Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials

 Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials

Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials

Ex 2.1

Question 1.
The graphs of y = p(x) are given in figures, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

  

Solution:

1. Graph does not intersect at any point on x-axis, hence there is no zeroes.
2. Graph intersect at one point on the x-axis, hence the number of zeroes is 1.
3. Graph intersect at three points on the x-axis, hence the number of zeroes is 3.
4. Graph intersect at two points on the x-axis, hence the number of zeroes is 2.
5. Graph intersect at four points on the x-axis, hence the number of zeroes is 4.
6. Graph intersect at three points on the x-axis, hence the number of zeroes is 3.

Ex 2.2

Question 1.
Find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficients

1. x2 – 2x – 8
2. 4s2 – 4s + 1
3. 6x3 – 3 – 7x
4. 4u2 + 8u
5. t2 – 15
6. 3x2 – x – 4

Solution:
1. p(x) = x2 – 2x – 8
= x2 – 4x + 2x – 8
[By splitting middle term]
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
For zeroes of p(x),
p(x) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0
or x + 2 = 0
⇒ x = 4
or x = -2
⇒ x = 4, – 2.
Zeroes of p(x) are 4 and -2.
⇒ α = 4, β = -2

2. p(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
[By splitting middle term]
= 2s(2s – 1) – 1 (2s – 1)
(2s – 1) (2s – 1)
For zeroes of p(s),
P(s) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1 = 0
or 2s – 1 = 0

3. p(x) = 6x2 – 3 – 7x
p(x) = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
For zeroes p(x) = 0
⇒ (2x – 3) (3x + 1) = 0
⇒ 2x – 3 = 0
or 3x + 1 = 0

4. p(u) = 4u2 + 8 u
p(u) = 4u (u + 2)
For zeroes, p(u) = 0
⇒ 4 u(u + 2) = 0
⇒ u(u + 2) = 0
⇒ u = 0
or u + 2 = 0
⇒ u = 0
⇒ u = -2
So, zeroes of p(u) are 0 and -2.
⇒ α = 0, β = -2
Verification of relationship between zeroes and coefficients:
Sum of the zeroes,
α + β = 0 + (-2) = -2

5. t2 – 15
P(t) = t2 – 15
For zeroes p(t) = 0
t2 – 15 = 0

6. p(x) = 3x2 – x – 4
p(x) = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)
For zeroes, p(x) = 0
⇒ (3x – 4)(x + 1) = 0
⇒ 3x – 4 = 0
or x + 1 = 0

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

  

6. 4, 1
Sum of the zeroes, S = 4
Product of the zeroes, P = 1
∴ Required polynomial is
p(x) = K[x2 – Sx + P]
p(x) = x2 – 4x + 1

Ex 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:

1. p(x) = x3 – 3x2 + 5x – 3; g(x) = x2 – 2
2. p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
3. p(x) = x4 – 5x + 6, g(x) = 2 – x2

Solution:
1. By applying the division algorithm to the given polynomials p(x) and g(x)

∴ Remainder = 7x – 9
degree of 7x – 9 = 1 < degree of (x2 – 2) = 2.
⇒ Quotient = x – 3
Remainder = 7x – 9
Verify: Quotient × Division + Remainder
= (x – 3) (x2 – 2) + (7x – 9)
= x3 – 2x – 3x2 + 6 + 7x – 9
= x3 – 3x2 + 5x – 3 = Dividend.

2. p(x) = x4 – 3x2 + 4x + 5
g(x) = x2 + 1 – x
Standard form of g(x) = x2 – x + 1.
Now, by applying the division algorithm to the given polynomials p(x) and g(x)

degree of 8 = 0 < degree of x2 – x + 1 = 2
So, Quotient = x2 + x – 3
Remainder = 8.
Verify: Quotient × Divisor + Remainder
= (x2 + x – 3) (x2 – x + 1) + 8
= x4 – x3 + x2 + x3 – x2 + x – 3x2 + 3x – 3 + 8
= x4 – 3x2 + 4x + 5
= Dividend

3. p(x) = x4 – 5x + 6
g(x) = 2 – x2
Standard form of g(x) = -x2 + 2
Now, by applying division algorithm to the given polynomial p(x) and g(x).

∵ deg. of -5x + 10 = 1 < deg. of -x2 + 2 = 2
So, Quotient = -x2 – 2
Remainder = -5x + 10.
Verify: Quotient × Divisor + Remainder
= (-x2 – 2) × (-x2 + 2) + (-5x + 10)
= (-x2)2 – 4 – 5x + 10
= x4 – 5x + 6 = Dividend

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

1. t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
2. x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
3. x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Solution:
By applying division algorithm to the given polynomials,

Since, the remainder is 0, hence the first polynomial is a factor of the second polynomial.

2. x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
By applying division algorithm to the given polynomials,

Since the remainder is 0, hence the first polynomial is a factor of the second polynomial.

3. x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
By applying division algorithm to the given polynomials,

Since the remainder is 2 (≠0), hence the first polynomial is not a factor of second polynomial.

So, 3x4 + 6x3 – 2x2 – 10x – 5
= [x2 – 5/3] (3x2 + 6x + 3)
Now, 3x2 + 6x + 3
= 3x2 + 3x + 3x + 3
= 3x(x + 1) + 3(x + 1)
= 3(x + 1) (x + 1)
So, its zeroes are -1 and -1.
⇒ Remaining two zeroes of the given polynomial are -1 and -1.

Question 4.
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:
p(x) = x3 – 3x2 + x + 2
q(x) = x – 2
r(x) = -2x + 4
By division algorithm for polynomials
p(x) = g(x) × q(x) + r(x)
⇒ x3 – 3x2 + x + 2
= (x – 2) g(x) + (-2x + 4)
⇒ (x – 2) g(x)
= x3 – 3x2 + x + 2 + 2x – 4
= x3 – 3x2 + 3x – 2
∴ g(x) = (x3 – 3x2 + 3x – 2) ÷ (x – 2)

∴ g(x) = x2 – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

1. deg p(x) = deg q(x)
2. deg q(x) = deg r(x)
3. deg r(x) =0.

Solution:
1. deg p(x) = deg q(x)
Let p(x) = 8x2 – 8x + 16
and g(x) = 8
∴ q(x) = x2 – x + 2
r(x) = 0
p(x) = g(x) × q(x) + r(x)

2. deg q(x) = deg r(x)
Let p(x) = x3 + x2 + x + 1
and g(x) = x2 – 1
∴ q(x) = x + 1
r(x) = 2x + 2
p(x) = g(x) × q(x) + r(x)

3. deg r(x) = 0
Let p(x) = x4 + x3 + x2 – 2x – 3
and g(x) = x2 – 2
∴ q(x) = x2 + x + 3
r(x) = 3
p(x) = g(x) × q(x) + r(x)

Ex 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:

2. Let p(x) = x3 – 4x2 + 5x – 2
Then, we have
p(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
p(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
Therefore, 2 and 1 are the two zeroes of x3 – 4x2 + 5x – 2. Hence (x – 2) (x – 1), i.e., x2 – 3x + 2 is a factors of the given polynomial. Now, we apply the division algorithm to the given polynomial and x2 – 3x + 2.

So, x3 – 4x2 + 5x – 2
= (x2 – 3x + 2) (x – 1)
⇒ x3 – 4x2 + 5x – 2 = (x – 2) (x – 1) (x – 1)
Hence, 2, 1 and 1 are the zeros of x3 – 4x2 + 5x – 2.
Comparing the given polynomial with ax3 + bx2 + cx + d, we get
a = 1
b = -4
c = 5
d = – 2
Let α = 2
β = 1
and γ = 1
Then, we have
α + β + γ = 2 + 1 + 1 = 4

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:

Question 3.
If the zeroes of the polynomial
x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Solution:
Then given polynomial is x3 – 3x2 + x + 1
Comparing with Ax3 + Bx2 + Cx + D, we get
A = 1
B = -3
C = 1
D = 1
Let α = a – b
β = a
γ = a + b
Then, we have

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:
Let us apply the division algorithm to the give Polynomial x4 – 6x3 + 16x2 – 25x + 10 and another Polynomial x2 – 2x + k.

Remainder
= (2k – 9)x – k (8 – k) + 10
But the remainder is given to be x + a.
Therefore, 2k – 9 = 1
⇒ 2k = 10 ⇒ k = 5
and -k(8 – k) + 10 = a
⇒ -5(8 – 5) + 10 = a
⇒ – 5 = a
⇒ a = -5