# Class 9 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.4

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

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

**Solution:**

p(x) = x^{3}+ x^{2}+ x + 1Let x+1 be the factor of p(x)

Then x = -1 will be the zero of p(x)

value of p(-1) should be 0

Checking,

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

=> -1 + 1 -1 + 1

=> 0

As p(-1)=0 so (x + 1) is a factor of p(x).

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

**Solution:**

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

Let x+1 be the factor of p(x)

Then x = -1 will be the zero of p(x)

value of p(-1) should be 0

Checking,

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

=> – 1 + 1 – 1 + 1 -1

=> -1

=> -1 ≠ 0

As p(-1) ≠ 0 so (x + 1) isnota factor of p(x).

**(iii) x ^{4}+3x^{3}+3x^{2}+x+1 **

**Solution:**

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

Let x+1 be the factor of p(x)

Then x = -1 will be the zero of p(x)

value of p(-1) should be 0

Checking,

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

=> 1 – 3 + 3 – 1 + 1

=> -1

=> -1 ≠ 0

As p(-1) ≠ 0 so (x + 1) isnota factor of p(x).

**(iv) x ^{3} – x^{2}– (2+√2)x +√2**

**Solution:**

p(x) = x^{3}– x^{2}– (2+√2)x +√2

Let x+1 be the factor of p(x)

Then x = -1 will be the zero of p(x)

value of p(-1) should be 0

Checking,

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

=> -1 – 1 + 2 + √2 + √2

=> 2√2

=> 2√2 ≠ 0

As p(-1) ≠ 0 so (x + 1) isnota factor of p(x).

### Question 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**

**Solution:**

p(x) = 2x^{3}+x^{2}– 2x–1g(x) = x + 1

By Factor Theorem we know that if x + 1 is a factor of p(x)

Then value of p(-1) should be 0

Checking,

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

=> -2 + 1 + 2 – 1

=> 0

As p(-1) = 0 therefore (x + 1) is a factor of 2x^{3}+ x^{2}– 2x – 1

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

**Solution:**

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

By Factor Theorem we know that if x + 2 is a factor of p(x)

Then value of p(-2) should be 0

Checking,

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

=> -8 + 12 – 6 + 1

=> -1

=> -1 ≠ 0

As p(-2) ≠ 0 therefore (x + 2) isnota factor of x^{3}+ 3x^{2}+3x + 1

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

**Solution:**

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

By Factor Theorem we know that if x – 3 is a factor of p(x)

Then value of p(3) should be 0

Checking,

=> p(3) = (3)^{3}– 4(3)^{2}+ 3 + 6

=> 27 – 36 + 3 + 6

=> 0

As p(3)=0 so (x – 3) is a factor of p(x).

### Question 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**

**Solution:**

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

By Factor Theorem,

As x-1 is a factor of p(x)

then x = 1 is the zero of p(x)

Therefore p(1) = 0

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

=> 1 + 1 + k = 0

=> 2 + k = 0

=>k = -2

**(ii) p(x) = 2x ^{2}+kx+√2**

**Solution:**

p(x) = 2x^{2 }+ kx + √2

By Factor Theorem,

As x-1 is a factor of p(x)

then x = 1 is the zero of p(x)

Therefore p(1) = 0

=> p(1) = 2(1)^{2}+ k(1) + √2

=> 2 + k + √2 = 0

=> 2 + √2 + k = 0

=>k = – (2 + √2)

**(iii) p(x) = kx ^{2}–√2x+1**

**Solution:**

p(x) = kx^{2}– √2x + 1

By Factor Theorem,

As x-1 is a factor of p(x)

then x = 1 is the zero of p(x)

Therefore p(1) = 0

=> p(1) = k(1)^{2}– √2(1) + 1

=> k – √2 + 1 = 0

=>k = √2 – 1

**(iv) p(x) = kx ^{2}–3x+k**

**Solution:**

p(x) = kx^{2}-3x + k

By Factor Theorem,

As x-1 is a factor of p(x)

Then x = 1 is the zero of p(x)

Therefore, p(1) = 0

=> p(1) = k(1)^{2}– 3(1) + k

=> k – 3 + k = 0

=> 2k – 3 = 0

=>k = 3/2

### Question 4. Factorize:

**(i) 12x ^{2}–7x+1**

**Solution:**

p(x) = 12x^{2}– 7x + 1

Using splitting the middle term method,

We need to find a pair of numbers whose sum is -7x

and product is 12x^{2}

-7x can be written as the sum of -3x and -4x

12x^{2}can be written as the product of -3x and -4x

=> 12x^{2}– 7x + 1

=> 12x^{2}-3x -4x +1

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

=>(3x – 1)(4x – 1)are the factors of 12x^{2}– 7x + 1

**(ii) 2x ^{2}+7x+3**

**Solution:**

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

Using splitting the middle term method,

We need to find a pair of numbers whose sum is 7x

and product is 6x^{2}

7x can be written as the sum of 1x and 6x

6x^{2}can be written as the product of 1x and 6x

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

=> 2x^{2}+ 1x + 6x + 3

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

=>(2x + 1)(x + 3)are the factors of 2x^{2}+ 7x + 3

**(iii) 6x ^{2}+5x-6**

**Solution:**

p(x) = 6x^{2}+ 5x – 6

Using splitting the middle term method,

We need to find a pair of numbers whose sum is 5x

and product is -36x^{2}

5x can be written as the sum of 9x and -4x

-36x^{2}can be written as the product of 9x and -4x

=> 6x^{2}+ 5x – 6

=> 6x^{2}+ 9x – 4x – 6

=> 3x(2x + 3) – 2(2x + 3)

=>(3x – 2)(2x + 3)are the factors of 6x^{2}+ 5x – 6

**(iv) 3x ^{2}–x–4**

**Solution:**

p(x) = 3x^{2}– x – 4

Using splitting the middle term method,

We need to find a pair of numbers whose sum is -x

and product is -12x^{2}

-x can be written as the sum of -4x and 3x

-12x^{2}can be written as the product of -4x and 3x

=> 3x^{2}– x – 4

=> 3x^{2}– 4x + 3x – 4

=> 3x(x + 1) – 4(x + 1)

=>(3x – 4)(x + 1)are the factors of 3x^{2}– x – 4

### Question 5. Factorize:

**(i) x ^{3}–2x^{2}–x+2**

**Solution:**

p(x) = x^{3}– 2x^{2}– x + 2

Factors of 2 are ±1 and ± 2

Using Hit and Trial Method

p(1) = (1)^{3}– 2(1)^{2}– (1) + 2

p(1) = 1 – 2 – 1 + 2

p(1) = 0

Therefore,(x – 1)is a factor of x^{3}– 2x^{2}– x + 2

PerformingLong Division :

Dividend = Divisor × Quotient + Remainder=>p(x) = (x – 1)(x^{2}– x – 2)

=> solving (x^{2}– x -2)

=> using Splitting the middle term method

=> x^{2}– 2x + x – 2

=> x(x – 2) + 1(x – 2)

=> (x + 1)(x – 2)

=>(x – 1)(x + 1)(x – 2)are the factors of p(x)

**(ii) x ^{3}–3x^{2}–9x–5**

**Solution:**

p(x) = x^{3}–3x^{2}–9x–5

Factors of -5 are ±1 and ± 5

Using Hit and Trial Method

let x = 1

p(1) = (1)^{3}– 3(1)^{2}– 9(1) – 5

p(1) = 1 – 3 – 9 -5

p(1) = -16

p(1) ≠ 0

let x = -1

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

p(-1) = -1 – 3 + 9 – 5

p(-1) = -9 + 9

p(-1) = 0

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

PerformingLong Division :

Dividend = Divisor × Quotient + Remainder=>p(x) = (x + 1)(x^{2 }– 4x – 5)

=> solving (x^{2}– 4x – 5)

=> using splitting the middle term method

=> x^{2}-5x + x – 5

=> x(x – 5) + 1(x – 5)

=> (x + 1)(x – 5)

=>(x + 1)(x + 1)(x – 5)are the factors of p(x)

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

**Solution:**

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

=> Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

Using Hit and Trial Method

let x = 1

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

p(1) = 1 + 13 + 32 + 20

p(1) = 66

p(1) ≠ 0

let x = -1

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

p(-1) = -1 + 13 – 32 + 20

p(-1) = -33 + 33

p(-1) = 0

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

PerformingLong Division :

Dividend = Divisor × Quotient + Remainder=> p(x) = (x + 1)(x^{2}+ 12x + 20)

=> solving x^{2}+ 12x + 20

=> using Splitting the middle term method

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

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

=> (x + 2)(x + 10) are the factors of x^{2}+ 12x + 20

=>(x + 1)(x + 2)(x + 10)are the factors of p(x)

**(iv) 2y ^{3}+y^{2}–2y–1**

**Solution:**

p(y) = 2y^{3}+y^{2}–2y–1

=> Factors of -1 are ±1

Using Hit and Trial Method

let x = 1

p(1) = 2(1)^{3}+ (1)^{2}– 2(1) – 1

p(1) = 2 + 1 -2 – 1

p(1) = 0

Therefore,(y – 1)is a factor of p(y)

PerformingLong Division :

Dividend = Divisor × Quotient + Remainder=> p(y) = (y – 1)(2y^{2}+ 3y + 1)

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

=> using splitting the middle term method

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

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

=>(2y + 1)(2y + 1) are the factors of 2y^{2}+ 3y + 1

=>(y – 1)(2y + 1)(2y + 1)are the factors of p(y)

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