NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.4

Refer page 1 - 9 for Exercise 2.4 in the PDF 

1. Determine which of the following polynomials has\[\left( {x + 1} \right)\] a factor:

i. \[{x^3} + {x^2} + x + 1\]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^3} + {x^2} + x + 1\]

Now, for \[x =  - 1\]

\[p\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1\]

\[p\left( { - 1} \right) =  - 1 + 1 - 1 + 1\]

\[p\left( { - 1} \right) = 0\]

Therefore, by the Factor Theorem, \[x + 1\] is a factor of \[{x^3} + {x^2} + x + 1\].

ii. \[{x^4} + {x^3} + {x^2} + x + 1\]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^4} + {x^3} + {x^2} + x + 1\]

Now, for \[x =  - 1\]

\[p\left( { - 1} \right) = {\left( { - 1} \right)^4} + {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1\]

\[p\left( { - 1} \right) = 1 - 1 + 1 - 1 + 1\]

\[p\left( { - 1} \right) = 1\]

Therefore, by the Factor Theorem, \[x + 1\] is not a factor of \[{x^4} + {x^3} + {x^2} + x + 1\].

iii. \[{x^4} + 3{x^3} + 3{x^2} + x + 1\]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^4} + 3{x^3} + 3{x^2} + x + 1\]

Now, for \[x =  - 1\]

\[p\left( { - 1} \right) = {\left( { - 1} \right)^4} + 3{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + \left( { - 1} \right) + 1\]

\[p\left( { - 1} \right) = 1 - 3 + 3 - 1 + 1\]

\[p\left( { - 1} \right) = 1\]

Therefore, by the Factor Theorem, \[x + 1\] is not a factor of \[{x^4} + 3{x^3} + 3{x^2} + x + 1\].

iv. \[{x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 \]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 \]

Now, for \[x =  - 1\]

\[p\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - \left( {2 + \sqrt 2 } \right)\left( { - 1} \right) + \sqrt 2 \]

\[p\left( { - 1} \right) =  - 1 + 1 + 2 - \sqrt 2  + \sqrt 2 \]

\[p\left( { - 1} \right) = 2 + 2\sqrt 2 \]

Therefore, by the Factor Theorem, \[x + 1\] is not a factor of\[{x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 \].

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

i. \[p\left( x \right) = 2{x^3} + {x^2} - 2x - 1\], \[g\left( x \right) = x + 1\]

Ans: Given that, 

\[p\left( x \right) = 2{x^3} + {x^2} - 2x - 1\]

\[g\left( x \right) = x + 1\]

We know that, 

Zero of \[g\left( x \right)\] is \[ - 1\]

Now,

\[p\left( { - 1} \right) = 2{\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - 2\left( { - 1} \right) - 1\]

\[p\left( { - 1} \right) =  - 2 + 1 + 2 - 1\]

\[p\left( { - 1} \right) = 0\]

Therefore, \[g\left( x \right) = x + 1\] is a factor of\[p\left( x \right) = 2{x^3} + {x^2} - 2x - 1\].

ii. \[p\left( x \right) = {x^3} + 3{x^2} + 3x + 1\], \[g\left( x \right) = x + 2\]

Ans: Given that, 

\[p\left( x \right) = {x^3} + 3{x^2} + 3x + 1\]

\[g\left( x \right) = x + 2\]

We know that, 

Zero of \[g\left( x \right)\] is \[ - 2\]

Now,

\[p\left( { - 2} \right) = {\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} + 3\left( { - 2} \right) + 1\]

\[p\left( { - 2} \right) =  - 8 + 12 - 6 + 1\]

\[p\left( { - 2} \right) =  - 1\]

Therefore, \[g\left( x \right) = x + 2\] is not a factor of \[p\left( x \right) = {x^3} + 3{x^2} + 3x + 1\].

iii. \[p\left( x \right) = {x^3} - 4{x^2} + x + 6\], \[g\left( x \right) = x - 3\]

Ans: Given that, 

\[p\left( x \right) = {x^3} - 4{x^2} + x + 6\]

\[g\left( x \right) = x - 3\]

We know that, 

Zero of \[g\left( x \right)\] is \[3\]

Now,

\[p\left( 3 \right) = {\left( 3 \right)^3} - 4{\left( 3 \right)^2} + \left( 3 \right) + 6\]

\[p\left( 3 \right) = 27 - 36 + 3 + 6\]

\[p\left( 3 \right) = 0\]

Therefore, \[g\left( x \right) = x - 3\] is a factor of \[p\left( x \right) = {x^3} - 4{x^2} + x + 6\].

3. Find the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right)\] in each of the following cases:

i. \[p\left( x \right) = {x^2} + x + k\]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = {x^2} + x + k\]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

\[ \Rightarrow p\left( 1 \right) = 0\]

\[ \Rightarrow p\left( 1 \right) = {\left( 1 \right)^2} + \left( 1 \right) + k = 0\]

\[ \Rightarrow 1 + 1 + k = 0\]

\[ \Rightarrow k =  - 2\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = {x^2} + x + k\] is \[ - 2\].

ii. \[p\left( x \right) = 2{x^2} + kx + \sqrt 2 \]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = 2{x^2} + kx + \sqrt 2 \]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

\[ \Rightarrow p\left( 1 \right) = 0\]

\[ \Rightarrow p\left( 1 \right) = 2{\left( 1 \right)^2} + k\left( 1 \right) + \sqrt 2  = 0\]

\[ \Rightarrow 2 + k + \sqrt 2  = 0\]

\[ \Rightarrow k =  - \left( {2 + \sqrt 2 } \right)\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = 2{x^2} + kx + \sqrt 2 \] is \[ - \left( {2 + \sqrt 2 } \right)\].

iii. \[p\left( x \right) = k{x^2} - \sqrt 2 x + 1\]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} - \sqrt 2 x + 1\]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

\[ \Rightarrow p\left( 1 \right) = 0\]

\[ \Rightarrow p\left( 1 \right) = k{\left( 1 \right)^2} - \sqrt 2 \left( 1 \right) + 1 = 0\]

\[ \Rightarrow k - \sqrt 2  + 1 = 0\]

\[ \Rightarrow k = \sqrt 2  - 1\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} - \sqrt 2 x + 1\] is $({\sqrt 2}-1)$.

iv. \[p\left( x \right) = k{x^2} + 3x + k\]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} + 3x + k\]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

\[ \Rightarrow p\left( 1 \right) = 0\]

\[ \Rightarrow p\left( 1 \right) = k{\left( 1 \right)^2} + 3\left( 1 \right) + k = 0\]

\[ \Rightarrow k - 3 + k = 0\]

\[ \Rightarrow k = \frac{3}{2}\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} + 3x + k\] is \[\frac{3}{2}\].

4. Factorise:

i. \[12{x^2} - 7x + 1\]

Ans: Given that, 

\[p\left( x \right) = 12{x^2} - 7x + 1\]

Splitting the middle term

\[ \Rightarrow 12{x^2} - 4x + 3x + 1\]

\[ \Rightarrow 4x\left( {3x - 1} \right) - 1\left( {3x - 1} \right)\]

\[ \Rightarrow \left( {3x - 1} \right)\left( {4x - 1} \right)\]

Therefore,  \[12{x^2} - 4x + 3x + 1 = \left( {3x - 1} \right)\left( {4x - 1} \right)\].

ii. \[2{x^2} + 7x + 3\]

Ans: Given that, 

\[p\left( x \right) = 2{x^2} + 7x + 3\]

Splitting the middle term

\[ \Rightarrow 2{x^2} + x + 6x + 3\]

\[ \Rightarrow x\left( {2x + 1} \right) + 3\left( {2x - 1} \right)\]

\[ \Rightarrow \left( {2x + 1} \right)\left( {x + 3} \right)\]

Therefore, \[2{x^2} + x + 6x + 3 = \left( {2x + 1} \right)\left( {x + 3} \right)\].

iii. \[6{x^2} + 5x - 6\]

Ans: Given that, 

\[p\left( x \right) = 6{x^2} + 5x - 6\]

Splitting the middle term

\[ \Rightarrow 6{x^2} + 9x - 4x - 6\]

\[ \Rightarrow 3x\left( {2x + 3} \right) - 2\left( {2x + 3} \right)\]

\[ \Rightarrow \left( {2x + 3} \right)\left( {3x - 2} \right)\]

Therefore,  \[6{x^2} + 9x - 4x - 6 = \left( {2x + 3} \right)\left( {3x - 2} \right)\].

iv. \[3{x^2} - x - 4\]

Ans: Given that, 

\[p\left( x \right) = 3{x^2} - x - 4\]

Splitting the middle term

\[ \Rightarrow 3{x^2} - 4x + 3x - 4\]

\[ \Rightarrow x\left( {3x - 4} \right) + 1\left( {3x - 4} \right)\]

\[ \Rightarrow \left( {3x - 4} \right)\left( {x + 1} \right)\]

Therefore,  \[3{x^2} - 4x + 3x - 4 = \left( {3x - 4} \right)\left( {x + 1} \right)\].

5. Factorise:

i. \[{x^3} - 2{x^2} - x + 2\]

Ans: Given that, 

\[p\left( x \right) = {x^3} - 2{x^2} - x + 2\]

Rearranging the above,

\[ \Rightarrow {x^3} - x - 2{x^2} + 2\]

\[ \Rightarrow x\left( {{x^2} - 1} \right) - 2\left( {{x^2} - 1} \right)\]

\[ \Rightarrow \left( {{x^2} - 1} \right)\left( {x - 2} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 2} \right)\]

Therefore, \[{x^3} - 2{x^2} - x + 2 = \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 2} \right)\].

ii. \[{x^3} - 3{x^2} - 9x - 5\]

Ans: Given that, 

\[p\left( x \right) = {x^3} - 3{x^2} - 9x - 5\]

\[ \Rightarrow {x^3} + {x^2} - 4{x^2} - 4x - 5x - 5\]

\[ \Rightarrow {x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) - 5\left( {x + 1} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} - 4x - 5} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} - 5x + x - 5} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left[ {x\left( {x - 5} \right) + 1\left( {x - 5} \right)} \right]\]

\[ \Rightarrow \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 5} \right)\]

Therefore, \[{x^3} - 3{x^2} - 9x - 5 = \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 5} \right)\].

iii. \[{x^3} + 13{x^2} + 32x + 20\]

Ans: Given that, 

\[p\left( x \right) = {x^3} + 13{x^2} + 32x + 20\]

\[ \Rightarrow {x^3} + {x^2} + 12{x^2} + 12x + 20x + 20\]

\[ \Rightarrow {x^2}\left( {x + 1} \right) + 12x\left( {x + 1} \right) + 20\left( {x + 1} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} + 12x + 20} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} + 2x + 10x + 20} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left[ {x\left( {x + 2} \right) + 10\left( {x + 2} \right)} \right]\]

\[ \Rightarrow \left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right)\] 

Therefore, \[{x^3} + 13{x^2} + 32x + 20 = \left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right)\].

iv. \[2{y^3} + {y^2} - 2y - 1\]

Ans: Given that, 

\[p\left( y \right) = 2{y^3} + {y^2} - 2y - 1\]

\[ \Rightarrow 2{y^3} - 2{y^2} + 3{y^2} - 3y + y - 1\]

\[ \Rightarrow 2{y^2}\left( {y - 1} \right) + 3y\left( {y - 1} \right) + 1\left( {y - 1} \right)\]

\[ \Rightarrow \left( {y - 1} \right)\left( {2{y^2} + 3y + 1} \right)\]

\[ \Rightarrow \left( {y - 1} \right)\left( {2{y^2} + 2y + y + 1} \right)\]

\[ \Rightarrow \left( {y - 1} \right)\left[ {2y\left( {y + 1} \right) + 1\left( {y + 1} \right)} \right]\]

\[ \Rightarrow \left( {y - 1} \right)\left( {2y + 1} \right)\left( {y + 1} \right)\]

Therefore, \[2{y^3} + {y^2} - 2y - 1 = \left( {y - 1} \right)\left( {2y + 1} \right)\left( {y + 1} \right)\].

Class 9 Maths Chapter 2 Exercise 2.4

Math is a framework of all creations, without which the world cannot move an inch, people need mathematics in their everyday life. It is a subject with the most astonishing and interesting facts and discoveries of all the time. The NCERT book of Class 9 Maths Chapter 2 exercise 2.4 contains the preliminary knowledge of the Polynomials Class 9 Exercise 2.4 which is included by CBSE to create a strong base of algebra in the students.

So, without any further ado, let’s have a glance at what Class 9 Maths Chapter 2 Exercise 2.4 course has in store for you:

Polynomial

A mathematical expression which has one or more algebraic terms each of which has a constant multiplied by one or more variables raised to a non-negative integral power.

The exercise 2.4 is all about factorization theorem, the theorem simply states that -

What Are the Zeros of Polynomials?

The zeroes of a polynomial are termed as the real value of the variable for which the value of the polynomial becomes zero. So,“a” and “b” are the zeros of polynomials p(x), if p(a) = 0 and p(b) = 0.

Factor Theorem:

If p(x) is a polynomial with degree n > 1 and a is any real number, then we have (x – a) is a factor of p(x), if p(a) = 0, and p(a) = 0, if (x – a) is a factor of p(x).

The factorisation of the polynomial can be done easily with the help of middle term splitting algorithm. For an expression ax² + bx + c, we first take the product of the coefficient which will give us ac and we need to split the middle term that is b in such a way that the sum of the two split parts of b is always equal to the product we obtained that is ac. The entire exercise is based on the concept of middle term split and the factor theorem. The CBSE has designed these problems of class 9 Chapter 2 exercise 2.4 in such a way that the students get a thorough practice of what they’ve learned. Exercise 2.4 contains a lot of questions on factor theorem so that the students have enough practice of the topic.

If you find the questions to these problems as tough then you can refer Vedantu’s NCERT Solution for Class 9 Maths Chapter 2 pdf. Our subject experts have penned the answers to these questions in a simple and stepwise manner that will make things easier for you.

NCERT Solutions for Class 9 Maths Chapter 2 Exercises

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