NCERT Solutions For Class 10 Maths Chapter 2 Polynomials

NCERT Solutions For Class 10 Maths Chapter 2 Polynomials Ex 2

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are prepared by specialised experienced teacher of mathematic. Maths are most important subject of board and with the help of this chapter-wise NCERT solution and little practices you can get very good marks in your respective board exam. It also help to build a foundation for topics that will be covered in the upcoming 11th and 12th.

Class 10 Maths Chapter 2 Polynomials contain topic degree of a polynomial, Zeros of Polynomial, Relationship between Zeros and Coefficients of a Polynomial, Division Algorithm for Polynomials and value of a polynomial at a given point, and the division algorithm. You can also check NCERT Solutions for Class 10 Maths Chapter 1.

Important Formulas:

Linear polynomial, ax + b

• Zero of a polynomial ax + b = -b/a

Quadratic Polynomial, ax² + bx + c

• Sum of zeros, α + β = -b/a
• Product of zeros, αβ = c/a

Cubic Polynomial, ax³ + bx² + cx + d

• α + β + γ = -b/a
• αβ + βγ + γα = c/a
• α β γ = -d/a

Exercise 2.1

1. 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.

Solution:

Exercise 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4

Solution:

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Solution:

Exercise 2.3

1. 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

Solution :

Given,

Dividend = p(x) = x³-3x²+5x–3

Divisor = g(x) = x²– 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

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

Solution:

Given,

Dividend = p(x) = x⁴ – 3x² + 4x +5

Divisor = g(x) = x² +1-x

Therefore, upon division we get,

Quotient = x² + x–3

Remainder = 8

(iii) p(x) =x⁴–5x+6, g(x) = 2–x²

Solution:

Given,

Dividend = p(x) =x⁴ – 5x + 6 = x⁴ +0x²–5x+6

Divisor = g(x) = 2–x² = –x²+2

Therefore, upon division we get,

Quotient = -x²-2

Remainder = -5x + 10

Question 2.
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) x² + 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:

Question 3. Obtain all other zeroes of 3x⁴+6x³-2x²-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Solution:

Question 4. 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).

Solution:

Given,

Question 5. 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)

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

(iii) deg r(x) = 0

Solution:

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Exercise 2.4

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:
(i) 2x³ + x² – 5x + 2;  14, 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1

Solution:

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.

Solution:

Let us consider the cubic polynomial is ax³+bx²+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

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

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

Solution:

We are given with the polynomial here,

p(x) = x³-3x²+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴px³+qx²+rx+s = x³-3x²+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b²

-1/1 = 1-b²

b² = 1+1 = 2

b = ±√2

Hence,1-√2, 1 ,1+√2 are the zeroes of x³-3x²+x+1.

Question 4. If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2 ±√3, find other zeroes.

Solution:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x⁴-6x³-26x²+138x-35

Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the above equation we get,

x²-4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x⁴-6x³-26x²+138x-35 = (x²-4x+1)(x² –2x−35)

Now, on further factorizing (x²–2x−35) we get,

x²–(7−5)x −35 = x²– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

Question 5.
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.
Solution: