**Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 – Polynomials****. This Exercise contains 5 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 2 or other Chapters, you can click the link at the end of this Note.**

**1. Determine which of the following polynomials has (x + 1) a factor :**

** ****(i) x ^{3} + x^{2} + x + 1**

(ii) x^{4} + x^{3} + x^{2} + x + 1

**(iii) x**

(iv) x

^{4}+ 3x^{3}+ 3x^{2}+ x + 1(iv) x

^{3}– x^{2}– (2 + √2**)x + √2**

**Solution:**

According to factor theorem, (x – a) is a factor of p(x), if p(a) = 0.

Therefore, the given linear polynomial i.e. (x+1), with zero equal to -1, is a factor of the given polynomial if p(-1) = 0.

**(i)** Given

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

Putting x = -1 in p(x)

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

= -1 + 1 – 1 + 1

= 0

Therefore, (x+1) **is a factor** of p(x) = x^{3} + x^{2} + x + 1.

**(ii)**Given

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

Putting x = -1 in p(x)

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

= 1 -1 + 1 – 1 + 1

= 1

Therefore, (x+1) **is not a factor** of p(x) = x^{4} + x^{3} + x^{2} + x + 1.

**(iii)** Given

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

Putting x = -1 in p(x)

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

= 1 -3 + 3 – 1 + 1

= 1

Therefore, (x+1) **is not a factor** of p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1.

**(iv)**Given

p(x) = x^{3} – x^{2} – (2 + √2)x + √2

Putting x = -1 in p(x)

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

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

= -1 -1 + 2 + √2 + √2

= 2√2

Therefore, (x+1) **is not a factor** of p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1.

**2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the ****following cases:**

**(i) p(x) = 2x ^{3} + x^{2} – 2x – 1, g(x) = x + 1**

**(ii) p(x) = x**

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

^{3}– 4x^{2}+ x + 6, g(x) = x – 3**Solution:**

According to factor theorem, (x-a) is a factor of p(x), if p(a) = 0.

**(i)** Given

p(x) = 2x^{3} + x^{2} – 2x – 1,

And

g(x) = x + 1

Equating g(x) to zero we get

x + 1 = 0

Or, x = -1

Applying factor theorem, g(x) is a factor of p(x) in this case if p(-1) = 0

Here,

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

= -2 + 1 + 2 – 1

= 0

Therefore, g(x) **is a factor** of p(x) in this case.

**(ii)** Given

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

And

g(x) = x + 2

Equating g(x) to zero we get

x + 2 = 0

Or, x = -2

Applying factor theorem , g(x) is a factor of p(x) in this case if p(-2) = 0

Here ,

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

= -8 + 12 – 6 + 1

= -1

Therefore, g(x) **is not a factor** of p(x) in this case.

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

Given

p(x) = x^{3} – 4x^{2} + x + 6,

And

g(x) = x – 3

Equating g(x) to zero we get

x – 3 = 0

Or, x = 3

Applying factor theorem , g(x) is a factor of p(x) in this case if p(3) = 0

Here ,

p(3) = (3)^{3} – 4(3)^{2} + (3) + 6

= 27 – 36 + 3 + 6

= 0

Therefore, g(x) **is a factor** of p(x) in this case.

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

** ****(i) p(x) = x ^{2} + x + k
(ii) p(x) = 2x^{2} + kx + √2
**

**(iii) p(x) = kx**

^{2}–**√2**

**x + 1**

(iv) p(x) = kx – 3x + k

(iv) p(x) = kx – 3x + k

** ****Solution:**

According to factor theorem, if (x-a) is a factor of p(x) then p(a) = 0. Here it is given that ( x – 1 ) , with its zero equal to 1, is a factor of p(x).

Applying factor theorem we get ,

p(1) = 0 ….(i)

Therefore,

**(i)** p(x) = x^{2} + x + k

Using equation (i) we get

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

Or, 2 + k = 0

Or, k = -2

Thus, we obtained k = -2.

**(ii)** p(x) = 2x^{2} + kx + √2

Using equation (i) we get

p(1) = 2(1)^{2} + k(1) + √2 = 0

Or, 2 + √2 + k = 0

Or, k = – (2 + √2 )

Thus, we obtained k = – (2 + √2).

**(iii)** p(x) = kx^{2} – √2x + 1

Using equation (i) we get

p(1) = k(1)^{2} – √2(1) + 1 = 0

Or , k – √2 + 1 = 0

Or , k = √2 – 1

Thus, we obtained k = √2 – 1.

**(iv)** p(x) = kx – 3x + k

Using equation (i) we get

p(1) = k(1) – 3(1) + k = 0

Or , 2k – 3 = 0

Or , k = 3/2

Thus, we obtained k = 3/2.

**4. Factorise :**

**(i) 12x ^{2} – 7x + 1
(ii) 2x^{2} + 7x + 3
**

**(iii) 6x**

(iv) 3x

^{2}+ 5x – 6(iv) 3x

^{2}– x – 4

**Solution :**

To factorize ax^{2} + bx + c using middle term splitting method we have to split b into two numbers such that their sum equals b and their product equals ac and then by further simplification we can factorize the expression.

**(i)** Let ,

p(x) = 12x^{2} – 7x + 1

We can write

-7 = (-3) + (-4)

so that

(-3)x(-4) = 12 x 1

Thus writing -7 as above we obtain

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

Taking 3x common from first two terms and -1 common from last two terms , we get

p(x) = 3x(4x – 1) -1(4x – 1)

Taking (4x – 1) common , we get

p(x) = (4x – 1)(3x – 1)

Thus, we get (4x – 1) and (3x – 1) as the factors of 12x^{2} – 7x + 1.

**(ii)**Let,

p(x) = 2x^{2} + 7x + 3

We can write

7 = (6) + (1)

so that

(6)x(1) = 2 x 3

Thus writing 7 as above we obtain

p(x) = 2x^{2} + 7x + 3 = 2x^{2} + 6x + x + 3

Taking 2x common from first two terms and 1 common from last two terms , we get

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

Taking (x + 3) common, we get

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

Thus, we get (x + 3) and (2x + 1) as the factors of 2x^{2} + 7x + 3.

**(iii)**Let,

p(x) = 6x^{2} + 5x – 6

We can write

5 = (9) + (-4)

so that

(9)x(-4) = (6) x (-6)

Thus writing 5 as above we obtain

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

Taking 3x common from first two terms and -2 common from last two terms, we get

p(x) = 3x(2x + 3) – 2(2x + 3)

Taking (2x + 3) common, we get

p(x) = (2x + 3)(3x – 2)

Thus, we get (2x + 3) and (3x – 2) as the factors of 6x^{2} + 5x – 6.

**(iv)** Let ,

p(x) = 3x^{2} – x – 4

We can write

-1 = (3) + (-4)

so that

(3)x(-4) = (3) x (-4)

Thus writing 1 as above we obtain

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

Taking 3x common from first two terms and -4 common from last two terms , we get

p(x) = 3x(x + 1) – 4(x + 1)

Taking (x + 1) common, we get

p(x) = (x + 1)(3x – 4)

Thus, we get (x + 1) and (3x – 4) as the factors of 3x^{2} – x – 4.

**5. Factorise :**

**(i) x ^{3} – 2x^{2} – x + 2**

(ii) x^{3} – 3x^{2} – 9x – 5

**(iii) x**

(iv) 2y

^{3}+ 13x^{2}+ 32x + 20(iv) 2y

^{3}+ y^{2}– 2y – 1

**Solution:**

**(i)** Let p(x) = x^{3} – 2x^{2} – x + 2

Factors of 2 are ±1, ±2

By trial , we can find that p(2) = 2^{3} – 2(2)^{2} – 2 + 2 = 0

Therefore , (x – 2) is a factor of p(x)

Now ,

we see that

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

Taking (x – 2) common

= (x – 2)(x^{2} – 1) [ Using a^{2} – b^{2} = (a + b)(a – b)]

= (x – 2)(x – 1)(x + 1)

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

**(ii)** Let p(x) = x^{3} – 3x^{2} – 9x – 5

Factors of -5 are ±1 , ±5

By trial , we can find that p(-1) = (-1)^{3} – 3(-1)^{2} – 9(-1) – 5

= -1 – 3 + 9 – 5

= 0

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

Now ,

we see that

x^{3} – 3x^{2} – 9x – 5 = x^{3} + x^{2} – 4x^{2} – 4x – 5x – 5

= x^{2}(x+1) -4x(x+1) -5(x+1)

Taking (x + 1) common

= (x +1)(x^{2} – 4x -5)

Splitting the middle term to factorize (x^{2} – 4x -5)

x^{2} – 4x – 5 = x^{2} – 5x + x – 5

= x(x – 5) + 1(x – 5)

Taking (x – 5) common we obtain

x^{2} – 4x – 5 = (x – 5 )(x + 1)

So, x^{3} – 3x^{2} – 9x – 5 = (x + 1)(x + 1)(x – 5).

**(iii)** Let p(x) = x^{3} + 13x^{2} + 32x + 20

Factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20

By trial , we can find that p(-1) = (-1)^{3} + 13(-1)^{2} + 32(-1) + 20

= -1 + 13 – 32 + 20

= 0

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

Now,

we see that

x^{3} + 13x^{2} + 32x + 20 = x^{3} + x^{2} + 12x^{2} + 12x + 20x + 20

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

Taking (x + 1) common

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

Splitting the middle term to factorize (x^{2} +12x + 20)

x^{2} +12x + 20 = x^{2} +10x + 2x + 20

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

Taking (x + 10) common we obtain

x^{2} +12x + 20 = (x +10 )(x + 2)

So, x^{3} + 13x^{2} + 32x + 20 = (x + 1)(x +10 )(x + 2).

**(iv)** Let p(y) = 2y^{3} + y^{2} – 2y – 1

Factors of -1 are ±1

By trial , we can find that p(1) = 2(1)^{3} + (1)^{2} – 2(1) – 1

= 2 + 1 – 2 – 1

= 0

Therefore , (y – 1) is a factor of p(y)

Now ,

we see that

2y^{3} + y^{2} – 2y – 1 = 2y^{3} – 2y^{2} + 3y^{2} – 3y + y – 1

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

Taking (y-1) common

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

Splitting the middle term to factorize (2y^{2} + 3y +1)

2y^{2} + 3y +1 =2y^{2} + 2y + y +1

= 2y(y + 1) + 1(y + 1)

Taking (y+1) common we obtain

2y^{2} + 3y +1 = (y + 1)(2y + 1)

So, 2y^{3} + y^{2} – 2y – 1 = (y – 1)(y + 1)(2y + 1).

**NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 – Polynomials, has been designed by the NCERT to test the knowledge of the student on the topic – Factorisation of Polynomials**

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5 – Number System can be accessed by clicking here.**

** Maths – NCERT Solutions Class 9**

**Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 – Polynomials**

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