Maths NCERT solutions for Class 10 Chapter 2 Polynomials

Table of Contents

Polynomials

Exercise 2.1

Q.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) 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.

Maths NCERT solutions for Class 10 Chapter 2 Polynomials

Exercise 2.2

Q.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8

(i) X2 – 2x – 8
= (x – 4) (x + 2)
The value of x2 – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x2

(ii) 4s2 – 4s + 1

(ii) 4s2 – 4s + 1 = (2s-1)2 The value of 4s2 – 4s + 1 is zero when 2s – 1 = 0, i.e., s = ½ Therefore, the zeroes of 4s2 – 4s + 1 are ½ and ½.

Sum of zeroes = ½ + ½ = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2
Product of zeroes = ½ × ½ = ¼ = Constant term/Coefficient of s2.

(iii) 6×2 – 3 – 7x

(iii) 6×2 – 3 – 7x
= 6×2 – 7x – 3
= (3x + 1) (2x – 3)
The value of 6×2 – 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e., x = -1/3 or x = 3/2
Therefore, the zeroes of 6×2 – 3 – 7x are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

(iv) 4u2 + 8u

(iv) 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 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.

(v) T2 – 15

(v) T2 – 15
= t2 – 0.t – 15
= (t – √15) (t + √15)
The value of t2 – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t2 – 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.

(vi) 3×2 – x – 4

(vi) 3×2 – x – 4
= (3x – 4) (x + 1)
The value of 3×2 – x – 4 is zero when 3x – 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3×2 – x – 4 are 4/3 and -1.
Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.

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

(i) 1/4 , -1 (ii) √2 , 1/3 (iii) 0, √5 (iv) 1,1 (v) -1/4 ,1/4 (vi) 4,1

(i) 1/4 , -1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4×2 – x -4.

(ii) √2 , 1/3
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3×2 -3√2x +1.

(iii) 0, √5
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.

(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 – x +1.

(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4×2 + x +1.

(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 – 4x +1.

Maths NCERT solutions for Class 10 Chapter 2 Polynomials

Exercise 2.3

Q.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 – 3×2 + 5x – 3, g(x) = x2 – 2 Quotient = x-3 and remainder 7x – 9

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

Quotient = x-3 and remainder 7x – 9

(ii) p(x) = x4 – 3×2 + 4x + 5, g(x) = x2 + 1 – x Quotient = x2 + x – 3 and remainder 8

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

Quotient = x2 + x – 3 and remainder 8

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2 Quotient = -x2 -2 and remainder -5x +10

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

Quotient = -x2 -2 and remainder -5x +10

Q. 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

t2 – 3 exactly divides 2t4 + 3t3 – 2t2 – 9t – 12 leaving no remainder. Hence, it is a factor of 2t4 + 3t3 – 2t2 – 9t – 12.

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

x2 + 3x + 1 exactly divides 3x4 + 5x3 – 7x2 + 2x + 2 leaving no remainder. Hence, it is factor of 3x4 + 5x3 – 7x2 + 2x + 2.

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

x3 – 3x + 1 didn’t divides exactly x5 – 4x3 + x2 + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x5 – 4x3 + x2 + 3x + 1.

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

p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3).

We factorize x2 + 2x + 1
= (x + 1)2
Therefore, its zero is given by x + 1 = 0
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 √(5/3) and – √(5/3), – 1 and – 1.

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

In the given question, Dividend = x3 – 3×2 + x + 2 Quotient = x – 2 Remainder = -2x + 4 Divisor = g(x) We know that, Dividend = Quotient × Divisor + Remainder ⇒ x3 – 3×2 + x + 2 = (x – 2) × g(x) + (-2x + 4)⇒ x3 – 3×2 + x + 2 – (-2x + 4) = (x – 2) × g(x) ⇒ x3 – 3×2 + 3x – 2 = (x – 2) × g(x) ⇒ g(x) = (x3 – 3×2 + 3x – 2)/(x – 2) ∴ g(x) = (x2 – x + 1)

Q.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)

(i) Let us assume the division of 6x2 + 2x + 2 by 2
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.

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

(ii) 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

(iii) Let us assume the division of x3+ 1 by 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.

Maths NCERT solutions for Class 10 Chapter 2 Polynomials