NCERT Solutions Class 10 Maths Chapter 2 Polynomials

Question

In NCERT Solutions Class 10 Maths Chapter 2 Polynomials we have provided a very organized and precise solution to all kinds of queries mention in this chapter. These solutions are prepared by a subject matter expert at Sarthaks. This NCERT solution is your one-stop solution for all kinds of doubt solving and for clearing concepts. We have made the solution with utmost clarity and easy to understand for all the learners. Innovative methods are used while preparing the website. Concepts are explained using a detailed diagram, flow charts, tips and tricks, shortcuts to make it intuitive for the students. We have provided solutions to exercise questions and in-text questions.

In this NCERT Solutions Class 10 Maths Chapter 2 Polynomial we can learn different topics related to Polynomial such as

• Polynomial
• Monomial, Binomial, Trinomials
• Degree of Polynomial
• Linear, Quadratic, and Cubic Polynomial
• Zeros of a Polynomials
• Analysis of Graphs
• Finding Number of Zeros using Graphs
• Zeros of Polynomial
• Relationship between Zeros and Coefficients
• Dividing Two Polynomials
• Division Algorithm

Our NCERT Solutions Class 10 Maths is the best way to assist the students in clearing the basics of this chapter, CBSE exam preparation, and as well as for the preparation for competitive exams like JEE Mains, JEE Advance, or any other similar exams. We have explained all kinds of equations and related topics to make concepts easy to understand for students. One can go through our solutions for easy understanding, clearing concepts, revising concepts, and tackling doubts. Our experts advise that ample practice of the question related to the topics will help students score good marks in the examination. These solutions are explained in the step-wise method.

Now all the solutions and practice questions are at your fingertip get started now.

Answer 1

NCERT Solutions Class 10 Maths Chapter 2 Polynomials

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

(i) 

(ii)

(iii) 

(iv)

(v)

(vi)

Answer: 

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point. 

(ii)The number of zeroes is 1 as the graph intersects the x-axis at only 1 point. 

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points. 

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points. 

(v)The number of zeroes is 4 as the graph intersects the x-axis at 4 points. 

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

2. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² - 2x - 8

(ii)​ 4s² - 4s + 1

(iii) 6x² - 3 - 7x

(iv) 4u² + 8u​​​​

(v) 3x² - x -4

Answer: 

(i) x² - 2x - 8 

= (x - 4) (x + 2)

The value of is zero when x² − 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = −2

Therefore, the zeroes of are 4 and −2.

Sum of zeroes = 4 - 2 = 2 = \(=\frac{-(-2)}{1} = \frac{-(Coefficient\,of\,x)}{Coefficient\,of\,x^2}\)

Product of zeroes = 4 x (-2) = - 8 = \(\frac{(-8)}{1}=\frac{Constant\,term}{Coefficient\,of\,x^2}\)

(ii)​ 4s² - 4s + 1

= (2s - 1)

The value of 4s² − 4s + 1 is zero when 2s − 1 = 0, i.e., \(s=\frac{1}{2}\) 

Therefore, the zeroes of 4s² − 4s + 1 are \(\frac{1}{2}\) and  \(\frac{1}{2}\).

Sum of zeroes = \(\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}\) = \(\frac{-(Coefficient\,of\,s)}{Coefficient\, of s^2}\)

Product of zeroes = \(\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}=\frac{Constant\,term}{Coefficient\, of\,s^2}\)

(iii) 6x² - 3 - 7x

= 6x² - 7x - 3 = (3x+1) (2x-3)

The value of 6x² − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0, i.e.,

\(x=\frac{-1}{3} \,\,or\,\, x = \frac{3}{2}\)

Therefore, the zeroes of 6x² − 3 − 7x are \(\frac{-1}{3}\,and\,\frac{3}{2}\).

Sum of zeroes = \(\frac{-1}{3 } + \frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-(Coefficient\, of \, x)}{Coefficeient \,of\,x^2}\)

Product of zeroes = \(\frac{-1}{3}\times\frac{3}{2}= \frac{-1}{2}=\frac{-3}{6}=\frac{Constant\, term}{Coefficient \,of\, x^2}\)

(iv) 4u² + 8u​​​​

4u² + 8u = 4u² + 8u +0

= 4u(u+2)

The value of 4u² + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = −2 

Therefore, the zeroes of 4u² + 8u are 0 and −2.

Sum of zeroes = 0 + (-2) = -2 = \(\frac{-(8)}{4}=\frac{-(Coefficient\,of\,u)}{Coefficient\,of\,u^2}\)

Product of zeroes = 0 x (-2) = 0 = \(\frac{0}{4}=\frac{Constant\,term}{Coefficient\,of\,u^2}\)

(v) 3x² - x -4

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

The value of 3x² − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0, i.e., when \(x = \frac{4}{3}\) or x = −1

Therefore, the zeroes of 3x² − x − 4 are 4/3 and −1.

Sum of zeroes = \(\frac{4}{3}+(-1)=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-(Coefficient\,of\,x)}{Coefficient\,of\,x^2}\)

Product of zeroes = \(\frac{4}{3}(-1)=\frac{-4}{3}=\frac{Constant\,term}{Coefficient\,of\,x^2}\)

3. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) \(\frac{1}{4},-1\)

(ii) \(\sqrt{2},\frac{1}{3}\)

(iii) \(0,\sqrt{5}\)

(iv) 1,1

(v) \(-\frac{1}{4},\frac{1}{4}\)

(vi) 4,1

Answer: 

 (i) \(\frac{1}{4},-1\) 

Let the polynomial be ax² + bx + c and its zeroes be \(\alpha\) and \(\beta\).

\(\alpha+\beta = \) \(\frac{1}{4}=\frac{-b}{a}\)

\(\alpha\beta=\)\(-1 = \frac{-4}{4}=\frac{c}{a}\)

If a = 4, then b = -1, c = -4

Therefore, the quadratic polynomial is 4x² − x − 4.

 (ii) \(\sqrt{2},\frac{1}{3}\)

Let the polynomial be ax² + bx + c and its zeroes be \(\alpha\) and \(\beta\).

\(\alpha+\beta = \)\(\sqrt{2}=\frac{3\sqrt{2}}{3}=\frac{-b}{a}\)

\(\alpha\beta=\)\(\frac{1}{3}=\frac{c}{a}\)

If a = 3, then b = \(-3\sqrt{2}\), c = 1

Therefore, the quadratic polynomial is 3x² \(-3\sqrt{2}\)x + 1.

 (iii) \(0,\sqrt{5}\)

Let the polynomial be ax² + bx + c and its zeroes be \(\alpha\) and \(\beta\).

\(\alpha+\beta = \) \(0 = \frac{0}{1}= \frac{-b}{a}\)

\(\alpha\times\beta=\) \(\sqrt{5}= \frac{\sqrt{5}}{1}=\frac{c}{a}\)

If a = 1, then b = 0, c = \(\sqrt{5}\)

Therefore, the quadratic polynomial is x² + \(\sqrt{5}\).

(iv) 1,1

Let the polynomial be ax² + bx + c and its zeroes be \(\alpha\) and \(\beta\).

\(\alpha+\beta = \) \(1 = \frac{1}{1}=\frac{-b}{a}\)

\(\alpha\times\beta=\) \(1 = \frac{1}{1}=\frac{c}{a}\)

If a = 1, then b = -1, c = 1

Therefore, the quadratic polynomial is x² - x + 1.

Answer 2

(v) \(-\frac{1}{4},\frac{1}{4}\)

Let the polynomial be ax² + bx + c and its zeroes be \(\alpha\) and \(\beta\).

\(\alpha+\beta = \) \(\frac{-1}{4}=\frac{-b}{a}\)

\(\alpha\times\beta=\) \(\frac{1}{4}=\frac{c}{a}\)

If a = 4, then b =1, c = 1

Therefore, the quadratic polynomial is 4x² + x + 1 .

 (vi) 4,1

Let the polynomial be ax² + bx + c and its zeroes be \(\alpha\) and \(\beta\).

\(\alpha+\beta = \) \(4 = \frac{4}{1}=\frac{-b}{a}\)

\(\alpha\times\beta=\) \(1 = \frac{1}{1}=\frac{c}{a}\)

If a = 1, then b = -4, c = 1

Therefore, the quadratic polynomial is x² - 4x + 1.

4. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ - 3x² + 5x - 3,                   g(x) = x² - 2

(ii) p(x) = x⁴ - 3x² + 4x + 5,                 g(x) = x² + 1 - x

(iii) p(x) = x⁴ - 5x +6,                          g(x) = 2 - x²

Answer: 

(i) p(x) = x³ - 3x² + 5x - 3,

g(x) = x² - 2

Quotient = x − 3 

Remainder = 7x − 9

(ii) p(x) = x⁴ - 3x² + 4x + 5 = x⁴ + 0.x³ -  3x² + 4x + 5

g(x) = x² + 1 - x = x² - x + 1

 

Quotient = x² + x − 3 

Remainder = 8

(iii) p(x) = x⁴ - 5x + 6,

g(x) = 2 - x² = -x² + 2

Quotient = −x² − 2 

Remainder = −5x +10

5. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t² - 3, 2t⁴ + 3t³ - 2t² - 9t - 12

(ii) x² + 3x + 1, 3x⁴ + 5x³ - 7x² + 2x + 2

(iii) x³ - 3x + 1, x⁵ - 4x³ + x² + 3x + 1

Answer: 

(i) t² - 3, 2t⁴ + 3t³ - 2t² - 9t - 12

t² - 3 = t² + 0.t - 3

Since the remainder is 0, 

Hence, t² - 3 is a factor of 2t⁴ + 3t³ - 2t² - 9t - 12.

(ii) x² + 3x + 1, 3x⁴ + 5x³ - 7x² + 2x + 2

Since the remainder is 0,

Hence, x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x² + 2x + 2.

(iii) x³ - 3x + 1, x⁵ - 4x³ + x² + 3x + 1

Since the remainder ≠ 0,

Hence, x³ - 3x + 1 is not a factor of x⁵ - 4x³ + x² + 3x + 1.

6. Obtain all other zeroes of 3x⁴ + 6x³ - 2x² - 10x - 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\).

Answer: 

p(x) = 3x⁴ + 6x³ - 2x² - 10x - 5

Since the two zeroes are   \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)

\(\therefore\,(x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}}) = (x^2-\frac{5}{3})\) is a factor of 3x⁴ + 6x³ - 2x² - 10x - 5

Therefore, we divide the given polynomial by \(x^2 -\frac{5}{3}\)

3x⁴ + 6x³ - 2x² - 10x - 5 

\(= (x^2-\frac{5}{3})(3x^2+6x+3)\)

\(=3 (x^2-\frac{5}{3})(x^2+2x+1)\)

We factorize x² + 2x + 1

= (x + 1)²

Therefore, its zero is given by x + 1 = 0 or x = −1 

As it has the term (x+1)² , therefore, there will be 2 zeroes at x = −1.

Hence, the zeroes of the given polynomial are \(\sqrt{\frac{5}{3}}\), \(-\sqrt{\frac{5}{3}}\)  −1 and −1.

7. On dividing x³ - 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x − 2 and − 2x + 4, respectively. Find g(x).

Answer:

p(x) = x³ - 3x² + x + 2  (Dividend)

g(x) = ?  (Divisor) 

Quotient = (x − 2) 

Remainder = (− 2x + 4) 

Dividend = Divisor × Quotient + Remainder

x³ - 3x² + x + 2 = g(x) x (x-2) + (-2x + 4)

x³ - 3x² + x + 2 + 2x - 4 = g(x) (x - 2)

x³ - 3x² + 3x - 2 = g(x) (x - 2)

g(x) is the quotient when we divide (x³ - 3x² + x + 2) by (x - 2) 

\(\therefore\) g(x) = (x² - x + 1)

Answer 3

8. Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and 

(i) deg p(x) = deg q(x) 

(ii) deg q(x) = deg r(x)

(iii)deg r(x) = 0

Answer:

According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that 

p(x) = g(x) × q(x) + r(x), 

where r(x) = 0 or degree of r(x) < degree of g(x) 

Degree of a polynomial is the highest power of the variable in the polynomial.

(i) deg p(x) = deg q(x) 

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant). 

Let us assume the division of 6x² + 2x + 2 by 2. 

Here, p(x) = 6x² + 2x + 2 

g(x) = 2 

q(x) = 3x² + x + 1and r(x) = 0 

Degree of p(x) and q(x) is the same i.e., 2. 

Checking for division algorithm, p(x) = g(x) × q(x) + r(x) 

6x² + 2x + 2 = (2) (3x² + x + 1) + 0 

Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x) 

Let us assume the division of x³ + x by x² , 

Here, p(x) = x³ + x g(x) = x² q(x) = x and r(x) = x 

Clearly, the degree of q(x) and r(x) is the same i.e., 1. 

Checking for division algorithm, p(x) = g(x) × q(x) + r(x)

x³ + x = (x²) × x + x x³ + x = x³ + x 

Thus, the division algorithm is satisfied.

(iii) deg r(x) = 0 

Degree of remainder will be 0 when remainder comes to a constant. 

Let us assume the division of x³ + 1by x². 

Here, p(x) = x³ + 1 g(x) = x² q(x) = x and r(x) = 1 

Clearly, the degree of r(x) is 0. 

Checking for division algorithm, 

p(x) = g(x) × q(x) + r(x) x³ + 1 = (x²) × x + 1 x³ + 1 = x³ + 1

Thus, the division algorithm is satisfied.

9. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³ +  x² - 5x + 2;   \(\frac{1}{2}\),1,-2

(ii) x³ - 4x² + 5x - 2;   2,1,1

Answer:

(i) 2x³ +  x² - 5x + 2

Zeros for this polynominal are   \(\frac{1}{2}\),1,-2

\(p(\frac{1}{2})=(2\frac{1}{2})^3+(\frac{1}{2})^2 -5(\frac{1}{2})+2\)

 \(=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\) 

= 0

p(1) = 2 x 1³ + 1² - 5 x 1 + 2

= 0

p(-2) = 2(-2)³ + (-2)² - 5(-2) + 2

= -16 + 4 + 10 + 2 

= 0

Therefore, \(\frac{1}{2}\) , 1, and −2 are the zeroes of the given polynomial. 

Comparing the given polynomial with ax³ + bx² + cx + d, 

we obtain a = 2, b = 1, c = −5, d = 2

We can take \(\alpha = \frac{1}{2},\beta=1,\gamma=-2\)

\(\alpha+\beta+\gamma=\frac{1}{2}+1+(-2)=-\frac{1}{2}=\frac{-b}{a}\)

\(\alpha\beta+\beta\gamma+\alpha\gamma=\frac{1}{2}\times1+1(-2)+\frac{1}{2}(-2)=\frac{-5}{2}=\frac{c}{a}\)

\(\alpha\beta\gamma=\frac{1}{2}\times1\times(-2)=\frac{-1}{1}=\frac{-(2)}{2}=\frac{-d}{a}\)

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii) x³ - 4x² + 5x - 2

Zeros for this polynomial are 2,1,1. 

p(2) = 2³ - 4(2²) + 5(2) - 2

= 8 - 16 + 10 - 2 = 0

p(1) = 1³ - 4(1)² + 5(1) - 2

= 1 - 4 + 5 - 2 = 0

Therefore, 2, 1, 1 are the zeroes of the given polynomial. 

Comparing the given polynomial with ax³ + bx² + cx + d, 

we obtain a = 1, b = −4, c = 5, d = −2. 

Verification of the relationship between zeroes and coefficient of the given polynomial

Sum of zeroes = 2 + 1 + 1 = 4 = \(\frac{-(-4)}{1}=\frac{-b}{a}\)

Multiplication of zeroes taking two at a time

= (2)(1) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 = \(\frac{(5)}{1}=\frac{c}{a}\)

Multiplication of zeroes = 2 × 1 × 1 = 2 = \(\frac{-(-2)}{1}=\frac{-d}{a}\)

Hence, the relationship between the zeroes and the coefficients is verified.

10. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively.

Answer:

Let the polynomial be ax³ + bx² + cx + d and the zeroes be \(\alpha,\beta \,and \,\gamma\)

It is given that 

\(\alpha+\beta+\gamma=\frac{2}{1}=\frac{-b}{a}\)

\(\alpha\beta+\beta\gamma+\alpha\gamma = \frac{-7}{1}=\frac{c}{a}\)

\(\alpha\beta\gamma=\frac{-14}{1}=\frac{-d}{a}\)

If a = 1, then b = −2, c = −7, d = 14 

Hence, the polynomial is x³ - 2x² - 7x + 14.

11. If the zeroes of polynomial , x³ - 3x² - x + 1 are a  - b, a, a + b. find a and b.

Answer:

p(x) = x³ - 3x² + x + 1

Zeroes are a − b, a + a + b 

Comparing the given polynomial with px³ + qx² + rx + t, 

we obtain p = 1, q = −3, r = 1, t = 1

Sum of zeroes = a - b + a + a + b

\(\frac{-q}{p}= 3a\)

\(\frac{-(-3)}{1}= 3a\)

3 = 3a

a = 1 

The zeros are 1 - b, 1, 1 + b

Multiplication of zeros = 1 (1 - b)(1 + b)

\(\frac{-t}{p}=1-b^2\)

\(\frac{-1}{1}= 1 - b^2\)

\( 1 - b^2 = -1\)

1 + 1 = b²

^{\(b = \pm\sqrt{2}\)}

Hence, a = 1 and b = \(\sqrt{2}\)  or \(-\sqrt{2}\)

Answer 4

12. It two zeroes of the polynomial , x⁴ - 6x³ - 26x² + 138x - 35 are \(2\pm\sqrt{3}\) find other zeroes.

Answer:

Given 2 + √3 and 2 – √3 are zeroes of the given polynomial. 

So, (2 + √3)(2 – √3) is a factor of polynomial.

Therefore,[x – (2 + √3)] [x – (2 – √3)] = x² + 4 − 4x − 3 

= x² − 4x + 1 is a factor of the given polynomial 

For finding the remaining zeroes of the given polynomial, we will find

Clearly, x⁴ - 6x³ - 26x² + 138x - 35 = (x² - 4x + 1) (x² - 2x - 35) (x² - 2x - 35) is also a factor of the given

It can be observed that polynomial (x² - 2x - 35) = (x - 7)(x + 5)

Therefore, the value of the polynomial is also zero when x - 7 = 0 or x + 5 = 0

Or x = 7 or -5

Hence, 7 and −5 are also zeroes of this polynomial.

13. If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 is divided by another polynomial, x² - 2x + k the remainder comes out to be x + a, find k and a.

Answer:

By division algorithm, 

Dividend = Divisor × Quotient + Remainder 

Dividend − Remainder = Divisor × Quotient

x⁴ - 6x³ + 16x² - 25x + 10 - x - a = x⁴ - 6x³ + 16x² - 26x + 10 - a 

will be divisible by x² - 2x + k

Let us divide x⁴ - 6x³ + 16x² - 26x + 10 - a  by x² - 2x + k

It can be observed that (-10 + 2k) x + (10 - a - 8k + k²) will be 0.

Therefore, (-10 + 2k) = 0 and (10 - a - 8k + k²) = 0

For (-10 + 2k) = 0, 2 k =10 And thus, k = 5 

For (10 - a - 8k + k²) = 0

10 − a − 8 × 5 + 25 = 0 

10 − a − 40 + 25 = 0

 − 5 − a = 0 

Therefore, a = −5 

Hence, k = 5 and a = −5