NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

This chapter will teach you about polynomials of any degree and their solutions using a graphical representation. The NCERT Solutions for Class 10 Maths Chapter 2 Polynomials provide a detailed explanation of questions from the Class 10 Maths NCERT Book.

You must refer to these Polynomials Class 10 notes to ensure that you have the correct answers and explanations for each question. Experts prepared these NCERT Solutions for Class 10 Maths Chapter 2 Polynomials, which are simple to understand. Other subjects’ NCERT solutions for Class 10 can be found here.

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths chapter 2 Polynomials Excercise: 2.1

Q1 (1) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the numbers of zeroes of p(x), in each case.

Answer: The number of zeroes of p(x) is zero as the curve does not intersect the x-axis.

Q1 (2) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

Answer: The number of zeroes of p(x) is one as the graph intersects the x-axis only once.

Q1 (3) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

Answer: The number of zeroes of p(x) is three as the graph intersects the x-axis thrice.

Q1 (4) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

Answer: The number of zeroes of p(x) is two as the graph intersects the x-axis twice.

Q1 (5) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

Answer: The number of zeroes of p(x) is four as the graph intersects the x-axis four times.

Q1 (6) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case

Answer: The number of zeroes of p(x) is three as the graph intersects the x-axis thrice.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Excercise: 2.2

Q1 (1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer:

x ²– 2x – 8 = 0

x ²– 4x + 2x – 8 = 0

x(x-4) +2(x-4) = 0

(x+2)(x-4) = 0

The zeroes of the given quadratic polynomial are -2 and 4

VERIFICATION

Sum of roots:

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-2}{1}\\ =2\\=\alpha +\beta

Verified

Product of roots:

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-8}{1}\\ =-8\\=\alpha \beta

Verified

Q1 (ii) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer:

\\4s^2 - 4s + 1 = 0 \\4s^2 - 2s - 2s + 1 = 0 \\2s(2s-1) - 1(2s-1) = 0 \\(2s-1)(2s-1) = 0

The zeroes of the given quadratic polynomial are 1/2 and 1/2

\\\alpha =\frac{1}{2}\\ \beta =\frac{1}{2}

VERIFICATION

Sum of roots:

\\\alpha +\beta =\frac{1}{2}+\frac{1}{2}=1

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-4}{4}\\ =1\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{1}{4}\\ =\alpha \beta

Verified

Q1 (3) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer:

6x ²– 3 – 7x = 0

6x ²– 7x – 3 = 0

6x ²– 9x + 2x – 3 = 0

3x(2x – 3) + 1(2x – 3) = 0

(3x + 1)(2x – 3) = 0

The zeroes of the given quadratic polynomial are -1/3 and 3/2

\\\alpha =-\frac{1}{3}\\ \beta =\frac{3}{2}

Sum of roots:

\\\alpha+\beta =-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-7}{6}\\ =\frac{7}{6}\\=\alpha +\beta

Verified

Product of roots:

\\\alpha \beta =-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-3}{6}\\ =-\frac{1}{2}\\=\alpha \beta

Verified

Q1 (4) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer: 4u ²+ 8u = 0

4u(u + 2) = 0

The zeroes of the given quadratic polynomial are 0 and -2

VERIFICATION

Sum of roots:

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{8}{4}\\ =-2\\=\alpha +\beta

Verified

Product of roots:

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{0}{4}\\=0\\ =\alpha \beta

Verified

Q1 (5) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer: t ²– 15 = 0

The zeroes of the given quadratic polynomial are $-\sqrt{15}$ and $\sqrt{15}$

VERIFICATION

Sum of roots:

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{0}{1}\\ =0\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =-\sqrt{15}\times \sqrt{15}=-15

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-15}{1}\\=-15\\ =\alpha \beta

Verified

Q1 (6) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Answer: 3x ²– x – 4 = 0

3x ²+ 3x – 4x – 4 = 0

3x(x + 1) – 4(x + 1) = 0

(3x – 4)(x + 1) = 0

The zeroes of the given quadratic polynomial are 4/3 and -1

VERIFICATION

Sum of roots:

\\\alpha +\beta =\frac{4}{3}+(-1)=\frac{1}{3}

\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-1}{3}\\ =\frac{1}{3}\\=\alpha +\beta

Verified

Product of roots:

\alpha \beta =\frac{4}{3}\times -1=-\frac{4}{3}

\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-4}{3}\\ =\alpha \beta

Verified

Q2 (1) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1/4 , -1

Answer:

\\\alpha +\beta =\frac{1}{4}\\ \alpha \beta =-1

The required quadratic polynomial is

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\frac{1}{4}x-1=0\\ 4x^{2}-x-4=0

Q2 (2) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively.

Answer:

\\\alpha +\beta =\sqrt{2}\\ \alpha \beta =\frac{1}{3}

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\sqrt{2}x+\frac{1}{3}=0\\ 3x^{2}-3\sqrt{2}x+1=0

The required quadratic polynomial is $3x^{2}-3\sqrt{2}x+1$

Q2 (3) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively.

Answer:

\\\alpha +\beta =0\\ \alpha \beta =\sqrt{5}

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-0x+\sqrt{5}=0\\ x^{2}+\sqrt{5}=0

The required quadratic polynomial is x ²+ $\sqrt{5}$ .

Q2 (4) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1,1

Answer:

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-1x+1=0\\ x^{2}-x+1=0

The required quadratic polynomial is x ²– x + 1

Q2 (5) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively.

Answer:

\\\alpha +\beta =-\frac{1}{4}\\ \alpha \beta =\frac{1}{4}

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\left (-\frac{1}{4} \right )x+\frac{1}{4}=0\\ 4x^{2}+x+1=0

The required quadratic polynomial is 4x ²+ x + 1

Q2 (6) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 4,1

Answer:

\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-4x+1=0\\

The required quadratic polynomial is x ²– 4x + 1

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Excercise: 2.3

Q1 (1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder
in each of the following :

Answer: The polynomial division is carried out as follows

The quotient is x-3 and the remainder is 7x-9

Q1 (2) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

Answer:

The division is carried out as follows

The quotient is $x^2+x-3$

and the remainder is 8

Q1 (3) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

Answer: The polynomial is divided as follows

The quotient is $-x^2-2$ and the remainder is $-5x+10$

Q2 (1) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Answer:

After dividing we got the remainder as zero. So $t^2 - 3$ is a factor of $2t^4 + 3t^3 - 2t^2 - 9t - 12$

Q2 (2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2

Answer: To check whether the first polynomial is a factor of the second polynomial we have to get the remainder as zero after the division

After division, the remainder is zero thus $x^2+3x+1$ is a factor of $3x^4 + 5x^3 - 7x^2 + 2x + 2$

Q2 (3) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$

Answer: The polynomial division is carried out as follows

The remainder is not zero, there for the first polynomial is not a factor of the second polynomial

Q3 Obtain all other zeroes of , if two of its zeroes are $\sqrt {\frac5{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}$

Answer:

Two of the zeroes of the given polynomial are $\sqrt {\frac5{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}$ .

Therefore two of the factors of the given polynomial are $x-\sqrt{\frac{5}{3}}$ and $x+\sqrt{\frac{5}{3}}$

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

$x^{2}-\frac{5}{3}$ is a factor of the given polynomial.

To find the other factors we divide the given polynomial with $3\times (x^{2}-\frac{5}{3})=3x^{2}-5$

The quotient we have obtained after performing the division is $x^{2}+2x+1$

\\x^{2}+2x+1\\ =x^{2}+x+x+1\\ =x(x+1)+(x+1)\\ =(x+1)^{2}

(x+1) ²= 0

x = -1

The other two zeroes of the given polynomial are -1.

Q4 On dividing by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

Answer: Quotient = x-2

remainder =-2x+4

\\dividend=divisor\timesquotient+remainder\\x^3-3x^2+x+2=g(x)(x-2)+(-2x+4)\\x^3-3x^2+x+2-(-2x+4)=g(x)(x-2)\\x^3-3x^2+3x-2=g(x)(x-2)\\

Carrying out the polynomial division as follows

Q5 (1) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm
and
(i) deg p(x) = deg q(x)

Answer: deg p(x) will be equal to the degree of q(x) if the divisor is a constant. For example

\\p(x)=2x^2-2x+8\\q(x)=x^2-x+4\\g(x)=2\\r(x)=0

Q5 (2) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg q(x) = deg r(x)

Answer: Example for a polynomial with deg q(x) = deg r(x) is given below

\\p(x)=x^3+x^2+x+1\g(x)=x^2-1\\q(x)=x+1\\r(x)=2x+2

Q 5 (3) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg r(x) = 0

Answer:

example for the polynomial which satisfies the division algorithm with r(x)=0 is given below

\\p(x)=x^3+3x^2+3x+5\\q(x)=x^2+2x+1\\g(x)=x+1\\r(x)=4

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Excercise: 2.4

Q1 (1) Verify that the numbers given alongside the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:
$2x^3 + x^2 - 5x +2 ; \frac{1}{2} , 1 , -2$

Answer:

p(x) = 2x ³+ x ²-5x + 2

\\p(\frac{1}{2})=2\times \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{2}-5\times \frac{1}{2}+2\\ p(\frac{1}{2})=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\\ p(\frac{1}{2})=0

p(1) = 2 x 1 ³+ 1 ²– 5 x 1 + 2

p(1) =2 + 1 – 5 + 2

p(1) = 0

p(-2) = 2 x (-2) ³+ (-2) ²– 5 x (-2) +2

p(-2) = -16 + 4 + 10 + 2

p(-2) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax ³+ bx ²+ cx + d, we have

a = 2, b = 1, c = -5, d = 2

The roots are $\alpha ,\beta \ and\ \gamma$

\\\alpha=\frac{1}{2}\\ \beta =1\\ \gamma =-2

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

Verified

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

Verified

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

Verified

Q1 (2) Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

Answer:

p(x) = x ³– 4x ²+ 5x – 2

p(2) = 2 ³– 4 x 2 ²+ 5 x 2 – 2

p(2) = 8 – 16 + 10 – 2

p(-2) = 0

p(1) = 1 ³– 4 x 1 ²+ 5 x 1 – 2

p(1) = 1 – 4 + 5 – 2

p(1) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax ³+ bx ²+ cx + d, we have

a = 1, b = -4, c = 5, d = -2

The roots are $\alpha ,\beta \ and\ \gamma$

\\\alpha+\beta +\gamma \\ =2+1+1\\ =4\\ =-\frac{-4}{1}\\=-\frac{b}{a}

Verified

\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =2\times 1+1\times 1+1\times 2\\ =5\\ =\frac{5}{1}\\ =\frac{c}{a}

Verified

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

Verified

Q2 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.