NCERT Solutions for Class 10 Maths Unit 2

NCERT Solutions for Class 10 Maths Unit 2

Polynomials Class 10

Exercise 2.1

Q1 :

The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the

number of zeroes ofp(x), in each case.

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 

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:
(i) x2 – 2x – 8 = x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
Either x + 2 = 0 or x – 4 = 0
⇒  x = -2 or x = 4
Hence, zeroes of this polynomial are -2 and 4.
Verification:
Sum of the zeroes = (-2) + (4) = 2
=  = 
Product of zeroes = (-2) (4) = -8 =  =  
Hence verified.

(ii) 4s2 – 4s + 1 = (2s – l)2 = (2s – l)(2s – 1)
Either 2s – 1 = 0 or 2s – 1 = 0
i.e. , s =   , 
Hence, the two Zeroes are  and 
Verification:

Hence verified.

(iii) 6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x (2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
Either 2x – 3 = 0 or 3x+1 = 0

Hence verified.

(iv) 4u2 + 8u  ⇒  4u(u + 2)
Either 4u = 0 or u + 2 = 0
⇒ u = 0 or u = -2
Hence, the two zeroes are 0 and -2.
Verification:
Sum of the zeroes = 0 + (-2) = -2
=  = 
Product of zeroes = 0 x (-2) = 0 =  
Hence verified.

(v) t2 – 15 = t2 – ()2
= (t + () (t- ()
Either t + ( = 0 or t – ( = 0
⇒ t = -( or t = (
Hence, the two zeroes are -( and + .
Verification:
Sum of the zeroes = -( +  = 0
= 
Product of zeroes = – x  = -15
= 
Hence verified.

(vi) 3x2 – x – 4 = 3x2 – 4x + 3x – 4
=  x(3x – 4) + l(3x 4)
= (x + 1) (3x – 4)
Either x + 1 = 0 or 3x-4 = 0
⇒  x = -1 or x =  
Verification:
Sum of the zeroes = -1 +  =   =  
Product of zeroes = -1 x  =   =  
Hence verified.

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

Solution:
(i) Let the zeroes of polynomial be α and β.
Then, α + β =  and αβ = -1
∴ Required polynomial is given by,
x2 – (α + β)x + αβ = x2 – x + (-1)
= x2 – x – 1
= 4x2 – x – 4

(ii) Let the zeroes of polynomial be α and β.
Then, α + β= √2 and αβ = 
∴  Required polynomial is:
x2 – (α + β)x + αβ = x2 – √2x + 
= 3x2 – 3√2x + 1

(iii) Let the zeroes of the polynomial be α and β.
Then, α + β = 0 and αβ = √5
∴ Required polynomial
= x2 – (α + β)x + αβ
= x2– 0 x x + √5 = x2 + √5

(iv) Let the zeroes of the polynomial be α and β.
Then, α + β = 1 and αβ = 1.
∴  Required polynomial
= x2 – (α + β)x + αβ
= x2 – x + 1

(v) Let the zeroes of the polynomial be α and β.
Then, α + β = –  and αβ = 
∴ Required polynomial
= x2 – (α + β)x + αβ
= x2 – (-  ) + 
=  4x2 + x + 1 = 0

(vi) Let the zeroes of the polynomial be α and β.
Then, α + β = 4 and αβ = 1.
∴ Required polynomial = x2 -(α + β)x + αβ
= x2 – 4x + 1

                   Exercise 2.3

Question 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:

 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:

∴  Remainder is 0, therefore, t2 – 3 is a factor of polynomial 2t4 + 3t3 – 2t2 -9t – 12.

∴ Remainder is 0, therefore, x2 + 3x + 1 is a factor of polynomial 3x4 + 5x3 – 7x2 + 2x + 2.

∴ Remainder = 2 ≠ 0, therefore, x3 – 3x + 1 is not a factor of polynomial x5 – 4x3 + x2+ 3x + 1.

 Question 3.
 Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are  and  and –
Solution:

=  x (3x2– 5).Since both  and(3x2– 5)are the factors, therefore 3x2 – 5 is a factor of the given polynomial.
Now, we divide the given polynomial by 3x2 – 5.

Hence, the other zeroes of the given polynomial are -1 and –1.

 Question 4.
On dividing x3 – 3x2 + x + 2bya polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
x3 – 3x2 + x + 2  = g(x) x (x – 2) + (-2x + 4) [By division algorithm]

 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:
(i) p(x) = 2X2 + 2x + 8,
q(x) = x2 + x + 4,
g(x) = 2 and r(x) = 0

(ii) p(x) = x3 + x2 + x + 1,
q(x) = x + 1,
g(x) = x2 – 1 and r(x) = 2x + 2

(iii) p(x) = x3 – x2 + 2x + 3,
g(x) = x2 + 2,
q(x) = x – 1 and r(x) = 5

                                      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;  , 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
∴  p(x) = 2x3 + x2 – 5x + 2


(ii) Compearing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x3 – 4x2 + 5x – 2
⇒  p(2) = (2)3 – 4(2)2 + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)3 – 4(1)2 + 5 x 1-2
= 1 – 4 + 1 – 2
= 6-6 = 0
Hence, 2, 1 and 1 are the zeroes of x3 – 4x2 + 5x – 2.
Hence verified.
Now we take α = 2, β = 1 and γ = 1.
α + β + γ = 2 + 1 + 1 =  = 
αβ + βγ + γα = 2 + 1 + 2 =  = 
αβγ = 2 x 1 x 1  =  = .
Hence verified.

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 α , β and  γ be the zeroes of the required polynomial.
Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14.
∴ Cubic polynomial
= x3 – (α + β + γ)x2 + (αβ + βγ + γα)x – αβγ
= x3 – 2x2 – 1x + 14
Hence, the required 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:
Let α , β and  γ be the zeroes of polynomial x3 – 3x2 + x + 1.
Then α =  a-b, β = a and γ = a + b.
∴ Sum of zeroes = α + β + γ
⇒   3 = (a – b) + a + (a + b)
⇒  (a – b) + a + (a + b) = 3
⇒  a-b + a + a + b = 3
⇒       3a = 3
⇒ a =   = 1 …(i)
Product of zeroes = αβγ
⇒ -1 = (a – b) a (a + b)
⇒ (a – b) a (a + b) = -1
⇒   (a2 – b2)a = -1
⇒  a3 – ab2 = -1   … (ii)
Putting the value of a from equation (i) in equation (ii), we get:
(1)3-(1)b2 = -1
⇒ 1 – b2 = -1
⇒ – b2 = -1 – 1
⇒  b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2.

 Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, finnd other zeroes.
Solution:
Since two zeroes are 2 + √3 and 2 – √3,
∴  [x-(2 + √3)] [x- (2 – √3)]
= (x-2- √3)(x-2 + √3)
= (x-2)2– (√3)2
x2 – 4x + 1 is a factor of the given polynomial.
Now, we divide the given polynomial by x2 – 4x + 1.

So, x4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) (x2 – 7x + 5x – 35)
= (x2-4x + 1) [x(x- 7) + 5 (x-7)]
= (x2 – 4x + 1) (x – 7) (x + 5)
Hence, the other zeroes of the given polynomial are 7 and -5.

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:
We have
p(x) = x4 – 6x3 + 16x2 – 25x + 10
Remainder = x + a   … (i)
Now, we divide the given polynomial 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.

Using equation (i), we get:
(-9 + 2k)x + 10-8 k + k2 = x + a
On comparing the like coefficients, we have:
-9 + 2k = 1
⇒ 2k = 10
⇒ k =  = 5  ….(ii)
and 10 -8k + k2– a   ….(iii)
Substituting the value of k = 5, we get:
10 – 8(5) + (5)2 = a
⇒   10 – 40 + 25 = a
⇒  35 – 40 =   a
⇒   a =   -5
Hence, k = 5 and a = -5.