PSEB Solutions for Class 10 Maths Chapter 2 Polynomials
PSEB 10th Class Maths Solutions Chapter 2 Polynomials
Ex 2.1
Question 1.
The graphs of y = p (x) are given in Fig. below, for some polynomials p(x). Find the number of zeroes of p (x), in each case.
Solution:
The graphs of y = p (x) are given in figure above, for some polynomials p (x). The number of zeroes of p (x) in each case are given below:
(i) From the graph, it is clear that it does not meet x-axis at any point.
Therefore, it has NIL; no. of zeroes.
(ii) From the graph, it is clear that it meets x- axis at only one point.
Therefore, it has only one no. of zeroes.
(iii) From the graph, it is clear that it meets x-axis at three points.
Therefore, it has three no. of zeroes.
(iv) From the graph, it is clear that it meets x- axis at two points.
Therefore, it has two no. of zeroes.
(v) From the graph, it is clear that it meets x- axis at four points.
Therefore, it has four no. of zeroes.
(vi) From the graph, it is clear that it meets pr-axis at three points.
Therefore, it has three no. of zeroes.
Ex 2.2
Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the 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) Given Quadratic polynomial,
x2 – 2x – 8
∵ [S = -2, P = -8]
= x2 – 4x + 2x – 8
= x (x – 4) + 2 (x – 4)
= (x – 4) (x + 2)
The value of x2 – 2x – 8 is zero,
iff (x – 4) = 0 or (x + 2) = 0
iff x = 4 or x = – 2
Therefore, zeroes of x2 – 2x – 8 are – 2 and 4.
Now, Sum of zeroes = (- 2) + (4) = 2
(iii) Given that, sum of zeroes and products of zeroes of given quadratic polynomial are 0 and √5 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 0
and αβ = Product of zeroes = √5
Now, ax2 + bx + c = k(x – α) (x – β)
where k is any constant
= k [x2 – (a + (α+ β)x + αβ)
= k[x2 – 0x + √5]
= k[x2 + √5]
for different values of k, we get different quadratic polynomial.
(iv) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 1 and 1 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 1 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β)
where k is any constant.
= k [x2 – (α + β)x + αβ]
= k [x2 – 0x + √5]
= it [x2 – x + √5]
for different values of k, we get different quadratic polynomial.
(vi) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 4 and 1 respectively. Let the quadratic polynomial be
ax2 + bx + c and its zeroes be α and β
α + β = sum of zeroes = 4 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β) where k is any constant
= k [x2 – (α + β) x + αβ]
= k [x2 – 4x + 1]
For different values of k, we get different quadratic polynomials.
Ex 2.3
Question 1.
Apply the division algorithm to find the quotient and remainder on dividing p (x) by g (x) as given below:
(i) p (x) = x3 – 3x2 + 5x – 3, g (x) = x2 – 2 (Pb. 2018 Set I, II, III)
(ii) p (x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p (x) = x4 – 5x + 6, g (x) = 2 – x2 [Pb. 2017 Set-B]
Solution:
(i) Given that p (x) = x3 – 3x2 + 5x – 3 and g (x) = x2 – 2,
By division algorithm,
x3 – 3x2 + 5x – 3 = (x – 3) (x2 – 2) + (7x – 9)
Hence, quotient = x – 3 and remainder = 7x – 9
(ii) Given that p (x) = x4 – 3x2 + 4x + 5
or p (x) = x4 + 0x3 – 3x2 + 4x + 5
and g (x) = x2 + 1 – x
or g (x) = x2 – x + 1
By Division Algorithm,
x4 – 3x2 + 4x + 5 = (x2 + x – 3) (x2 – x + 1) + 8
Hence, Quotient = x2 + x – 3 and remainder = 8
(iii) Given that p (x) = x4 – 5x + 6
or p (x) = x4 + 0x3 + 0x2 – 5x + 6
and g (x) = 2 – x2
or g (x) = – x2 + 2
By division algorithtm,
x4 – 5x + 6 = (- x2 – 2) (- x2 + 2) + (- 5x + 10)
Hence, quotient = -x2 – 2.
remainder = – 5x + 10
Question 2.
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
(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:
∵ remainder is zero
∵ By division algorithm,
t2 – 3 is factor of 2t4 + 3t3 – 2t2 – 9t – 12
(ii)
∵ remainder is zero
∴ By division algorithm, x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2
(iii)
∵ remainder is not zero.
∴ By division algorithm, x3 – 3x + 1 is not a factor of x5 – 4x3 + x2 + 3x + 1.
Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the dhriskm algorithm and
(i) deg p (x) = deg q (x)
(ii) deg r (x) = 0
(iii) deg q (x) = deg r (x)
Solution:
(i) Let p(x) = 5x2 – 5x +10; g(x) = 5 q(x) = x2 – x + 2; r(x) = 0
∴ By division algorithm,
5x2 – 5x + 10 = 5(x2 – x + 2) + 0
or p(x) = g(x) q(x) + r(x)
Also, deg p(x) = deg q(x) = 2
(ii) Let p(x) = 7x3 – 42x + 53
g(x) = x3 – 6x + 7
q(x) = 7; r(x) = 4
∴ By division algorithm,
7x3 -42x + 53 = 7(x3 – 6x+ 7)+ 4
or p(x) = q(x) g(x) + r(x)
Also, deg q(x) = 0 = deg r(x)
(iii) Let p(x) = 4x3 + x2 + 3x + 6;
g(x) = x2 + 3x + 1;
q(x) = 4x – 11;
r(x) = 32x + 17
∴ By division algorithm,
4x3 + x2 + 3x + 6 = (4x – 11) (x2 + 3x + 1) + (32x + 17)
or p(x) = q(x) . g(x) + r(x)
Also, deg q(x) = deg r(x)
Ex 2.4
Question 1.
Verify that the number given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
(ii) Let p (x) = x3 – 4x2 + 5x – 2.
Compare it with ax3 + bx2 + cx + d
∴ a = 1, b = – 4, c = 5, d = – 2
Now p(2) = (2)3 – 4 (2)2 + 5 (2) – 2
= 8 – 16 + 10 – 2
= 18 – 18 = 0 2 is zero of p (2).
and p( 1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2 = 6 – 6 = 0
From above discussion it is clear that 2, 1, 1 are the zeroes of given polynomial. Ans.
Let these zeroes are
α = 2, β = 1, γ = 1
Now, α + β + γ = 2 + 1 + 1 = 4
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:
The general expression of cubic polynomial are
ax3 + bx2 + cx + d.
Let α, β, γ be its zeroes
α + β + γ = Sum of zeroes = 2
αβ + βγ + γα = Sum of product of zeroes = – 7
αβγ = Product of zeroes = – 14
∴ ax3 + bx2 + cx + d
= k [(x – α) (x – β) (x – γ)] where k is any constant.
= k [x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ]
= k [x3 – 2x2 – 7x + 14] [Using (1)]
For different values of k, we get different cubic polynomial.
Question 3.
If the zeroes of the polynomial x3 – 3x2 + x +1 are a – b, a, a + b, find a and b.
Solution:
Let p (x) = x3 – 3x2 + x + 1 and its zeroes are a – b, a, a + b.
a – b is zero of p (x) ………(Given)
∴ p (a – b) = 0
or (a – b)3 – 3 (a – b)2 + (a – b) + 1 = 0 or
[a3 – b3 – 3a2b + 3ab2] – 3 [a2 + b2 – 2ab] + a – b + 1 = 0
or a3 – b3 – 3a2b + 3ab2 – 3a2 – 3b2
+ 6ab + a – b + 1 = 0 ….(1)
and a is zero of p (x) …………(Given)
∴ p (a) = 0
or a3 – 3a3 + a + 1 = 0 ………….(2)
Also, a + b is zero of p (x) …(Given)
∴ p (a + b) = 0
or (a + b)3 – 3 (a + b)2 + (a + b) + 1 = 0
or (a3 + b3 + 3a2b + 3ab2) – 3 (a2 + b3 + 2ab) + a + b – 1 = 0 or
a3 + b3 + 3a2b + 3ab2 – 3a2 – 3b3 – 6ab + a + b + 1 = 0 …………….(3)
Adding (1) and (3), we get :
2a3 + 6ab2 – 6a2 – 6b2 + 2a + 2 = 0 or
a3+ 3ab2 – 3a2 – 3b2 + a + 1 = 0 or
(a3 – 3a3 + a + 1) + (3ab2 – 3b2) = 0 or
0 + 3b2 (a – 1) = 0 [Using (2)]
or a – 1 = 0
or a = 1 ………(4)
From (3) and (4), we get:
(1)3 + b3 + 3(1)2b + 3(1 )b2 – 3(1)2 – 3b2 – 6 (1) b + 1 + b + 1 = 0 or
1 + b3+ 3b + 3b2 – 3 – 3b2 – 6b + b + 2 = 0
or b3 – 2b = 0 or
b (b2 – 2) = 0 or
b2 – 2 = 0 or
b2 = 2
or b = ±√2
Hence a = 1, b = ±√2
Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, find other zeroes.
Solution:
Given that two zeroes are (2 + √3) and (2 – √3)
∴ [x – (2 + √3)] [x – (2 – √3)] are factors of given polynomial.
Now, [x – (2 + √3)] [x – (2 – √3)]
= x2 – [2 – √3 + 2 + √3]x + [(2 + √3) + (2 – √3)]
= x2 – 4x + [(2)2 – (√3)2]
= x2 – 4x + 1
∴ (x2 – 4x + 1) is factor of given polynomial. Now, apply division algorithm to given polynomial and (x2 – 4x + 1)
∴ x4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) [x2 + 5x – 7x – 35]
S = – 2, P = – 35
= (x2 – 4x + 1) [x (x + 5) – 7 (x + 5)]
= (x2 – 4x + 1) (x + 5).(x – 7)
Now, other zeroes of polynomials are given by :
x + 5 = 0 Or x – 7 = 0
x = – 5 Or x = 7
∴ the zeroes of the given fourth degree polynomial are :
2 + √3, 2 – √3, -5, 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:
Given that, polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k then remainder comes out to be x + a
So, first of all we divide,
x4 – 6x3 + 16x2 – 25x + 10 with x2 – 2x + k and find quotient and remainder.
∴ by division algorithm for Polynomials x4 – 6x3 + 16x2 – 25x +10
= (x2 – 2x + k) [x2 – 4x + (8 – k) + [(- 9 + 2k) x + (10 – 8k + k2]
Quotient = x2 – 4x + (8 – k) and
Remainder = (- 9 + 2k) x + (10 – 8k + k2)
But, remainder = x + a ….(Given)
∴ (- 9 + 2k) x + (10 -8k + k2) = x + a
Compare the like coefficients, we get :
-9 + 2k = 1 Or
2k = 1 + 9
2k = 10
k = 10/2 = 5 or
10 – 8k + k2 = a
Putting the value of k, we get
10 – 8 × 5 + (5)2 = a
10 – 40 + 25 = a
-5 = a
a = -5
Hence k = 5 and a = -5