CBSE Class 10 NCERT Solutions for Maths Chapter 2, Polynomials

CBSE Class 10 NCERT Solutions for Maths Chapter 2, Polynomials to check each and every concept and their solutions are also provided in this website. In this solutions are useful for the Central Board Of Secondary Education Students to prepare our examinations well. So, we can can check once in this article and go our preparation. You can download the CBSE Class 10 NCERT Solutions for Maths Chapter 2, Polynomials PDF is also available in this article.

Download the CBSE Class 10 NCERT Solutions for Maths chapter 2, Polynomials

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CBSE Class 10 Maths NCERT Problems & Solutions

Question 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)

(v)

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.

PAGE NO 33:

Question 1:

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

 

 

ANSWER:

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

Therefore, the zeroes of are 4 and −2.

Sum of zeroes = 

Product of zeroes 

The value of 4s2 − 4s + 1 is zero when 2s − 1 = 0, i.e.,

Therefore, the zeroes of 4s2 − 4s + 1 areand.

Sum of zeroes = 

Product of zeroes 

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

Therefore, the zeroes of 6x2 − 3 − 7x are.

Sum of zeroes = 

Product of zeroes = 

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 = 

Product of zeroes = 

The value of t2 − 15 is zero when  or , i.e., when 

Therefore, the zeroes of t2 − 15 are  and.

Sum of zeroes =

Product of zeroes = 

The value of 3x2 − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0, i.e., when  or x = −1

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

Sum of zeroes = 

Product of zeroes 

PAGE NO 33:

Question 2:

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

  

  

ANSWER:

Let the polynomial be , and its zeroes be and .

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

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is 3x2 − x + 1.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be .

Therefore, the quadratic polynomial is.

PAGE NO 36:

Question 1:

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:

ANSWER:

(i) 

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

Comparing the given polynomial with , we obtain a = 2, b = 1, c = −5, d = 2

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

(ii) 

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

Comparing the given polynomial with , we obtain a = 1, b = −4, c = 5, d = −2.

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

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5 

Multiplication of zeroes = 2 × 1 × 1 = 2 

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

PAGE NO 36:

Question 2:

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 and the zeroes be .

It is given that

If a = 1, then b = −2, c = −7, d = 14

Hence, the polynomial is .

PAGE NO 36:

Question 3:

Obtain all other zeroes of , if two of its zeroes are .

ANSWER:

Since the two zeroes are ,

is a factor of .

Therefore, we divide the given polynomial by .

We factorize 

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

x = −1

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

Hence, the zeroes of the given polynomial are, −1 and −1.

PAGE NO 36:

Question 4:

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

ANSWER:

g(x) = ? (Divisor)

Quotient = (x − 2)

Remainder = (− 2x + 4)

Dividend = Divisor × Quotient + Remainder

g(x) is the quotient when we divide by

PAGE NO 36:

Question 5:

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 by 2.

Here, p(x) = 

g(x) = 2

q(x) =  and 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)

= 2()

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 = (x) × 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 = (x) × x + 1

x3 + 1 = x3 + 1

Thus, the division algorithm is satisfied.

PAGE NO 37:

Question 3:

If the zeroes of polynomial  are, find a and b.

ANSWER:

Zeroes are a − ba + a + b

Comparing the given polynomial with , we obtain

p = 1, q = −3, r = 1, t = 1

The zeroes are .

Hence, a = 1 and b =  or .

PAGE NO 37:

Question 4:

]It two zeroes of the polynomial  are, find other zeroes.

ANSWER:

Given that 2 + and 2­­ are zeroes of the given polynomial.

Therefore, 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 the quotient by dividing  by x2 ­− 4x + 1.

Clearly, = 

It can be observed that is also a factor of the given polynomial.

And 

Therefore, the value of the polynomial is also zero when or 

Or x = 7 or −5

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

PAGE NO 37:

Question 5:

If the polynomial  is divided by another polynomial, the remainder comes out to be x + a, find k and a.

ANSWER:

By division algorithm,

Dividend = Divisor × Quotient + Remainder

Dividend − Remainder = Divisor × Quotient

 will be perfectly divisible by .

Let us divide  by 

It can be observed that will be 0.

Therefore, = 0 and = 0

For = 0,

2 k =10

And thus, k = 5

For = 0

10 − a − 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0

− 5 − a = 0

Therefore, a = −5

Hence, k = 5 and a = −5

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