## MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.4

**Class 9 Maths Chapter 2 Exercise 2.4 MP Board Question 1.**

Determine which of the following polynomials has (x + 1) as factor:

- x
^{3}+ x^{2}+ x + 1 - x
^{4}+ x^{3}+ x^{2}+ x + 1 - x
^{4}+ 3x^{2}+ 3x^{2}+ x + 1 - x
^{3}– x^{2}– (2 + √2 ) x + √2

Solution:

1. p(x) = x^{3} + x^{2} + x + 1

x + 1 = 0 ∴ x = – 1

p(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1

= – 1 + 1 – 1 + 1 = 0

(x + 1) is a factor of p(x).

2. p(x) = x^{4} + x^{3} + x^{2} + x + 1

x + 1 = 0, x = – 1

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

= 1 – 1 + 1 – 1 + 1 = 1

(x + 1) is not a factor of p(x).

3. p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

x + 1 = 0,

x = – 1

p(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{3} + (-1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

(x + 1) is not a factor of p(x).

4. p(x) = x^{3} – x^{2} – (2 + √2 ) x + √2

x + 1 = 0, x = – 1

p(-1) = (-1)^{3} – (-1)^{2} – (2 + √2) x (-1) + √2

= – 1 – 1 + 2 + √2 + √2 = 2√2

(x + 1) is not a factor of p(x).

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**Class 9 Maths Chapter 2 Exercise 2.4 Solutions Question 2.**

Use the factor theorem to determine whether g(xr) is a factor of p(x) in each of the following cases:

- p(x) = 2x
^{3}+ x^{2}– 2x – 1, g(x) = x + 1 - p(x) = x
^{3}+ 3x^{2}+ 3x + 1, g(x) = x + 2 - p(x) = x
^{2}– 4x^{2}+ x + 6, g(x) = x – 3

Solution:

1. p(x) = 2x^{3} + x^{2} – 2x – 1, g(x) = 0

x + 1 = 0 x = – 1

p(-1) = 2(-1)^{3} + (-1)^{2} – 2(-1) – 1

= – 2 + 1 + 2 – 1 = 0

(x + 1) is a factor of p(x).

2. p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

g(x) = 0

x + 2 = 0, x = – 2

zero of g(x) is – 2.

Now, p(-2)

= (-2)^{3} + 3 (-2)^{2} + 3(-2) + 1 ,

= – 8 + 12 – 6 + 1 = – 1 ≠ 0

By factor theorem, g(x) is not a factor of p(x).

3. p(x) = x^{3} – 4x^{2} + x + 6

g(x) = x – 3 = 0, x = 3

p(3) = (3)^{3} – 4(3)^{2} + (3) + 6

= 27 – 36 + 3 + 6

= 36 – 36 = 0

(x – 3) is a factor of p(x).

**Maths Class 9 Chapter 2 Polynomials Exercise 2.4 Question 3.**

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

- p(x) = x
^{2}+ x + k - p(x) = 2x
^{2}+ kx + √2 - p(x) = kx
^{2}– √2x + 1 - p(x) = kx
^{2}– 3x + k.

Solution:

1. p(x) = x^{2} + x + k

If x – 1 is a factor of p(x), then p( 1) = 0

(∴ By factor theorem)

(1)^{2} + (1) + k = 0

1 + 1 + k = 0

2 + k = 0

k = – 2

2. p(x) = 2x^{2} + kx + √2

g(x) = x – 1, x = 1

P(1) = 2(1)^{2} + k(1) + √2 = 0

2 + k + √2 = 0

k = – (2 + √2)

3. p(x) = kx^{2} – √2 x + 1

g(x) = x – 1,

x = 1

p(1) = k(1)^{2} – √2(1) + 1

k – √2 + 1 = 0

k = √2 – 1

4. p(x) = kx^{2} – 3x + k

g(x) = x – 1, x = 1

p(1) = k(1)^{2} – 3(1) + k = 0

k – 3 + k = 0

2k – 3 = 0

**Class 9 Chapter 2 Exercise 2.4 Question 4.**

Factorise:

- 12x
^{2}– 7x + 1 - 2x
^{2}+ 7x + 3 - 6x
^{2}+ 5x – 6 - 3x
^{2}– x – 4

Solution:

1. 12x^{2} – 7x + 1

12x^{2} – 7x + 1 = 12x^{2} – 4x – 3x + 1

= 4x(3x – 1) – 1 (3x – 1)

= (3x – 1) (4x – 1)

2. 2x^{2} + 7x + 3

2x^{2} + 7x + 3 = 2x^{2} + 6x + x + 3

= 2x (x + 3) + 1 (x + 3)

= (x + 3) (2x + 1)

3. 6x^{2} + 5x – 6

6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3) (3x – 2)

4. 3x^{2} – x – 4

3x^{2} – x – 4 = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4) (x + 1)

**9th Class Maths Polynomials Exercise 2.4 Question 5.**

Factorize:

- x
^{3}– 2x^{2}– x + 2 - x
^{3}– 3x^{2}– 9x – 5 - x
^{2}+ 13x^{2}+ 32x + 20 - 2y
^{3}+ y^{2}– 2y – 1.

Solution:

1. x^{3} – 2x^{2} – x + 2

Let p(x) = x^{3} – 2x^{2} – x + 2

We find that

p(1) = (1)^{3} – 2(1)^{2} – (1) + 2

= 1 – 2 – 1 + 2 = 0

By factor theorem, (x – 1) is a factor of p(x).

Now,

x^{3} – 2x^{2} – x + 2 = x^{2}(x – 1) – x(x – 1) – 2(x – 1)

= (x – 1) (x^{2} – x – 2)

= (x – 1) (x^{2} – 2x + x – 2)

= (x – 1) {x(x – 2) + 1(x – 2)}

= (x – 1) (x – 2) (x + 1).

2. p(x) = x^{3} – 3x^{2} – 9x – 5

x = – 1

p(-1) = (-1)^{3} – 3(-1)^{2} – 9(- 1) – 5

= – 1 – 3 + 9 – 5 = 0

(x + 1) is a factor of p(x).

Now, p(x) = (x + 1) (x^{2} – 4x – 5)

= (x + 1) (A2 – 5A + A – 5)

= (x + 1) [x(x – 5) + 1 (x – 5)]

= (x + 1) (x – 5) (x + 1)

3. p(x) = x^{3} + 13x^{2} + 32x + 20

x = – 1,

p(-1) = (-1)^{3} + 13(-1)^{2} + 32(-1) + 20

= – 1 + 13 – 32 + 20 = 0

(x + 1) is a factor of p(x)

p(x) = (x + 1) (x^{2} + 12x + 20)

= (x + 1) (x^{2} + 10x + 2x + 20)

= (x + 1) [x (x + 10) + 2 (x + 10)]

= (x + 1) (x + 2) (x + 10)

4. p(y) = 2y^{3} + y^{2} – 2y – 1

y = 1

P(1) = 2(1)^{3} + (1)^{2} – 2(1) – 1

= 2 + 1 – 2 – 1

= 0

(y – 1) is a factor of p(y)

p(x) = (y – 1) (2y^{2} + 3y + 1)

= (y – 1) (2y^{2} + 2y + 1y + 1)

= (y – 1)[2y(y + 1) + 1 (y + 1)]

= (y – 1) (y + 1) (2y + 1)

Algebraic Identities:

- (x + y)
^{2}= x^{2}+ 2xy + y^{2} - (x – y)
^{2}= x^{2}– 2xy + y^{2} - x
^{2}– y^{2}= (x + y) (x – y) - (x + a) (x + b) = x
^{2}+ (a + b) x + ab