2. Polynomials Mathematics class 10 exercise Exercise 2.4
2. Polynomials Mathematics class 10 exercise Exercise 2.4 ncert book solution in english-medium
NCERT Books Subjects for class 10th Hindi Medium
Exercise 2.1
Chapter 2. Polynomials
Exercise 2.1
Q1. 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(i):
Number of zeroes of p(x) = 0;
Solution(ii):
Number of zeroes of p(x) = 1;
Solution (iii):
Number of zeroes of p(x) = 3;
Solution (iv):
Number of zeroes of p(x) = 2;
Solution (v):
Number of zeroes of p(x) = 4;
Solution (vi):
Number of zeroes of p(x) = 3;
Exercise 2.2
Chapter 2. Polynomials
Exercise 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
Solution:
Using Middle term splitting method:
x2 – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0 x + 2 = 0
⇒ x = 4, x = – 2
Zeroes; α = 4 β = – 2
Verifying the relationship between the zeroes and the coefficients.
a = 1, b = – 2, and c = – 8
(ii) 4s2 – 4s + 1
Solution:
4s2– 4s + 1 = 0
⇒ 4s2 – 2s – 2s + 1 = 0
⇒ 2s (2s – 1) –1(2s – 1) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1= 0, 2s – 1= 0
⇒ 2s = 1, 2s = 1
Coefficients
a = 4, b = – 4, c = 1
Relationship between zeroes and coefficients
L.H.S = R.H.S
L.H.S = R.H.S
Hence verified the relationship between coefficients and the zeroes in both cases.
(iii) 6x2 – 3 – 7x
Solution:
⇒ 6x2 – 3 – 7x = 0
⇒ 6x2 – 7x – 3 = 0 (After rearranging the equation)
⇒ 6x2 – 9x + 2x – 3 = 0
⇒ 3x (2x – 3) +1 (2x – 3) = 0
⇒ (2x – 3) (3x + 1) = 0
⇒ 2x – 3 = 0, 3x + 1 = 0
⇒ 2x = 3, 3x = - 1
Hence verified the relationship between coefficients and the zeroes in both cases.
(iv) 4u2 + 8u
Solution: 4u(u + 2) = 0
4u = 0, u + 2 = 0
u = 0, u = – 2
Zeroes: a = 0, b = – 2
Coefficients:
a = 4, b = 8, c = 0
Verifying relationship between zeroes and coefficients
⇒ 0 = 0
⇒ L.H.S = R.H.S
Hence verified, the relationship between coefficients and zeroes in both cases.
(v) t2 – 15
Solution:
t2 – 15 = 0
t2 = 15
t = ±√15
t = √15, t = – √15
Zeroes: a = √15, b = – √15
Coefficients a = 1, b = 0, c = – 15
Verifying relationship between zeroes and Coefficients
Hence verified, the relationship between coefficients and zeroes in both cases.
(vi) 3x2 – x – 4
Solution:
3x2 – x – 4 = 0
⇒ 3x2 + 3x – 4x – 4 = 0
⇒ 3x (x + 1) – 4 (x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0, 3x – 4 = 0
⇒ x = –1 and 3x = 4
L.H.S= R.H.S
Hence verified, the relationship between coefficients and zeroes in both cases.
Q2. . Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
Hence, required quadratic polynomial is 4x2 – x – 4
Hence, required quadratic polynomial is 3x2 –3 x + 1
Hence, required quadratic polynomial is x2 + √5
Hence, required quadratic polynomial is x2 – x + 1
Hence, required quadratic polynomial is 4X2 + x + 1
Hence, required quadratic polynomial is x2 – 4 + 1
Exercise 2.3
Exercise 2.3 class 10 maths chapter 2. Polynomials
Q1. 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: (i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Quotients q(x) = x – 3 and Remainder = 7x – 9
Solution: (ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Quotients q(x) = x2 + x – 3 and Remainder = 8
Solution: (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Quotients q(x) = – x2 – 2 and 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:
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution: (i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
Hence Remainder r(x) is 0
Therefore, t2 – 3 is the factor of 2t4 + 3t3 – 2t2 – 9t – 12
Solution: (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
Hence Remainder r(x) is 0
Therefore, x2 + 3x + 1 is the factor of 3x4 + 5x3 – 7x2 + 2x + 2
Solution: (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Hence Remainder r(x) = 2
Therefore, x3 – 3x + 1, is not a factor of x5 – 4x3 + x2 + 3x + 1
Q3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are
Solution:
Given that : p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Or 3x2 - 5 = 0
Therefore, 3x2 - 5 is the factor of p(x)
Now Dividing 3x4 + 6x3 - 2x2 - 10x - 5 by 3x2 - 5
Therefore, p(x) = (3x2 – 5) (x2 + 2x + 1)
Now, factorizing and getting zeroes x2 + 2x + 1 -
= x2 + x + x + 1 = 0
= x(x + 1) + 1(x + 1) = 0
= (x + 1) (x + 1) = 0
Or x + 1 = 0, x + 1 = 0
Or x = – 1, x = – 1
Therefore, two zeroes are – 1 and – 1.
Q4. 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 that: Dividend p(x) = x3 – 3x2 + x + 2
Quotient q(x) = x – 2,
Remainder r(x) = – 2x + 4
Divisor g(x) =?
Dividend = divisor × quotient + remainder
p(x) = g(x) × q(x) + r(x)
x3 – 3x2 + x + 2 = g(x) (x – 2) + (– 2x + 4)
x3 – 3x2 + x + 2 + 2x – 4 = g(x) (x – 2)
g(x) (x – 2) = x3 – 3x2 + 3x – 2
Dividing x3 – 3x2 + 3x – 2 by x - 2 we obtain g(x)-
Therefore, Divisor g(x) = x2 – x + 1
Q5. 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:
Using Euclid’s Division algorithm:
p(x) = g(x) × q(x) + r(x) where q(x) – 0
(i) deg p(x) = deg q(x)
The deg of dividend p(x) and quotient q(x) can be equal when deg of divisor is 0 or any number.
Example : Let p(x) = 2x2 - 6x + 3
And let g(x) = 2
On dividing
p(x) = 2x2 - 6x + 2 + 1
= 2(x2 - 3x + 1) + 1
Now comparing 2(x2 - 3x + 1) + 1 by p(x) = g(x) × q(x) + r(x) we get:
So, q(x) = x2 - 3x + 1and r(x) = 1
By which we obtain deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
Solution: This situation comes when deg p(x) and deg g(x) is equal-
Let p(x) = 2x2 + 6x + 7 and g(x) = x2 + 3x + 2
On dividing: q(x) = 2 and r(x) = 3
Therefore, deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
r(x) = 0 is obtained when p(x) is completely divisible by g(x):
Let p(x) = x2 – 1 and g(x) = x + 1
On dividing we obtain:
q(x) = x – 1 and r(x) = 0
Exercise 2.4
Exercise 2.4 Class 10 maths chapter 2. Polynomials
Q1. 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; α = ½, β = 1 and γ = – 2;
(ii) x3 – 4x2 + 5x – 2; α = 2, β = 1, and γ = 1
Solution: (i) 2x3 + x2 - 5x + 2; α = ½, β = 1 and γ = – 2;
Given: p(x) = 2x3 + x2 - 5x + 2
p(1) = 2(1)3 + (1)2 - 5(1) + 2
= 2 + 1 - 5 + 2
= 5 - 5 = 0
p(x) = 0
Therefore, 1 is the zero of p(x)
अब, p(-2) = 2(-2)3 + (-2)2 - 5(-2) + 2
= -16 + 4 + 10 + 2
= 16 - 16 = 0
p(x) = 0
Therefore, -2 is the zero of p(x)
So, α = ½, β = 1 and γ = – 2 are zeroes.
And coefficients a = 2, b = 1, c = - 5 and d = 2
Verification of relation between zeroes and coefficients:
Verified by equation (1) (2) and (3)
Solution: (ii) x3 – 4x2 + 5x – 2; α = 2, β = 1, and γ = 1
Given that: p(x) = x3 – 4x2 + 5x – 2
p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 4.4 + 10 – 2
= 8 – 16 + 10 – 2
= 18 – 18 = 0
Hence, α = 2 is the zero of p(x)
Now for β = 1
p(x) = x3 – 4x2 + 5x – 2
p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4.1 + 5 – 2
= 1 – 4 + 5 – 2
= 6 – 6 = 0
Hence, β = 1 is the zero of p(x)
Now, for γ = 1
p(x) = x3 – 4x2 + 5x – 2
p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4.1 + 5 – 2
= 1 – 4 + 5 – 2
= 6 – 6 = 0
Hence, γ = 1 is the zero of p(x)
So, α = 2, β = 1 and γ = 1 are zeroes.
And coefficients a = 1, b = – 4, c = 5 and d = – 2
Verification of relation between zeroes and coefficients:
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