NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 Polynomials

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

Conclusion

Class 9 Maths Ex 2.3 of Chapter 2 - Polynomials, is crucial for a solid foundation in math. Understanding the concept of factoring quadratic expressions is a key takeaway. Vedantu's NCERT solutions can guide you in conquering quadratic equations using factorization. Regular practice with NCERT solutions provided by platforms like Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward. Consistent practice and understanding will lead to proficiency in AP-related problems

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