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NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials


Free NCERT Solutions for Class 10 Maths Chapter 2 Polynomials solved by Expert Teachers as per NCERT (CBSE) Book guidelines and brought to you by Toppers Bulletin. These Polynomials Exercise Questions with Solutions for Class 10 Maths covers all questions of Chapter 2 Polynomials Class 10 and help you to revise complete Syllabus and Score More marks as per CBSE Board guidelines from the latest NCERT book for class 10 maths. You can read and download NCERT Book Solution to get a better understanding of all topics and concepts. Following topics and sub-topics in Chapter 2 Polynomials are covered.
2.1 Introduction,
2.2 Geometrical Meaning of the Zeroes of a Polynomial,
2.3 Relationship between Zeroes and Coefficients of a Polynomial,
2.4 Division Algorithm for Polynomials and
2.5 Summary

Polynomials NCERT Solutions – Class 10 Maths

Exercise 2.1 : Solutions of Questions on Page Number : 28


Q1 : 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.


Exercise 2.2 : Solutions of Questions on Page Number : 33


Q1 : 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 x + 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 2x – 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


Q2 : 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 beand.

Therefore, the quadratic polynomial is x2 – 1+ 1.


Exercise 2.3 : Solutions of Questions on Page Number : 36


Q1 : Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i)
(ii)
(iii)
Answer :


Quotient = x – 3
Remainder = 7x – 9


Quotient = x2 + x – 3
Remainder = 8


Quotient = – x2 – 2
Remainder = – 5x +10


Q2 : Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
Answer :

=

Since the remainder is 0,
Hence, is a factor of .


Since the remainder is 0,
Hence, is a factor of .


Since the remainder ,
Hence, is not a factor of


 


Q3 : 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.


Q4 : On dividing by a polynomial g(x), the quotient and remainder were x – 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


Q5 : 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 = (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.


Exercise 2.4 : Solutions of Questions on Page Number : 37


Q 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.


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.
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 .


Q3 : If the zeroes of polynomial are, find a and b.
Answer :
Zeroes are a – b, a + 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- .


Q4 : 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.


Q5 : 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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