Contents
GSEB Solutions for Class 10 mathematics – Polynomials (English Medium)
Exercise-2.1
Question 1:
Identify the type of the following polynomials : (on base of power)
- p(x) = x2 – 5x + 6
- p(x) = x2 – x3 + x + q
- p(x) = 5x2 + 8x + 3
- p(x) = x3
Solution 1:
- Here p(x) = x2 – 5x + 6
Degree of this polynomial is 2. So it is a quadratic polynomial in x.
- Here p(x) = x2 – x3 + x + 1
∴ p(x) = -x3 + x2 + x + 1
Degree of this polynomial is 3. So it is a cubic polynomial in x.
- Here p(x) = 5x2 + 8x + 3
Degree of this polynomial is 2. So it is a quadratic polynomial in x.
- Here p(x) = x3
Degree of this polynomial is 3. So it is a cubic polynomial in x.
Question 2:
Obtain the degree of the following polynomials :
- p(x) = 3x – x4 + x2 +2x3 + 7
- p(x) = x3 – 3x – x2 + 6
- p(x) = 3x – 9
- p(x) = 2x2 – x + 1
Solution :
- Here p(x) = 3x – x4 + x2 + 2x3 + 7
∴p(x) = -x4 + 2x3 + x2 + 3x + 7
The term containing the highest power in this polynomial is -x4. So the degree of the polynomial is 4. - 2. p(x) = x3 – 3x – x2 + 6
∴ p(x) = x3 – x2 – 3x + 6
The term containing the highest power in this polynomial is x3. So the degree of the polynomial is 3. - 3. Here, p(x) = 3x – 9
The term containing the highest power in this polynomial is 3x. So the degree of the polynomial is 1. - p(x) = 2x2 – x + 1
The term containing the highest power in this polynomial 2x2. So the degree of the polynomial is 2.
Question 3:
Find the coefficients of the underlined terms :
- p(x) = 10x3 + 7x2 – 3x + 5
- p(x) = 7 – 5x5 + 3x4 + x2 – x
- p(x) = 25 – 125x
- p(x) = x3 – x2 + x +7
Solution :
- Here, p(x) = 10x3 + 7x2 – 3x + 5
Coefficient of the underlined term x3 is 10. - Here, p(x) = 7- 5x5 + 3x4 + x2 – x
Coefficient of the underlined term x5 is -5. - p(x) = 25 – 125x
Coefficient of the underlined term x is -125. - p(x) = x3 – x2 + x + 7
Coefficient of the underlined term x2 is-1.
Question 4:
Obtain the value of the following polynomials at the given values of x :
- p(x) = 2x3 + 3x2 + 7x + 9 ; at x = 0, 1
- p(x) = 3x2 + 10x + 7 ; at x = -3, 1
- p(x) = x2 – 2x + 5 ; at x = -1, 5
- p(x) = 2x4 – 3x3 + 7x + 5 ; at = -2, 2
Solution :
- Here, p(x) = 2x3 + 3x2 + 7x + 9
∴ p(0) = 2(0)3 + 3(0)2 + 7(0) + 9
= 0 + 0 + 0 + 9
= 9
Next, p(1) = 2(1)3 + 3(1)2 + 7(1) + 9
= 2 + 3 + 7 + 9
= 21 - Here, p(x) = 3x2 + 10x + 7
∴ p(-3) = 3(-3)2 + 10(-3) + 7
= 3(9) + 10(-3) + 7
= 27- 30 + 7
∴ p(-3) = 4
Next, p(1) = 3(1)2 + 10(1) + 7
= 3 + 10 + 7
∴ p(1) = 20 - Here, p(x) = x2 – 2x + 5
∴p(-1) = (-1)2 – 2(-1) + 5
= 1 + 2 + 5
∴ p(-1) = 8
Next, p(5) = (5)2 – 2(5) + 5
= 25 – 10 + 5
∴ p(5) = 20 - Here, p(x) = 2x4 – 3x3 + 7x + 5
∴p(-2) = 2(-2)4 – 3(-2)3 + 7(- 2) + 5
= 2(16)- 3(-8) + 7(-2) + 5
= 32 + 24 – 14 + 5
∴p(-2) = 47
Next, p(2) = 2(2)4 – 3(2)3 + 7(2) + 5
= 2(16) – 3(8) + 7(2) + 5
= 32- 24 + 14 + 5
∴ p(2)= 27
Question 5:
Examine the validity of the following statements :
- (x + 1) is a factor of p(x) = 3x3 + 2x2 + 7x + 8
- (x + 2) is a factor of p(x) = x3 + x2 + x + 2
- (x – 1) is a factor of p(x) = x4 – 2x3 + 3x – 2
- (x – 3) is a factor of p(x) = x2 – 2x – 3
Question 5(1):
Solution :
If (x + 1) is factor of p(x), then p(-1) = 0.
Now, p(-1) = 3(-1)3 + 2(-1)2 + 7(-1) + 8
= 3(-1) + 2(1) + 7(-1) + 8
= -3 + 2 – 7 + 8
= 0
Thus, (x + 1) is a factor of the given polynomial.
∴ The given statement is valid.
Question 5(2):
Solution :
If(x + 2) is a factor of p(x), then p(-2) = 0.
Now, p(-2) = (-2)3 + (-2)2 + (-2) + 2
= (-8) + (4) + (-2) + 2
=-8 + 4 – 2 + 2
= -4 ≠ 0
Thus, (x + 2) is not a factor of the given polynomial.
∴ The given statement is invalid.
Question 5(3):
Solution :
If(x – 1) is a factor of p(x), then p(1) = 0.
Now, p(1) = (1)4 – 2(1)3 + 3(1) – 2
= 1 – 2 + 3 – 2
= 0
Thus, (x – 1) is a factor of the given polynomial.
∴ The given statement is valid.
Question 5(4):
Solution :
If(x – 3) is a factor of p(x) then p(3) = 0.
Now, p(3) = (3)2 – 2(3) – 3
= 9 – 6 – 3
= 0
Thus, (x -3) is a factor of the given polynomial.
∴ The given statement is valid.
Question 6:
Factorize the following polynomials :
- p(x) = x3 – x2 – x + 1
- p(x) = 5x2 + 11x + 6
- p(x) = x3 – 3x2 + 9x – 27
- p(x) = x3 + 2x2 + 3x + 2
Question 6(1):
Solution :
x3 – x2 – x + 1 = x2(x – 1) – 1(x – 1)
= (x – 1)(x2 – 1)
= (x – 1)(x – 1)(x + 1)
= (x – 1)2(x + 1)
Question 6(2):
Solution :
5x2 + 11x + 6 = 5x2 + 5x + 6x + 6
= 5x(x + 1) + 6(x + 1)
= (x + 1)(5x + 6)
Question 6(3):
Solution :
x3 – 3x2 + 9x – 27 = x2(x – 3) + 9(x – 3)
= (x – 3)(x2 + 9)
Question 6(4):
Solution :
x3 + 2x2 + 3x + 2 = x3 + x2 + x2 + x + 2x + 2
= x2(x + 1) + x(x + 1) + 2(x + 1)
= (x + 1)(x2 + x + 2)
Question 7:
Prove that x – 2 is a factor of p(x) = x3 – 2x2
Solution :
If (x – 2) is a factor of p(x), then p(2)=0.
Now, p(2)= (2)3 – 2(2)2
= 8 – 2(4)
= 8 – 8
= 0
Thus, (x – 2) is a factor of p(x).
Exercise-2.2
Question 1:
Find the number of zeros of the following polynomials .
- p(x) = x2 – x
- p(x) = x – x2 – 1
- P(x) = 3x – 2
- p(x) = x3 – x
Question 1(1):
Solution :
Here p(x)= x2 – x
To find the zeros of p(x), let p(x) = 0
∴ x2 – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
∴ 0 and 1 are zeroes of p(x).
Hence, the number of zeroes of the given polynomial is 2.
Question 1(2):
Solution :
Here p(x) = x – x2 – 1
To find the zeroes of p(x), let p(x) = 0
∴ x – x2 – 1 = 0
∴ x2 – x + 1= 0
(∵ The square of a real number cannot be negative.)
Hence, the number of zeroes of the given polynomial is 0.
Question 1(3):
Solution :
Question 1(4):
Solution :
Here p(x) = x3 – x
To find the zeroes of p(x), let p(x) = 0
∴ x3 – x = 0
∴ x(x2 – 1) = 0
∴ x(x + 1)(x – 1) = 0
∴ x = 0 or x = -1 or x = 1
∴ 0, -1 and 1 are the zeros of p(x).
Thus, the number of zeroes of the given polynomial is 3.
Question 2:
Find the number of zeros and real zeros of p(x) = x3 + 1. Show them by a graph.
Solution :
For p(x) = x3 + 1, to find the zeroes of p(x), let p(x) = 0.
∴ x3 + 1 = 0
∴ (x + 1)(x2 – x + 1) = 0
∴ x + 1 = 0 or x2 – x + 1 = 0
∴ x = -1 and real zeroes of x2 – x + 1 are not possible.
∴ -1 is the only real zero.
Thus, number of zeros is for the given equation is 1 and that real zero is 0.
For p(x) = x3 + 1, taking x = -2, -1, 0 and 2.
We can obtain the following table:
x | -2 | -1 | 0 | 2 |
p(x) = x3 + 1 | -7 | 0 | 1 | 9 |
Plotting these points on a graph paper as shown in the figure, we can see that the graph intersects the x-axis at (-1, 0) so, -1 is the zero of p(x).
Question 3:
Draw the graph of p(x) = x2 + 1 and find the real zeros of this polynomial.
Solution :
Taking values as x = -3, -2, -1, 0, 1, 2, 3 in p(x), we obtain the following table:
X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
p(x) = x2 + 1 | 10 | 5 | 2 | 1 | 2 | 5 | 10 |
Plotting all these points on a graph paper and joining them as shown in the figure.
We see that the graph does not intersect the x-axis at any point. So p(x) has no real zero. The graph lies in the upper half-plane of the x-axis
Question 4:
From the figure 2.10 find the number of zeros of y = p(x).
Solution :
- The graph of y = p(x) intersects the x-axis at one point. So, the number of real zeroes of p(x) is 1.
- The graph of y = p(x) does not intersect the x-axis at any point. So, the number of real zeroes of p(x) is zero or p(x) had no real 0.
- The graph of y = p(x) intersects the x-axis at three distinct points. So the number of real zeroes of p(x) is 3.
- The graph of y = p(x) intersects the x-axis at two distinct points. So, the number of real zeroes of p(x) is 2.
- The graph y = p(x) intersect the x-axis at four distinct points. So, the number of real zeroes of p(x) is 4.
- The graph of y = p(x) intersects the x-axis at three distinct points. So, number of real zeroes of p(x) is 3.
Question 5:
Find the number of zeros and zeros of p(x) = x2 – 4. Represent them graphically.
Solution :
Here, p(x) = x2 – 4
To find the zeroes of p(x), let p(x) = 0
∴ x2 – 4 = 0
(x – 2)(x + 2) = 0
x = 2 or x = -2
The real zeroes are 2 and -2, which are 2 in all.
To draw the graph of this polynomial, we take some different values of x and prepare following table:
X | -3 | -2 | 0 | 2 | 3 |
p(x) = x2 – 4 | 5 | 0 | -4 | 0 | 5 |
Plot these points on a graph paper as shown in the figure. Here we join all these point (-3, 5), (-2, 0), (0, -4), (2, 0) and (3, 5) we get the shape of the graph as a ∪ as shown, i.e. opening upwards.
We can see that this graph intersects the x-axis at two distinct points (-2, 0) and (2, 0). The x-coordinates of these point are considered as zeroes of this polynomial.
Thus, -2 and 2 are the zeroes of p(x).
Exercise-2.3
Question 1:
Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x2 – 5x + 4. Also verifr the
relationship between the zeros and the coefficients of p(x).
Solution :
Here p(x) = x2 – 5x + 4
∴ p(4) = (4)2 – 5(4) + 4
= 16 – 20 + 4
= 0
Also p(1) = (1)2 – 5(1)+ 4
= 1- 5 + 4
= 0
Hence, 4 and 1 are the zeroes of the quadratic polynomial p(x) = x2 – 5x + 4.
In p(x) = x2 – 5x + 4
a = 1, b = -5, c = 4
Now, sum of the zeroes
= 4 + 1
= 5
Question 2:
Find the zeros of the following quadratic polynomials
- p(x) = x2 + 4x – 21
- p(x) = 6×2 – 11x + 5
- p(x) = 4×2 + 9x + 5
- p(x) = 3×2 + 5x – 8
- p(x) = x2 – 81
- p(x) = x2 – x – 6
Question 2(1):
Solution :
Here p(x) = x2 + 4x – 21, then to find its zeroes let p(x) = 0,
x2 + 4x – 21 = 0
∴ x2 + 7x – 3x – 21 = 0
∴ x(x + 7) – 3(x + 7) = 0
∴ (x + 7)(x – 3) = 0
∴ x = -7 or x = 3-7 and 3 are the zeroes of the quadratic polynomial p(x).
Question 2(2):
Solution :
To find the zeroes of p(x), let p(x) = 0,
6x2 – 11x + 5 = 0
∴ 6x2 – 6x – 5x + 5 = 0
∴ 6x(x – 1) – 5(x – 5) = 0
∴ (x – 1)(6x – 5) = 0
Question 2(3):
Solution :
To find the zeroes of p(x), let p(x) = 0,
4x2 + 9x + 5 = 0
∴4x2 + 4x + 5x + 5 = 0
∴4x(x + 1) + 5(x + 1) = 0
∴ (x + 1)(4x + 5) = 0
Question 2(4):
Solution :
To find the zeroes of p(x), let p(x) = 0,
3x2 + 5x – 8 = 0
∴ 3x2 + 8x – 3x – 8 = 0
∴ x(3x + 8) – 1(3x + 8) = 0
∴ (3x + 8)(x – 1) = 0
Question 2(5):
Solution :
To find the zeroes of p(x), let p(x) = 0,
x2 – 81= 0
∴ (x)2 – (9) 2 = 0
∴ (x – 9)(x + 9) = 0
∴ x = 9 or x = -9
-9 and 9 are the zeroes of the quadratic polynomial p(x).
Question 2(6):
Solution :
To find the zeroes of p(x), let p(x) = 0,
x2 – x – 6 = 0
∴ x2 – 3x + 2x – 6 = 0
∴ x(x – 3) + 2 (x – 3) = 0
∴ (x – 3)(x + 2) = 0
∴ x = 3 or x = -2
3 and -2 are the zeroes of the quadratic polynomial p(x).
Question 3:
Find the zeros, the sum of the zeros and the product of the zeros of the quadratic polynomial p(x) = 3x2 – x – 4
Solution :
Here, p(x) = 3x2 – x – 4
= 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
=(3x – 4)(x + 1)
To find the zeroes of p(x), let p(x) = 0,
(3x – 4)(x + 1) = 0
Question 4:
Obtain a quadratic polynomial with the following conditions :
- The sum of zeros = 2; the product of zeros = -3
- The sum of zeros = -3; the product of zeros = -4
- The sum of zeros = \(\frac{1}{3}\); the product of zeros = \(\frac{1}{2}\)
Question 4(1):
Solution :
The sum of zeroes = 2; the product of zeroes = -3
Suppose a and b are the zeroes of the quadratic polynomial p(x).
∴c = -3, a = -3(k) = -3k
Now, p(x) = ax2 + bx + c
= kx2 – 2kx – 3k (∵ Substituting values of a, b andc)
p(x) = k(x2 – 2x – 3), k ≠ 0
Hence p(x) is the required quadratic polynomial.
Question 4(2):
Solution :
The sum of zeroes = -3; product of zeroes = -4
Suppose a and b are zeroes of the quadratic polynomial p(x).
∴ The sum of zeroes = a + b
∴c = -4a = -4k
Now, p(x) = ax2 + bx + c
= kx2 + 3kx – 4k (∵ Substituting values of a, b and c)
= k(x2 – 3x – 4), k ≠ 0
Hence p(x) is the required quadratic polynomial.
Question 4(3):
Solution :
Question 5:
Obtain the quadratic or the cubic polynomial as the case may be in the standard form with the following coefficients :
- a = 6, b = 17, c = 11
- a = 1, b = -1, c = -1, d = 1
- a = 5, b = 7, c = 2,
- a = 1, b = -3, c = -1, d = 3
- a = 3, b = -5, c = -11, d = -3
Solution :
- The quadratic polynomial is,
p(x) = ax2 + bx + c; a ≠ 0, a, b, c ∊ R.
∴ Substituting a = 6, b = 17 and c = 11 we get the required quadratic polynomial
p(x) = 6x2 + 17x + 11.
- The cubic polynomial is,
p(x) = ax3 + bx2 + cx + d; a ≠ 0, a, b, c, d ∊ R.
∴ Substituting a = 1, b = 1 and c = 1 we get the required cubic polynomial
p(x) = x3 – x2 – x + 1.
- The quadratic polynomial is,
p(x) = ax2 + bx + c; a ≠ 0, a, b, c ∊ R.
∴ Substituting a = 5, b = 7 and c = 2 we get the required quadratic polynomial
p(x) = 5x2 + 7x + 2.
- The cubic polynomial is,
p(x) = ax3 + bx2 + cx + d; a ≠ 0, a, b, c, d ∊ R.
∴ Substituting a = 1, b = -3, c = -1 and d = 3, we get the required cubic polynomial
p(x) = x 3 - 3x2 – x + 3.
- The cubic polynomial is,
p(x) = ax3 + bx2 + cx + d; a ≠ 0, a, b, c, d ∊ R.
∴ Substituting a = 3, b = -5, c = -11 and d = -3, we get the required cubic polynomial
p(x) = 3x3 - 5x2 – 11x – 3.
Exercise-2.4
Question 1:
Divide the following polynomial p(x) by the polynomial s(x).
- p(x) = 2x3 – 13x2 + 23x – 12, s(x) = 2x – 3
- p(x) = \(\frac{2}{3}\) x2 + 5x + 6, s(x) = x + 6
- p(x) = 40x2 + 11x – 63, s(x) = 8x – 9
- p(x) = 2x3 + 9x2 + 13x + 6, s(x) = 2x2 + 5x + 3
- p(x) = x4 + 4x3 + 5x2 – 7x – 3, s(x) = x2 – 1
Question 1(1):
Solution :
Here,
Dividend = p(x) = 2x3 – 13x2 + 13x + 23x – 12 and
Divisor = s(x) = 2x – 3
Thus, the quotient polynomial q(x) = x2 – 5x + 4 and the reminder polynomial r(x) = 0.
Question 1(2):
Solution :
Question 1(3):
Solution :
Here,
Dividend = p(x) = 40x2 + 121x – 63 and
Divisor = s(x) = 8x – 9
Thus, the quotient polynomial q(x) = 5x + 7 and
the reminder polynomial r(x) = 0.
Question 1(4):
Solution :
Here,
Dividend = p(x)= 2x3 + 9x2 + 13x + 6 and
Divisor = s(x)= 2x2 + 5x + 3
Thus, the quotient polynomial q(x) = x + 2
and the reminder polynomial r(x) = 0
Question 1(5):
Solution :
Here,
Dividend polynomial = p(x) = x4 + 4x3 + 5x2 – 7x – 3 Divisor polynomial = s(x) = x2 – 1.
Thus, the quotient polynomial q(x) = x2 + 4x + 6 and the remainder polynomial r(x) = 3 – 3x.
Question 2:
Find the remainder polynomial when the cubic polynomial x3 – 3x2 + 4x + 5 is divided by x – 2.
Solution :
Here we do the calculation by the synthetic division method.
Dividend polynomial p(x) = x3 – 3x2 + 4x + 5 and the divisor polynomial s(x) = x – 2
Coefficients of x3, x2, x and x0 in the dividend polynomial are 1, -3, 4 and 5 respectively.
Taking x – 2 = 0 we get x = 2.
Thus, the reminder obtained when the cubic polynomial x3 – 3x2 + 4x + 5 is divided by x – 2 is r(x) = 9.
Question 3:
3 is a zero of p(x) = 3x3 – x2 – ax – 45. Find ‘a’.
Solution :
Given that, 3 is a zero of polynomial p(x)
⇒ p(3) = 0
For p(x) = 3x3 – x2 – ax – 45
∴ 3(3)3 – (3)2 – a(3) – 45 = 0
∴ 3(27) – 9 – 3a – 45 = 0
∴ 81- 54 – 3a = 0
∴ 3a = 27
∴ a = 9
Hence, if 3 is a zero of the polynomial p(x), then a = 9.
Question 4:
The product of two polynomials is 6×3 + 29×2 + 44x + 21 and one of the polynomials is 3x + 7. Find the other polynomial.
Solution :
Question 5:
If polynomial p(x) is divided by x2 + 3x + 5, the quotient polynomial and the remainder polynomials are 2×2 + x + 1 and x – 3 respectively. Find p(x).
Solution :
Here, Divisor polynomial = s(x) = x2 + 3x + 5
Quotient polynomial = q(x) = 2x2 + x + 1
Remainder polynomial = r(x) = x – 3.
We know that,
Dividend = Divisor × Quotient + Remainder
∴ p(x) = s(x).q(x) + r(x)
= (x2 + 3x + 5)(2x2 + x + 1) + (x – 3)
= 2x4 + x3 + x2 + 6x3 + 3x2 + 3x + 10x2 + 5x + 5 + x – 3
=2x4 + 7x3 + 14x2 + 9x + 2
∴ Required polynomial is p(x) = 2x4 + 7x3 + 14x2 + 9x + 2
Question 6:
Divide p(x) = x3 – 4x2 + 5x – 2 by x – 2 Find r(x).
Solution :
Here we do the calculation by the synthetic division method.
Here, dividend polynomial = p(x) = x3 – 4x2 + 5x – 2.
Divisor polynomial = s(x) = x – 2.
Coefficients of x3, x2 x and x0 in the dividend polynomial are q, -4, 5 and -2 and taking x – 2 = 0 we get x = 2.
Thus, polynomial p(x) = x3 – 4x 2 + 5x – 2 is divided by x – 2, the remainder polynomial r(x) = 0.
Question 7:
There are x4 + 57x + 15 pens to be distributed in a class of x2 + 4x + 2 students. Each student should get the maximum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed (x ∈ N).
Solution :
Here, we divide the number of pens x4 + 57x + 15 by the number of students x2 + 4x + 2. The quotient polynomial obtained will be the number of pens each student receives and the remainder polynomial gives the number of pens left undistributed.
Here, p(x) = x4 + 57x + 15 and
s(x) = x2 + 4x + 2
Each student will get q(x) = x2 – 4x + 14 (the quotient polynomial) number of pens and the number of pens undistributed are r(x) = 9x – 13 (the remainder polynomial).
Question 8:
A trader bought 2×2 – x + 2 TV sets for ₹ 8×4 + 71 – 6. Find the price of one TV set.
Solution :
2x2 – x + 2 TV sets can be bought in Rs. 8x4+7x-6.
∴ The price of one TV set is same as the quotient polynomial obtained by dividing 8x4 + 7x – 6 by 2x2 – x + 2
Thus, the price of one TV set is Rs. 4x2 + 2x – 3.
Question 9:
\(-\sqrt{2}\) and \(\sqrt{2}\) are two of the zeros of p(x) = 2×4 + 7×3 – 8×2 – 14x + 8. Find the remaining zeros of p(x).
Solution :
∴ The quotient polynomial is q(x) = 2x2 + 7x – 4 and the remainder polynomial is r(x) = 0
∴ p(x) = 2x4 + 7x3 – 8x2 – 14x + 8
= (x2 – 2)(2x2 + 7x – 4)
∴ It is clear that, the zeroes of the second factor 2x2 + 7x – 4 are the remaining zeroes of the polynomial p(x).
Now, to find the zeroes of q(x) = 2x2 + 7x – 4 we take q(x) = 0.
Exercise-2
Question 1:
State whether the following statements are true or false :
- \(\frac{7}{5}\) is a zero of the linear polynomial p(x) = 5x + 7.
- p(x) = x2 + 2x + 1 has two distinct zeros.
- The cubic polynomial p(x) = x3 + x2 – x – 1 has two distinct zeros.
- The graph of the cubic polynomial p(x) = x3 meets the X—axis at only one point.
- Any quadratic polynomial p(x) has at least one zero, x ∈ R
Solution :
2. False.
Reason: To find the zeros of p(x), let p(x) = 0
∴ x2 + 2x + 1 = 0
∴ (x + 1)2 = 0
∴ x + 1 = 0 or x = -1
∴ x = -1 or x = -1
Here, both the zeroes are equal, i.e. -1, and hence not distinct.
3. True.
Reason: To find the zeroes of p(x), let p(x) = 0
∴x3 + x2 – x – 1 = 0
∴x2(x + 1) – 1(x + 1) = 0
∴ (x + 1)(x2 – 1) = 0
∴ (x + 1)(x + 1)(x – 1) = 0
∴ x + 1 = 0 or x + 1 = 0 or x – 1 = 0
∴ x = -1 o r x = -1 or x = 1
∴ Two distinct zeroes of p(x) are 1 and -1.
Hence, p(x) has at the most two distinct zeroes.
4. True.
Reason: The graph of p(x) = x3 is the x-axis then p(x) = 0
∴ x3 = 0
∴ x = 0
∴ y = p(0) = 0
Thus, the graph of p(x) = x3 meets the x-axis at only one point (0, 0).
5. False
If the graph of the quadratic polynomial p(x) does not intersect the x-axis at any point, then the quadratic polynomial does not have any real zero.
Example: polynomial p(x) = x2 + 4x + 5, does not have any real zeros.
Question 2:
Find the zeros and number of zeros of p(x) = x2 + 9x + 18. Show them on a graph.
Solution :
Here, p(x) = x2 + 9x + 18
To find the zeroes of p(x), let p(x) = 0
∴ x2 + 9x + 18 = 0
∴ x2 + 6x + 3x + 18 = 0
∴ x(x + 6) +3(x + 6) = 0
∴ (x + 6)(x + 3) = 0
∴ x = -6 or x = -3
Thus, the distinct zeroes -6 and -3, i.e. 2 in all.To draw the graph of this polynomial, we take different values of x and prepare the following table:
x | -7 | -6 | -5 | -4 | -3 | -2 |
p(x)= x2 + 9x + 18 | 4 | 0 | -2 | -2 | 0 | 4 |
On plotting these points on a graph paper and joining the points (-7, 4), (-6, 0), (-5, -2), (-4, -2), (-3, 0) and (-2, 4). As shown in the figure the shape of the graph is a U, as shown i.e., it opens upwards.
Also the graph intersects the x-axis at two distinct points (-6, 0) and (-3, 0). Their x-coordinates are considered as the zeroes of this polynomial.
Thus, -6 and -3 are the zeroes of p(x).
Question 3:
Find the zeros, the sum and the product of zeros of p(x) = 4x2 + 12x + 5.
Solution :
For 4x2 + 12x + 5
= 4x2 + 10x + 2x + 5
= 2x(2x + 5) + 1(2x + 5)
= (2x + 5)(2x + 1)
To find the zeroes of p(x) = 0
∴ (2x + 5)(2x + 1) = 0
Question 4:
-4 and 9 are the sum and product of the zeros respectively of a quadratic polynomial. Find the quadratic polynomial.
Solution :
Suppose, a and b are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c.
Thus, c = 9a = 9k
Now, p(x) = ax2 + bx + c
∴ p(x) = kx2 + 4kx + 9k
= k(x2 + 4x + 9), k ≠ 0
∴ Required quadratic polynomial is k(x2 + 4x + 9), k ≠ 0.
Question 5:
Find q(x) and r(x), for the quadratic polynomial p(x) = 11x – 21 + 2x2 when divided by 1 + 2x
Solution :
Here, Dividend = p(x) = 11x – 21 + 2x2
= 2x2 + 11x – 21 (∵ rearranging the terms)
Divisor = s(x) = 1 + 2x = 2x+1
∴ Quotient polynomial is q(x) = x + 5 and the remainder polynomial is r(x) = -26.
Question 6:
Divide 2x3 + 3x2 – 11x – 6 by x2 + x – 6
Solution :
Here, Dividend = p(x) = 2x3 + 3x2 – 11x – 6
Divisor = s(x) = x2 + x – 6
∴ Quotient polynomial q(x) = 2x + 1 and the
remainder polynomial r(x) = 0.
Question 7:
4 is a zero of the cubic polynomial p(x) = x3 – 3x2 – 6x + 8. Find the remaining zeros of p(x)
Solution :
We will do the calculation by the synthetic division method.
Here 4 is the zero of polynomial p(x).
So (x – 4) is the factor of p(x).
∴ Dividend = p(x) = x3 – 3x2 – 6x + 8 and
Divisor polynomial = s(x) = x – 4.
Coefficients of x3, x2, x and x° are 1, -3, -6 and 8 respectively. Taking x – 4 = 0
∴ x = 4
∴ p(x) = x3 – 3x2 – 6x + 8
= (x – 4) (x2 + x – 2)
= (x – 4)(x + 2)(x – 1)
To find the remaining zeroes, let p(x)=0
(x – 4)(x + 2)(x – 1) = 0
∴ The remaining zeroes of p(x) are -2 and 1.
Question 8:
The product of two polynomials is 3x4 + 5x3 – 21x2 – 53x — 30. If one of them is x2 – x – 6, find the other polynomial.
Solution :
∴ The quotient polynomial q(x) = 3x2 + 8x + 5 and the remainder polynomial r(x) = 0.
∴ The polynomial 3x2 + 8x + 5 is a factor of 3x4 + 5x3 – 21x2 – 53x – 30.
Question 9:
2 + \( \sqrt{3}\) and 2 – \(\sqrt{3}\) are the zeros of p(x) = x4 – 6x3 – 26x2 + 138x – 35. Find the remaining zeros of p(x).
Solution :
Question 10:
Select a proper option (a), (b), (c) or (d) from given options :
Question 10(1):
The linear polynomial p(x) = 7x (1) – 3 has the zero ..
Solution :
Question 10(2):
The cubic polynomial p(x) = x3 – x has….. zeros.
Solution :
d. 3
p(x) = x3 – x
= x(x2 – 1) = x(x – 1)(x + 1)
∴ x = 0, x = 1, x = -1
∴ The cubic polynomial has 3 zeroes.
Question 10(3):
The graph of p(x) = 3x – 2 – x2 intersects the X-axis in ….. points.
Solution :
c. 2
The zeroes of p(x) are the intersections points of the equation, p(x) = 3x – 2 – x2 with the x-axis.
∴ Whatever the number of distinct zeros of p(x) that many number of distinct intersection points can be found on the x-axis.
To find zeros of p(x)= 3x – 2 – x2,
Let p(x) = 0
∴ 3x – 2 – x2 = 0
∴ x2 – 3x + 2 = 0
∴ x2 – 2x – x + 2 = 0
∴ x(x – 2) – 1(x – 2) = 0
∴ (x – 1)(x – 2) = 0
∴x = 1 or x = 2
∴ The graph of p(x) intersect the x-axis in 2 distinct points which are (1, 0) and (2, 0).
Question 10(4):
The sum of the zeros of 3x2 + 5x – 2 is …..
Solution :
Question 10(5):
The graph of p(x) = 3x + 5 represents …….
Solution :
a. a straight line
p(x) = 3x + 5 is a linear polynomial. So its graph is a straight line.
Question 10(6):
quadratic polynomial has no zero. Its graph ……
Solution :
d. is in any one half plane of the x-axis
The number of intersection points of a quadratic polynomial is the number of real zeroes of that quadratic polynomial.
Here, the quadratic polynomial has no zero.
∴ It is in any one half plane of the x-axis.
Question 10(7):
For the graph in figure 2.11 y = p(x) has … zeros.
Solution :
d. 4
Here, the graph intersects the x-axis in four points.
∴ p(x) has 4 zeroes.
Question 10(8):
The product of the zeros of x2 – 4x + 3 is …..
Solution :
Question 10(9):
a = 3, b = 5, c = 7, d = 11 in the standard notation gives the cubic polynomial …..
Solution :
d. 3x3 + 5x2 + 7x + 11
The standard form of a cubic polynomial is
p(x) = ax3 + bx2 + cx + d
∴ For a = 3, b = 5, c = 7 and d = 11,
p(x) = 3x3 + 5x2 + 7x + 11.