# NCERT Solutions for class 9 Maths Chapter 2- Polynomials Exercise 2.4

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4 are provided here. These NCERT Maths solutions are prepared by our subject experts which makes it easy for students to learn. The students use it for reference while solving the exercise problems. The fourth exercise in Polynomials- Exercise 2.4 discusses the Factorisation of Polynomials. They provide a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook for class 9. The NCERT solutions are always prepared by following NCERT guidelines so that it should cover the whole syllabus accordingly. These are very helpful in scoring well in first and second term examinations.

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### Access other exercise solutions of class 9 Maths Chapter 2- Polynomials

Exercise 2.1 Solutions 5 Questions
Exercise 2.2 Solutions 4 Questions
Exercise 2.3 Solutions 3 Questions
Exercise 2.5 Solutions 16 Questions

### Access Answers of Maths NCERT class 9 Chapter 2 â€“ Polynomials Exercise 2.4

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

(i) x3+x2+x+1

Solution:

Let p(x) =Â x3+x2+x+1

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(âˆ’1) = (âˆ’1)3+(âˆ’1)2+(âˆ’1)+1

= âˆ’1+1âˆ’1+1

= 0

âˆ´By factor theorem, x+1 is a factor ofÂ x3+x2+x+1

(ii) x4+x3+x2+x+1

Solution:

Let p(x)=Â x4+x3+x2+x+1

The zero of x+1 is -1. . [x+1= 0 means x = -1]

p(âˆ’1) = (âˆ’1)4+(âˆ’1)3+(âˆ’1)2+(âˆ’1)+1

= 1âˆ’1+1âˆ’1+1

= 1 â‰  0

âˆ´By factor theorem, x+1 is not a factor ofÂ x4Â +Â x3Â +Â x2Â +Â xÂ + 1

(iii) x4+3x3+3x2+x+1Â

Solution:

Let p(x)=Â x4+3x3+3x2+x+1

The zero of x+1 is -1.

p(âˆ’1)=(âˆ’1)4+3(âˆ’1)3+3(âˆ’1)2+(âˆ’1)+1

=1âˆ’3+3âˆ’1+1

=1 â‰  0

âˆ´By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1

(iv) x3 â€“ x2â€“ (2+âˆš2)x +âˆš2

Solution:

Let p(x) =Â x3â€“x2â€“(2+âˆš2)x +âˆš2

The zero of x+1 is -1.

p(âˆ’1) = (-1)3â€“(-1)2â€“(2+âˆš2)(-1) + âˆš2 = âˆ’1âˆ’1+2+âˆš2+âˆš2

= 2âˆš2 â‰  0

âˆ´By factor theorem, x+1 is not a factor ofÂ x3â€“x2â€“(2+âˆš2)x +âˆš2

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

(i) p(x) = 2x3+x2â€“2xâ€“1,Â g(x) =Â x+1

Solution:

p(x) = 2x3+x2â€“2xâ€“1,Â g(x) =Â x+1

g(x) = 0

â‡’ x+1 = 0

â‡’ x = âˆ’1

âˆ´Zero of g(x) is -1.

Now,

p(âˆ’1) = 2(âˆ’1)3+(âˆ’1)2â€“2(âˆ’1)â€“1

= âˆ’2+1+2âˆ’1

= 0

âˆ´By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x3+3x2+3x+1,Â g(x) =Â x+2

Solution:

p(x) = x3+3x2+3x+1,Â g(x) =Â x+2

g(x) = 0

â‡’ x+2 = 0

â‡’ x = âˆ’2

âˆ´ Zero of g(x) is -2.

Now,

p(âˆ’2) = (âˆ’2)3+3(âˆ’2)2+3(âˆ’2)+1

= âˆ’8+12âˆ’6+1

= âˆ’1 â‰  0

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

(iii) p(x)=x3â€“4x2+x+6,Â g(x) =Â xâ€“3

Solution:

p(x) = x3â€“4x2+x+6,Â g(x) =Â xÂ -3

g(x) = 0

â‡’ xâˆ’3 = 0

â‡’ x = 3

âˆ´ Zero of g(x) is 3.

Now,

p(3) = (3)3âˆ’4(3)2+(3)+6

= 27âˆ’36+3+6

= 0

âˆ´By factor theorem, g(x) is a factor of p(x).

3. Find the value ofÂ k, ifÂ xâ€“1 is a factor ofÂ p(x) in each of the following cases:

(i) p(x) = x2+x+k

Solution:

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

By Factor Theorem

â‡’ (1)2+(1)+k = 0

â‡’ 1+1+k = 0

â‡’ 2+k = 0

â‡’ k = âˆ’2

(ii) p(x) = 2x2+kx+âˆš2

Solution:

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

â‡’ 2(1)2+k(1)+âˆš2 = 0

â‡’ 2+k+âˆš2 = 0

â‡’ k = âˆ’(2+âˆš2)

(iii) p(x) = kx2â€“âˆš2x+1

Solution:

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

By Factor Theorem

â‡’ k(1)2-âˆš2(1)+1=0

â‡’ k = âˆš2-1

(iv) p(x)=kx2â€“3x+k

Solution:

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

By Factor Theorem

â‡’ k(1)2â€“3(1)+k = 0

â‡’ kâˆ’3+k = 0

â‡’ 2kâˆ’3 = 0

â‡’ k= 3/2

4. Factorize:

(i) 12x2â€“7x+1

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1Ã—12 = 12

We get -3 and -4 as the numbers [-3+-4=-7 and -3Ã—-4 = 12]

12x2â€“7x+1= 12x2-4x-3x+1

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

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

(ii) 2x2+7x+3

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2Ã—3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6Ã—1 = 6]

2x2+7x+3 = 2x2+6x+1x+3

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

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

(iii) 6x2+5x-6Â

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 5 and product = 6Ã—-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4Ã—9 = -36]

6x2+5x-6 = 6x2+9xâ€“4xâ€“6

= 3x(2x+3)â€“2(2x+3)

= (2x+3)(3xâ€“2)

(iv) 3x2â€“xâ€“4Â

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -1 and product = 3Ã—-4 = -12

We get -4 and 3 as the numbers [-4+3 = -1 and -4Ã—3 = -12]

3x2â€“xâ€“4Â = 3x2â€“4x+3xâ€“4

= x(3xâ€“4)+1(3xâ€“4)

= (3xâ€“4)(x+1)

5. Factorize:

(i) x3â€“2x2â€“x+2

Solution:

LetÂ p(x) = x3â€“2x2â€“x+2

Factors of 2 are Â±1 and Â± 2

Now,

p(x) = x3â€“2x2â€“x+2

p(âˆ’1) = (âˆ’1)3â€“2(âˆ’1)2â€“(âˆ’1)+2

= âˆ’1âˆ’2+1+2

= 0

Therefore, (x+1) is the factor ofÂ p(x)

Now, Dividend = Divisor Ã— Quotient + Remainder

(x+1)(x2â€“3x+2) = (x+1)(x2â€“xâ€“2x+2)

= (x+1)(x(xâˆ’1)âˆ’2(xâˆ’1))

= (x+1)(xâˆ’1)(x-2)

(ii) x3â€“3x2â€“9xâ€“5

Solution:

LetÂ p(x) =Â x3â€“3x2â€“9xâ€“5

Factors of 5 are Â±1 and Â±5

By trial method, we find that

p(5)Â = 0

So,Â (x-5)Â is factor ofÂ p(x)

Now,

p(x) =Â x3â€“3x2â€“9xâ€“5

p(5) =Â (5)3â€“3(5)2â€“9(5)â€“5

= 125âˆ’75âˆ’45âˆ’5

= 0

Therefore, (x-5) is the factor ofÂ Â p(x)

Now, Dividend = Divisor Ã— Quotient + Remainder

(xâˆ’5)(x2+2x+1) = (xâˆ’5)(x2+x+x+1)

= (xâˆ’5)(x(x+1)+1(x+1))

= (xâˆ’5)(x+1)(x+1)

(iii) x3+13x2+32x+20

Solution:

LetÂ p(x) =Â x3+13x2+32x+20

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

By trial method, we find that

p(-1)Â = 0

So,Â (x+1)Â is factor ofÂ p(x)

Now,

p(x)=Â x3+13x2+32x+20

p(-1) =Â (âˆ’1)3+13(âˆ’1)2+32(âˆ’1)+20

= âˆ’1+13âˆ’32+20

= 0

Therefore, (x+1) is the factor ofÂ p(x)

Now, Dividend = Divisor Ã— QuotientÂ +Remainder

(x+1)(x2+12x+20) = (x+1)(x2+2x+10x+20)

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

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

(iv) 2y3+y2â€“2yâ€“1

Solution:

LetÂ p(y) =Â 2y3+y2â€“2yâ€“1

Factors =Â 2Ã—(âˆ’1)= -2 are Â±1 and Â±2

By trial method, we find that

p(1)Â = 0

So,Â (y-1)Â is factor ofÂ p(y)

Now,

p(y) =Â 2y3+y2â€“2yâ€“1

p(1) =Â 2(1)3+(1)2â€“2(1)â€“1

= 2+1âˆ’2

= 0

Therefore, (y-1) is the factor ofÂ p(y)

Now, Dividend = Divisor Ã— QuotientÂ + Remainder

(yâˆ’1)(2y2+3y+1) = (yâˆ’1)(2y2+2y+y+1)

= (yâˆ’1)(2y(y+1)+1(y+1))

= (yâˆ’1)(2y+1)(y+1)

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