Bihar Board Class 10th Maths Chapter 2 Polynomials Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.1

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

Solution:

1. There are no zeros as the graph does not intersect the x-axis.
2. The number of zeroes is one as the graph intersects the x-axis at one point only.
3. The number of zeroes is three as the graph intersects the x-axis at three points.
4. The number of zeroes is two as the graph intersects the x-axis at two points.
5. The number of zeroes is four as the graph intersects the x-axis at four points.
6. The number of zeroes is three as the graph intersects the x-axis three points.

BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:

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

Solution:
1. We have:
x2 – 2x – 8 = x2 + 2 – 4x – 8
= x(x + 2) – 4(x + 2)
= (x + 2)(x – 4)
The value of x2 – 2x – 8 is 0, when the value of (x + 2)(x – 4) is 0, i.e., when x + 2 = 0 or x – 4 = 0, i.e., when x = – 2 or x = 4.
∴ The zeroes of x2 – 2x – 8 are – 2 and 4.
Therefore, sum of the zeroes = (- 2) + 4 = 2 = −(−2)1

and product of zeroes = (- 2)(4) = – 8 = −81

2. We have:
4s2 – 4s + 1 = 4s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
The value of 4s2 – 4s + 1 is 0, when the value of (2s – 1)(2s – 1) is 0, i.e., when 2s – 1 = 0 or 2s – 1 = 0, i.e.,
when s = 12 or s = 12.
∴ The zeroes of 4s2 – 4s + 1 are 12 and 12.
Therefore, sum of the zeroes = 12 + 12 = 1 = −(−4)4

3. We have:
6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
The value of 6x2 – 3 – 7x is 0, when the value of (3x + 1)(2x – 3) is 0, i.e; when 3x + 1 = 0 or 2x – 3 = 0, i.e;
when x = – 13 or x = 32.
∴ The zeros of 6x2 – 3 – 7x are – 13 and 32.
Therefore, sum of the zeros = – 13 + 32 = 76 = −(−7)6

4. We have:
4u2 + 8u is 0, when the value of 4u(u + 2) is 0, i.e; when u = 0 or u + 2 = 0, i.e; when u = 0 or u = – 2.
∴ The zeroes of 4u2 + 8u are o and – 2.
Therefore, sum of the zeroes = 0 + (- 2) = – 2 = −84

and product of zeroes = (0)(- 2) = 0 = 04

5. We have:
t – 15 = (t – 15−−√)(t + 15−−√)
The value of t2 – 15 is 0, when the value of (t – 15−−√)(t + 15−−√) is 0, i.e; when t – 15−−√ = 0 or t + 15−−√ = 0,
i.e; when t = 15−−√ or t = – 15−−√.
∴ The zeroes of t2 – 15 are 15−−√ and – 15−−√.
Therefore, sum of the zeroes = 15−−√ + (- 15−−√) = 0

and product of the zeroes = (15−−√)(-15−−√)
= – 15 = −151

6. We have:
3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1)(3x – 4)
The value of 3x2 – x – 4 is 0, when the value of (x + 1)(3x – 4) is 0, i.e; when x + 1 = 0 or 3x – 4 = 0, i,e; when x = – 1 or x = 43.
∴ The zeroes of 3x2 – x – 4 are – 1 and 43.
Therefore, sum of the zeroes = – 1 + 43 = −3+43
= 13 = −(−1)3

and product of the zeroes = (- 1)(43) = – 43 = −43

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

1. 14, – 1
2. 2–√, 13
3. 0, 5–√
4. 1, 1
5. – 14, 14
6. 4, 1

Solution:
1. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,

If a = 4, then b = – 1 and c = – 4.
∴ One quadratic polynomial which fits the given conditions is 4x2 – x – 4.

2. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,

If a = 3, then b = – 32–√ and c = 1.
∴ One quadratic polynomial which fits the given conditions is 3x2 – 32x−−√ + 1.

3. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,

If a = 1, then b = 0 and c = 5–√.
∴ One quadratic polynomial which fits the given conditions is x2 – 0. x + 5–√, i.e; x2 + 5–√.

4. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,

If a = 1, then b = – 1 and c = 1.
∴ One quadratic polynomial which fits the given conditions is x2 – x + 1.

5. Let the polynomial be ax2 + bx + c and its zeroes be α and β. Then

If a = 4, then b = 1 and c = 1.
∴ One quadratic polynomial which fits the given conditions is 4x2 + x + 1.

6. Let the polynomial be ax2 + bx + c and its zeroes be α and β. Then,

If a = 1, then b = – 4, and c = 1.
∴ One quadratic polynomial which fits the given conditions is x2 – 4x + 1.

BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials 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) = r3 – 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. Here, dividend and divisor are both in standard forms. So, we have:

2. Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as
x2 – x + 1.
We have:

∴ The quotient is x2 + x – 3 and the remainder is 8.

3. To carry out the division, we first write divisor in the standard form.
So, divisor = – x2 + 2
We have:

∴ The quotient is – x2 – 2 and the remainder is – 5x + 10.

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:
1. Let us divide 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3.
We have:

Since the remainder is 0, therefore t2 – 3 is a factor of 2t4 + 3t3 – 2t2 – 9t – 12.

2. Let us divide 3x4 + 5x3 – 7x2 + 2x + 2 by x2 + 3x + 1.
We get,

Since, the remainder is 0, therefore x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2

3. Let us divide x5 – 4x3 + x2 + 3x + 1 by x3 – 3x + 1. We get,

Here, remainder is 2(≠ 0). Therefore, x3 – 3x + 1 is not a factor of
x5 – 4x3 + x2 + 3x + 1.

Question 3.
Obtain all the zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are
53−−√ and – 53−−√.
Solution:
Since two zeroes are 53−−√ and – 53−−√, so (x – 53−−√) and (x + 53−−√) are the factors of the given polynomial.
Now, (x – 53−−√) (x + 53−−√) = x2 – 53.

So, (3x2 – 5) is a factor of the given polynomial.
Applying the division algorithm to the given polynomial and 3x2 – 5, we have:

∴ 3x4 + 6x3 – 2x2 – 10x – 5 = (3x2 – 5)(x2 + 2x + 1)
Now, x2 + 2x + 1 = x2 + x + x + 1
= x(x + 1) + 1(x + 1)
= (x + 1)(x + 1)
So, its other zeroes are – 1 and – 1.
Thus, all the zeroes of the given fourth degree polynomial are 53−−√, – 53−−√, – 1 and – 1.

Question 4.
On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quoitent and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Solution:
Since on dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quoitent and remainder were (x – 2) and (- 2x + 4) respectively, therefore
Quoitent × Divisor + Remainder = Dividend
or (x – 2) × g(x) + (- 2x + 4) = x3 – 3x2 + x + 2
or (x – 2) × g(x) = x3 – 3x2 + x + 2 + 2x – 4
or g(x) = x3−3x2+3x−2x−2 …………… (1)
Let us divide x3 – 3x2 + 3x – 2 by x – 2. We get

∴ (1) gives 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:
There can be several examples for each of (i), (ii) and (iii).
However, one example for each case may be taken as under:

1. p(x) = 14, g(x)= 2, q(x) = x2 – x + 7, r(x) = 0
2. p(x) = x3 + x2 + x + 1, g(x) = x2 – 1, q(x) = x + 1, r(x) = 2x + 2
3. p(x) = x3 + 2x2 – x + 2, g(x) = – 1, q(x) = x + 2, r(x) = 4

BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeros.
Also verify the relationship between the zeroes and the coefficients in each case:

1. 2x3 + x2 – 5x + 2; 12, 1, – 2
2. x3 – 4x2 + 5x – 2; 2, 1, 1

Solution:
1. Comparing the given polynomial with ax3 + bx2 + cx + d, we get
a = 2, b = 1, c = – 5 and d = 2.

p(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
p(- 2) = 2(- 2)3 + (- 2)2 – 5(- 2) + 2
= 2(- 8) + 4 + 10 + 2
= – 16 + 16 = 0
∴ 12, 1 and – 2 are the zeroes of 2x3 + x2 – 5x + 2.
So, α = 12, β = 1 and γ = – 2.

2. Comparing the given polynomial with ax3 + bx2 + cx + d, we get
a = 1, b = – 4, c = 5 and d = – 2.
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
∴ 2, 1 and 1 are the zeroes of x3 – 4x2 + 5x – 2.
So, α = 2, β = 1 and γ = 1.
Therefore, α + β + γ = 2 + 1 + 1 = 4 = −(−4)1 = −ba
αβ + βγ + γα = (2)(1) + (1)(1) + (1)(2)
= 2 + 1 + 2 + 5 = 51 = ca
and αβγ = (2)(1)(1) = 2 = −(−2)1 = −da.

Question 2.
Find a cubic polynomial with the sum, sum of the products of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
Solution:
Let the cubic polynomial be ax3 + bx2 + cx + d, and its zeroes be α, β and γ.
Then, α + β + γ = 2 = −(−2)1 = −ba

αβ + βγ + γα = – 7 = −71 = ca
and αβγ = – 14 = −141 = −da
If a = 1, then b = – 2, c = – 7 and d = 14.
So, one cubic polynomial which fits the given conditions x3 – 2x2 – 7x + 14.

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a and a + b, find a and b.
Solution:
Since (a – b), a and (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1, therefore
So, (a – b) + a + (a + b) = −(−3)1 = 3
So, 3a = 3 or a = 1
(a – b)a + a(a + b) + (a + b)(a – b) = 11 = 1
or a2 – ab + a2 + ab + a2 – b2 = 1
or 3a2 – b2 = 1
So, 3(1)2 – b2 = 1 [∵ a = 1]
or 3 – b2 = 1
or b2 = 2 or b = ± 2–√
Hence, a = 1 and b = ± 2–√.

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± 3–√, find other zeroes.
Solution:
We have:
2 ± 3–√ are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35
Let x = 2 ± 3–√. So, x – 2 = ± 3–√
Squaring, we get
x2 – 4x + 4 = 3, i.e; x2 – 4x + 1 = 0
Let us divide p(x) by x2 – 4x + 1 to obtain other zeroes.

∴ p(x) = 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 + 5)(x – 7)
So, (x + 5) and (x – 7) are other factors of p(x).
∴ – 5 and 7 are other zeroes of the given polynomial.

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 divide x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.

∴ Remainder = (2k – 9)x – (8 – k)k + 10
But the remainder is given as x + a.
On comparing their coefficients, we have:
2k – 9 = 1 or 2k = 10 or k = 5
and – (8 – k)k + 10 = a
So, a = – (8 – 5)5 + 10
= – 3 × 5 + 10 = – 15 + 10 = – 5
Hence, k = 5 and a = – 5.