KC Sinha: Exercise 2.3 - Mathematics Solution Class 10 Chapter 2 Polynomials

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Question 1 

Divide 2x³ + 3x + 1 by x + 2 and find the quotient and the reminder. Is q(x) a factor of 2x³ + 3x + 1 ?

Sol :
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient = 2x² – 4x + 11
Remainder = – 21
No, 2x² – 4x + 11 is not a factor of 2x³ + 3x + 1 because remainder ≠ 0

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Question 2 

Divide 3x³ + x² + 2x + 5 by 1 + 2x + x² and find the quotient and the remainder. Is 1 + 2x + x² a factor of 3x³ + x² + 2x + 5

Sol :
Dividend = 3x³ + x² + 2x + 5
Divisor = x² + 2x + 1
Now, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient = 3x – 5
Remainder = 9x + 10
No, x² + 2x + 1 is not a factor of 3x³ + x² + 2x + 5 because remainder ≠ 0

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Question 3 A 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = x³ – 3x² + 4x + 2 , g(x) = x – 1
Sol :
p(x) = x³ – 3x² + 4x + 2,g(x) = x – 1
Dividend = x³ – 3x² + 4x + 2
Divisor = x – 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient [q(x)]= x² – 2x + 2
Remainder [r(x)]= 4

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Question 3 B 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = 4x³ – 3x² + 2x + 3 , g(x) = x + 4
Sol :

p(x) = 4x³ – 3x² + 2x + 3,g(x) = x + 4
Dividend = 4x³ – 3x² + 2x + 3
Divisor = x + 4
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient = 4x² – 13x + 54
Remainder = – 213

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Question 3 C 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = 2x⁴ + 3x³ + 4x² + 19x + 45, g(x) = x – 2
Sol :
p(x) = 2x⁴ + 3x³ + 4x² + 19x + 45, g(x) = x – 2
Dividend = 2x⁴ + 3x³ + 4x² + 19x + 45
Divisor = x – 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = 2x³+7x²+18x+55
r(x) = 155

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Question 3 D 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = x⁴ + 2x³ – 3x² + x – 1, g(x) = x – 2
Sol :
p(x) = x⁴ + 2x³ – 3x² + x – 1, g(x) = x – 2
Dividend = x⁴ + 2x³ – 3x² + x – 1
Divisor = x – 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = x³+4x²+5x+11
r(x) = 21

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Question 3 E 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = x³ – 3x²– x + 3, g(x) = x² – 4x + 3
Sol :
p(x) = x³ – 3x²– x + 3, g(x) = x² – 4x + 3
Dividend = x³ – 3x²– x + 3
Divisor = x² – 4x + 3
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,

q(x)= x + 1
r(x) = 0

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Question 3 F 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = x⁶ + x⁴ + x³ + x² + 2x + 2, g(x) = x³ + 1
Sol :
p(x) = x⁶ + x⁴ + x³ + x² + 2x + 2, g(x) = x³ + 1
Dividend = x⁶ + x⁴ + x³ + x² + 2x + 2
Divisor = x³ + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = x³ + x
r(x) = x² + x + 2

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Question 3 G 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = x⁶ + 3x² + 10 and g(x) = x³ + 1
Sol :
p(x) = x⁶ + 3x² + 10 , g(x) = x³ + 1
Dividend = x⁶ + 3x² + 10
Divisor = x³ + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = x³ – 1
r(x) = 3x² + 11

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Question 3 H 

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :

p(x) = x⁴ + 1, g(x) = x + 1
Sol :
p(x) = x⁴ + 1, g(x) = x + 1
Dividend = x⁴ + 1,
Divisor = x + 1

q(x) = x³ – x² + x – 1
r(x) = 0

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Question 4 

By division process, find the value of k for which x – 1 is a factor of x³ – 6x² + 11x + k.

Sol :
On dividing x³ – 6x² + 11x + k by x – 1 we get,

Since x – 1 is a factor of x³ – 6x² + 11x + k,
This means x – 1 divides the given polynomial completely.
→ 6 x + k = 0
→ k = – 6x

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Question 5 

By division process, find the value of c for which 2x + 1 is a factor of 4x⁴ – 3x² + 3x + c.

Sol :
On dividing 4x⁴ – 3x² + 3x + c by 2x + 1 we get,

Since 2x + 1 is a factor of 4x⁴ – 3x² + 3x + c,
This means 2x + 1 divides the given polynomial completely,
→ 4x + c = o
→ c = – 4x

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Question 6 A 

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:

p(x) = 2x² + 3x + 1, g(x) = x + 2
Sol :
p(x) = 2x² + 3x + 1, g(x) = x + 2
Dividend = 2x² + 3x + 1,
Divisor = x + 2
Apply the division algorithm we get,

q(x) = 2x – 1
r(x) = 3

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Question 6 B 

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:

p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
Sol :
p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
Dividend = x³ – 3x² + 5x – 3,
Divisor = x² – 2
On applying division algorithm, we get,

q(x) = x – 3
r(x) = 7x – 9

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Question 6 C 

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:

p(x) = x⁴ – 1 , g(x) = x + 1
Sol :
p(x) = x⁴ – 1,g(x) = x + 1
Dividend = x⁴ – 1
Divisor = x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient = x³ – x² + x – 1
Remainder = 0

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Question 6 D 

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:

p(x) = x³ – 3x² + 4x + 2 , g(x) = x – 1
Sol :
p(x) = x³ – 3x² + 4x + 2,g(x) = x – 1
Dividend = x³ – 3x² + 4x + 2
Divisor = x – 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient = x – 5
Remainder = 13x + 7

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Question 6 E 

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:

p(x) = x³ – 6x² + 11x – 6 , g(x) = x² – 5x + 6
Sol :
p(x) = x³ – 6x² + 11x – 6,g(x) = x² – 5x + 6
Dividend = x³ – 6x² + 11x – 6
Divisor = x² – 5x + 6
Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = x – 1

Remainder = 6x

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Question 6 F 

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:

p(x) = 6x³ + 13x² + x – 2 , g(x) = 2x + 1
Sol :
p(x) = 6x³ + 13x² + x – 2,g(x) = 2x + 1
Dividend = 6x³ + 13x² + x – 2
Divisor = 2x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process

Quotient = 3x² + 5x – 2
Remainder = 0

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Question 7 A 

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:

x – 2, x³ + 3x³ – 12x + 4
Sol :
Let us divide x³ + 3x² – 12x + 4 by x – 2
The division process is

Here, the remainder is 0, therefore x – 2 is a factor of x³ + 3x² – 12x + 4

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Question 7 B 

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:

x² + 3x + 1,3x⁴ + 5x³ – 7x² + 2x + 2
Sol :
Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1
The division process is

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

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Question 7 C 

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:

x² – 3x + 4,2x⁴ – 11x³ + 29x² – 30x + 29
Sol :
Let us divide 2x⁴ – 11x³ + 29x² – 30x + 29 by x² – 3x + 4
The division process is

Here, the remainder is 58x – 115 , therefore x² – 3x + 4 is not a factor of 2x⁴ – 11x³ + 29x² – 30x + 29

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Question 7 D 

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:

x² – 4x + 3,x³ – x³ – 3x⁴ – x + 3
Sol :
Let us divide x³ – 3x² – x + 3 by x² – 4x + 3
The division process is

Here, the remainder is 0, therefore x² – 4x + 3 is a factor of x³ – 3x² – x + 3

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Question 7 E 

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:

t – 1, t³ + t² – 2t + 1
Sol :
Let us divide t³ + t² – 2t + 1 by t – 1
The division process is

Here, the remainder is 1, therefore t – 1 is not a factor of t³ + t² – 2t + 1

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Question 7 F 

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:

t² – 5t + 6,t² + 11t – 6
Sol :
Let us divide t² + 11t – 6 by t² – 5t + 6
The division process is

Here, the remainder is 16t – 12,
Therefore, t² – 5t + 6 is not a factor of t² + 11t – 6

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Question 8 

Give examples of polynomials p(x), g(x), q(x) and r(x) satisfying the Division Algorithm

p(x) = g(x).q(x) + r(x),deg r(x)<deg g(x)
And also satisfying
(i) deg p(x) = deg q(x) + 1
(ii)deg q(x) = 1
(iii) deg q(x) = deg r(x) + 1
Sol :
(i)Let p(x) = 12x² + 8x + 25, g(x) = 4,
q(x) = 3x² + 2x + 6 , r(x) = 0
Here, degree p(x) = degree q(x) = 2
Now, g(x).q(x) + r(x) = (3x² + 2x + 6)×4 + 1
= 12x² + 8x + 24 + 1
= 12x² + 8x + 25
(ii) Let p(x) = t³ + t² – 2t, g(x) = t² + 2t,
q(x) = t – 1 , r(x) = 0
Here, degree q(x) = 1
Now, g(x).q(x) + r(x) = (t² + 2t)×(t – 1) + 0
= t³ – t² + 2t² – 2t
= t³ + t² – 2t
(iii) Let p(x) = x³ + x² + x + 1 , g(x) = x² – 1,
q(x) = x + 1 , r(x) = 2x + 2
Here, degree q(x) = degree r(x) + 1 = 1
Now, g(x).q(x) + r(x) = (x² – 1)×(x + 1) + 2x + 2
= x³ + x² – x – 1 + 2x + 2
= x³ + x² + x + 1

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Question 9 A 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

x³ – 6x² + 11x – 6;3
Sol :
Given zeroes is 3
So, (x – 3) is the factor of x³ – 6x² + 11x – 6
Let us divide x³ – 6x² + 11x – 6 by x – 3
The division process is

Here, quotient = x² – 3x + 2
= x² – 2x – x + 2
= x(x – 2) – 1(x – 2)
= (x – 1)(x – 2)
So, the zeroes are 1 and 2
Hence, all the zeroes of the given polynomial are 1, 2 and 3.

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Question 9 B 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

x⁴ – 8x³ + 23x² – 28x + 12;1,2
Sol :
Given zeroes are 1 and 2
So, (x – 1)and (x – 2) are the factors of x⁴ – 8x³ + 23x² – 28x + 12
⟹ (x – 1)(x – 2) = x² – 3x + 2 is a factor of given polynomial.
Consequently, x² – 3x + 2 is also a factor of the given polynomial.
Now, let us divide x⁴ – 8x³ + 23x² – 28x + 12 by x² – 3x + 2
The division process is

Here, quotient = x² – 5x + 6
= x² – 2x – 3x + 6
= x(x – 2) – 3(x – 2)
= (x – 3)(x – 2)
So, the zeroes are 3 and 2
Hence, all the zeroes of the given polynomial are 1, 2 ,2 and 3.

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Question 9 C 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

x³ + 2x² – 2; – 2
Sol :
Given zeroes is – 2
So, (x + 2) is the factor of x³ + 2x² – x – 2
Let us divide x³ + 2x² – x – 2 by x + 2
The division process is

Here, quotient = x² – 1
= (x – 1)(x + 1)
So, the zeroes are – 1 and 1
Hence, all the zeroes of the given polynomial are – 1, – 2 and 1.

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Question 9 D 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

x³ + 5x² + 7x + 3; – 3
Sol :
Given zeroes is – 3
So, (x + 3) is the factor of x³ + 5x² + 7x + 3
Let us divide x³ + 5x² + 7x + 3 by x + 3
The division process is

Here, quotient = x² + 2x + 1
= (x + 1)²
So, the zeroes are – 1 and – 1
Hence, all the zeroes of the given polynomial are – 1, – 1 and – 3.

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Question 9 E 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

x⁴ – 6x³ – 26x² + 138x – 35;2±√3
Sol :
x⁴ – 6x³ – 26x² + 138x – 35;2±√3
Given zeroes are 2 + √3 and 2 – √3
So, (x – 2 – √3)and (x – 2 + √3) are the factors of x⁴ – 6x³ – 26x² + 138x – 35
⟹ (x – 2 – √3)(x – 2 + √3)
= x² – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3
= x² – 4x + 1 is a factor of given polynomial.
Consequently, x² – 4x + 1 is also a factor of the given polynomial.
Now, let us divide x⁴ – 6x³ – 26x² + 138x – 35 by x² – 4x + 1

The division process is

Here, quotient = x² – 2x – 35
= x² – 7x + 5x – 35
= x(x – 7) + 5(x – 7)
= (x + 5)(x – 7)
So, the zeroes are – 5 and 7
Hence, all the zeroes of the given polynomial are – 5, 7, 2+√3 and 2-√3

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Question 9 F 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

x⁴ + x³ – 34x² – 4x + 120;2, – 2.
Sol :
x⁴ + x³ – 34x² – 4x + 120;2, – 2.
Given zeroes are – 2 and 2
So, (x + 2)and (x – 2) are the factors of x⁴ + x³ – 34x² – 4x + 120
⟹ (x + 2)(x – 2) = x² – 4 is a factor of given polynomial.
Consequently, x² – 4 is also a factor of the given polynomial.
Now, let us divide x⁴ + x³ – 34x² – 4x + 120 by x² – 4
The division process is

Here, quotient =x² + x – 30
= x² + 6x – 5x – 30
= x(x + 6) – 5(x + 6)
= (x + 6)(x – 5)
So, the zeroes are – 6 and 5
Hence, all the zeroes of the given polynomial are – 2 , – 6, 2 and 5.

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Question 9 G 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

2x⁴ + 7x³ – 19x² – 14x + 30;√2, – √2
Sol :
2x⁴ + 7x³ – 19x² – 14x + 30;√(2, – √2)
Given zeroes are √2 and – √2
So, (x – √2)and (x + √2) are the factors of 2x⁴ + 7x³ – 19x² – 14x + 30
⟹ (x – √2)(x + √2) = x² – 2 is a factor of given polynomial.
Consequently, x² – 2 is also a factor of the given polynomial.
Now, let us divide 2x⁴ + 7x³ – 19x² – 14x + 30 by x² – 2
The division process is

Here, quotient =2x² + 7x – 15
= 2x² + 10x – 3x – 15
= 2x(x + 5) – 3(x + 5)
= (2x – 3)(x + 5)
So, the zeroes are – 5 and $\frac{3}{2}$
Hence, all the zeroes of the given polynomial are – 5, – √2, √2 and $\frac{3}{2}$

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Question 9 H 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

2x⁴ – 9x³ + 5x² + 3x – 1;2±√3
Sol :
2x⁴ – 9x³ + 5x² + 3x – 1;2±√3
Given zeroes are 2 + √3 and 2 – √3
So, (x – 2 – √3)and (x – 2 + √3) are the factors of 2x⁴ – 9x³ + 5x² + 3x – 1
⟹ (x – 2 – √3)(x – 2 + √3)
= x² – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3
= x² – 4x + 1 is a factor of given polynomial.
Consequently, x² – 4x + 1 is also a factor of the given polynomial.
Now, let us divide 2x⁴ – 9x³ + 5x² + 3x – 1 by x² – 4x + 1
The division process is

Here, quotient = 2x² – x – 1
= 2x² – 2x + x – 1
= 2x(x – 1) + 1(x – 1)
= (2x + 1)(x – 1)
So, the zeroes are $-\frac{1}{2}$ and 1
Hence, all the zeroes of the given polynomial are $-\frac{1}{2}$, 1, 2 + √3 and 2 – √3

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Question 9 I 

Find all the zeroes of the polynomial given below having given numbers as its zeroes.

2x³ – 4x – x² + 2;√2, – √2

Sol :
2x³ – 4x – x² + 2;√2, – √2
Given zeroes are √2 and – √2
So, (x – √2) and (x + √2) are the factors of 2x³ – 4x – x² + 2
⟹ (x – √2)(x + √2) = x² – 2 is a factor of given polynomial.
Consequently, x² – 2 is also a factor of the given polynomial.
Now, let us divide 2x³ – 4x – x² + 2by x² – 2
The division process is

Here, quotient = 2x – 1

So, the zeroes is $\frac{1}{2}$
Hence, all the zeroes of the given polynomial are -√2,√2 and $\frac{1}{2}$

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Question 10 

Verify that 3,-1, $-\frac{1}{3}$ are the zeroes of the cubic polynomial p(x) = 3x² – 5x² – 11x – 3 and then verify the relationship between the zeroes and the coefficients.

Sol :
Let p(x) = 3 x³ – 5 x² – 11x – 3
Then, p( – 1) = 3( – 1)³ – 5( – 1)² – 11( – 1) – 3
= – 3 – 5 + 11 – 3
= 0

$p\left(-\frac{1}{3}\right)=3\left(-\frac{1}{3}\right)^{3}-5\left(-\frac{1}{3}\right)^{2}-11\left(-\frac{1}{3}\right)-3$

$=\left(-\frac{1}{9}\right)-\left(\frac{5}{9}\right)+\left(\frac{11}{3}\right)-3$

$=\left(\frac{-1-5+33-27}{9}\right)$

= 0

p(3) = 3(3)³ – 5(3)² – 11(3) – 3
= 81 – 45 – 33 – 3
= 0
Hence, we verified that 3, – 1 and $-\frac{1}{3}$ are the zeroes of the given polynomial.
So, we take α = 3, β = – 1, $\gamma=-\frac{1}{3}$
Verification
α + β + γ $=3+(-1)+\left(-\frac{1}{3}\right)=\left(\frac{5}{3}\right)$

$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{5}{3}$

αβ + βγ + γα $=(3)(-1)+(-1)\left(-\frac{1}{3}\right)+\left(-\frac{1}{3}\right)(3)$
$=\left(-\frac{11}{3}\right)$
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{-11}{3}=\left(-\frac{11}{3}\right)$
and αβγ $=3 \times-1 \times\left(-\frac{1}{3}\right)$
= 1
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-3}{3}=1$
Thus, the relationship between the zeroes and the coefficients is verified.

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Question 11 A 

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

x³ – 4x² + 5x – 2;2, 1, 1

Sol :
Let p(x) = x³ – 4x² + 5x – 2
Then, 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
Hence, 2, 1 and 1 are the zeroes of the given polynomial x³ – 4x² + 5x – 2.
Now, Let α = 2 , β = 1 and γ = 1
Then, α + β + γ = 2 + 1 + 1 = 4
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-4}{1}=4$
αβ + βγ + γα = (2)(1) + (1)(1) + (1)(2)
= 2 + 1 + 2
= 5
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{5}{1}=5$
and αβγ = 2 × 1 × 1
= 2
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is verified.

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Question 11 B 

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

x³ – 6x² + 11x – 6 ;1, 2, 3
Sol :
Let p(x) = x³ – 6x² + 11x – 6
Then, p(1) = (1)³ – 6(1)² + 11(1) – 6
= 1 – 6 + 11 – 6
= 0
p(2) = (2)³ – 6(2)² + 11(2) – 6
= 8 – 24 + 22 – 6
= 0
p(3) = (3)³ – 6(3)² + 11(3) – 6
= 27 – 54 + 33 – 6
= 0
Hence, 1, 2 and 3 are the zeroes of the given polynomial x³ – 6x² + 11x – 6.
Now, Let α = 1 , β = 2 and γ = 3
Then, α + β + γ = 1 + 2 + 3 = 6
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-6}{1}=6$
αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{11}{1}=11$
and αβγ = 1 × 2 × 3
= 6
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-6}{1}=6$
Thus, the relationship between the zeroes and the coefficients is verified.

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Question 11 C 

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

x³ + 2x² – x – 2; – 2 – 2, 1
Sol :
Let p(x) = x³ + 2x² – x – 2
Then, p( – 2) = ( – 2)³ + 2( – 2)² – ( – 2) – 2
= – 8 + 8 + 2 – 2
= 0
p(1) = (1)³ + 2(1)² – (1) – 2
= 1 + 2 – 1 – 2
= 0
Hence, – 2, – 2 and 1 are the zeroes of the given polynomial x³ + 2x² – x – 2.
Now, Let α = – 2 , β = – 2 and γ = 1
Then, α + β + γ = – 2 + ( – 2) + 1 = – 3
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{2}{1}=-2$
αβ + βγ + γα = ( – 2)( – 2) + ( – 2)(1) + (1)( – 2)
= 4 – 2 – 2
= 0
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{-1}{1}=-1$
and αβγ = ( – 2) × ( – 2) × 1
= 4
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is not verified.

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Question 11 D 

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

x³ + 5x² + 7x + 3 ; – 3, 2 – 1, – 1
Sol :
Let p(x) = x³ + 5x² + 7x + 3.
Then, p( – 1) = ( – 1)³ + 5( – 1)² + 7( – 1) + 3
= – 1 + 5 – 7 + 3
= 0
p( – 3) = ( – 3)³ + 5( – 3)² + 7( – 3) + 3
= – 27 + 45 – 21 + 3
= 0
Hence, – 1, – 1 and – 3 are the zeroes of the given polynomial x³ + 5x² + 7x + 3.
Now, Let α = – 1 , β = – 1 and γ = – 3
Then, α + β + γ = – 1 + ( – 1) + ( – 3) = – 5
$=-\frac{\text { Coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{5}{1}=-5$
αβ + βγ + γα = ( – 1)( – 1) + ( – 1)( – 3) + ( – 3)( – 1)
= 1 + 3 + 3
= 7
$=\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}=\frac{7}{1}=7$
and αβγ = ( – 1) × ( – 1) × ( – 3)
= – 3
$=-\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{3}}=-\frac{3}{1}=-3$
Thus, the relationship between the zeroes and the coefficients is verified.

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Question 12 

Find a cubic polynomial having 1, 2, 3 as its zeroes.

Sol :
Let the zeroes of the cubic polynomial be
α = 1, β = 2 and γ = 3
Then, α + β + γ = 1 + 2 + 3 = 6
αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
and αβγ = 1 × 2 × 3
= 6
Now, required cubic polynomial
= x³ – (α + β + γ) x² + (αβ + βγ + γα)x – αβγ
= x³ – (6) x² + (11)x – 6
= x³ – 6 x² + 11x – 6
So, x³ – 6 x² + 11x – 6 is the required cubic polynomial which satisfy the given conditions.

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Question 13 

Find a cubic polynomial having – 3, – 2, 2 as its zeroes.

Sol :
Let the zeroes of the cubic polynomial be
α = – 3, β = – 2 and γ = 2
Then, α + β + γ = – 3 + ( – 2) + 2
= – 3 – 2 + 2
= – 3
αβ + βγ + γα = ( – 3)( – 2) + ( – 2)(2) + (2)( – 3)
= 6 – 4 – 6
= – 4
and αβγ = ( – 3) × ( – 2) × 2
= 6 × 2
= 12
Now, required cubic polynomial
= x³ – (α + β + γ) x² + (αβ + βγ + γα)x – αβγ
= x³ – ( – 3) x² + ( – 4)x – 12
= x³ + 3 x² – 4x – 12
So, x³ + 3x^2 – 4x – 12 is the required cubic polynomial which satisfy the given conditions.

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Question 14 

Find a cubic polynomial with the sum of its zeroes are 0, – 7 and – 6 respectively.

Sol :
Let the zeroes be α, β and γ.
Then, we have
α + β + γ = 0
αβ + βγ + γα = – 7
and αβγ = – 6
Now, required cubic polynomial
= x³ – (α + β + γ) x² + (αβ + βγ + γα)x – αβγ
= x³ – (0) x² + ( – 7)x – ( – 6)
= x³ – 7x + 6
So, x³ – 7x + 6 is the required cubic polynomial.

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Question 15 A 

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:

2, – 7, – 14
Sol :
Let the zeroes be α, β and γ.
Then, we have
α + β + γ = 2
αβ + βγ + γα = – 7
and αβγ = – 14

Now, required cubic polynomial

=x³-(α+β+γ)x²+(αβ+βγ+γα)-αβγ

=x³-(2)x²+(-7x)-(-14)

Sox³ – 2x² – 7x + 14 is the required cubic polynomial.

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Question 15 B 

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:

$-4 \frac{1}{2}, \frac{1}{3}$
Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=-4$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2}$
$\alpha \beta \gamma=\frac{1}{3}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-(-4) x^{2}+\left(\frac{1}{2}\right) x-\left(\frac{1}{3}\right)$
$=\frac{6 x^{3}+24 x^{2}+3 x-2}{6}$
So, 6x³ + 24x² + 3x – 2 is the required cubic polynomial which satisfy the given conditions.

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Question 15 C 

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:

$\frac{5}{7}, \frac{1}{7}, \frac{1}{7}$

Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=\frac{5}{7}$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{7}$
$\alpha \beta \gamma=\frac{1}{7}$

Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-\left(\frac{5}{7}\right) x^{2}+\left(\frac{1}{7}\right) x-\left(\frac{1}{7}\right)$
$=\frac{7 x^{3}-5 x^{2}+x-1}{7}$
So, 7x³ – 5x² + x – 1 is the required cubic polynomial which satisfy the given conditions.

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Question 15 D 

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:

$\frac{2}{5}, \frac{1}{10}, \frac{1}{2}$

Sol :
Let the zeroes be α, β and γ.
Then, we have
$\alpha+\beta+\gamma=\frac{2}{5}$
$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{10}$
$\alpha \beta \gamma=\frac{1}{2}$
Now, required cubic polynomial
$=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$
$=x^{3}-\left(\frac{2}{5}\right) x^{2}+\left(\frac{1}{10}\right) x-\left(\frac{1}{2}\right)$
$=\frac{10 x^{3}-4 x^{2}+x-5}{10}$
So, 10x³ – 4x² + x – 5 is the required cubic polynomial which satisfy the given conditions.