given that f(3) = g(3)

That is 9a + 54 = a + 30

9a – a = 30 – 54 ⇒ 8a = -24 ∴ a = -3

Substituting a = -3 in f(3), we get

f(3) = 9(-3) + 54 = -27 + 54

f(3) = 27

∴ The remainder is 27.

**9th Class Math 3.3 Exercise Solution Question 7.**

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

(i) x^{3} + 5x^{2} – 10x + 4

(ii) x^{4} + 5x^{2} – 5x + 1

Solution:

(i) Let P(x) = x^{3} + 5x^{2} – 10x + 4

By factor theorem (x – 1) is a factor of P(x), if P(1) = 0

P(1) = 1^{3} + 5(1^{2}) – 10(1) + 4 = 1 + 5 – 10 + 4

P(1) = 0

∴ (x – 1) is a factor of x^{3} + 5x^{2} – 10x + 4

(ii) Let P(x) = x^{4} + 5x^{2} – 5x + 1

By/actor theorem, (x – 1) is a factor of P(x), if P( 1) = 0

P(1) = 1^{4} + 5 (1^{2}) – 5(1) + 1 = 1 + 5 – 5 + 1 = 2 ≠ 0

∴ (x – 1) is not a factor of x^{4} + 5x^{2} – 5x + 1

**Class 9th Maths Exercise 3.3 Question 8.**

Using factor theorem, show that (x – 5) is a factor of the polynomial 2x^{3} – 5x^{2} – 28x + 15

Solution:

LetP(x) = 2x^{3} – 5x^{2} – 28x + 15

By factor theorem, (x – 5) is a factor of P(x), if P(5) = 0

P(5) = 2 (5)^{2} – 5 (5)^{2} – 28 (5) + 15

= 2 × 125 – 5 × 25 – 140 + 15

= 250 – 125 – 140 + 15 = 265 – 265 = 0

∴ (x – 5) is a factor of 2x^{3} – 5x^{2} – 28x + 15

**Class 9 Maths Exercise 3.3 Solutions Question 9.**

Determine the value of m, if (x + 3) is a factor of x^{3} – 3x^{2} – mx + 24.

Solution:

Let P(x) = x^{3} – 3x^{2} – mx + 24

By using factor theorem,

(x + 3) is a factor of P(x), then P (-3) = 0

P(-3) = (-3)^{3} -3 (-3)^{2} – m (-3) + 24 = 0

⇒ -27 – 3 × 9 + 3m + 24 = 0 ⇒ 3m = 54 – 24

⇒ m = \(\frac{30}{3}\) = 10

**Exercise 3.3 Class 9 Maths 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:

Let P(x) = ax^{2} + 5x + b

(x – 2) is a factor of P(x), if P(2) = 0

P(2) = a(2)^{2} + 5(2) + b = 0

4a + 10 + b = 0

4 a + b = – 10 …………….. (1)

(x – \(\frac{1}{2}\)) is a factor of P(x), P(\(\frac{1}{2}\)) = 0

2a + 8b = -20

a + 4b = – 10 ……………… (2)

From (1) and (2)

4a + b = – 10 …………. (1)

a + 4b = – 10 ………… (2)

(1) and (2) ⇒ 4a + b = a + 4b

3a = 3b

∴ a = b. Hence it is proved.

**Ex 3.3 Class 9 Maths Solutions Question 11.**

If (x – 1)divides the polynomial kx^{3} – 2x^{2} + 25x – 26 without remainder, then find the value of k.

Solution:

Let P(x) = kx^{3} – 2x^{2} + 25x – 26

By factor theorem, (x – 1) divides P(x) without remainder, P (1) = 0

P(1) = k(1)^{3} -2 (1)^{2} + 25 (1) – 26 = 0

k – 2 + 25 – 26 = 0

k – 3 =0

k = 3

**Class 9th Math Exercise 3.3 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 P(x) = x^{2} – 2x^{2} – 8

By using factor theorem,(x + 2) is a factor of P(x), if P (-2) = 0

P(-2) = (-2)^{2} – 2 (-2) – 8 = 4 + 4 – 8 = 0

and also (x -4) is a factor of P (x), if P (4) = 0

p (4) = 4^{2} – 2(4) – 8 = 16 – 8 – 8 = 0

∴ (x + 2), (x – 4) are the sides of a rectangle whose area is x^{2} – 2x – 8.