Polynomials (Exercise 2.2 & 2.3) RD Sharma Solutions | Advance Learner Course: Mathematics (Maths) Class 9 PDF Download

                                                         Exercise 2.2

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

(i) f(x) = 2x³+ x²– 5x + 2; 1/2, 1, – 2
(ii)  g(x) = x³– 4x² + 5x – 2; 2, 1, 1
Solution: 

(i) f(x) = 2x³ + x² – 5x + 2;  1/2, 1, – 2
(a) By putting x = 1/2 in the above equation, we will get

(b) By putting x = 1 in the above equation, we will get
f(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
(c) By putting x = -2 in the above equation, we will get
f(−2) = 2(−2)³ + (−2)² – 5(−2) + 2
= -16 + 4 + 10 + 2 = – 16 + 16 = 0
Now,
Sum of zeroes = α + β + γ = - b/a

Product of the zeroes = αβ + βγ + αγ = c/a

Hence, verified.

(ii) g(x) = x³ – 4x² + 5x – 2; 2, 1, 1
(a) By putting x = 2 in the given equation, we will get
g(2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 18 – 18 = 0
(b) By putting x = 1 in the given equation, we will get
g(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
Now,
Sum of zeroes= α + β + γ =-b/a
⇒ 2 + 1 + 1 = −(−4) 
4 = 4
Product of the zeroes = αβ + βγ + αγ = c/a
2 × 1 + 1 × 1 + 1 × 2 = 5
2 + 1 + 2 = 5
5 = 5
αβγ  = –(−2)
2 × 1 × 1 = 2
2 = 2
Hence, verified.

Question: 2
Find a cubic polynomial with the sum, sum of the product of its zeroes is taken two at a time, and product of its zeroes as 3, – 1 and – 3 respectively.
Solution:
Any cubic polynomial is of the form ax³ + bx² + cx + d:
= x³ – (sum of the zeroes) x² + (sum of the products of its zeroes) x – (product of the zeroes)
= x³ – 3x² + (−1)x + (−3)
= k[x³ – 3x² – x – 3]
k is any non-zero real numbers.

Question: 3
If the zeroes of the polynomial f(x) = 2x³ – 15x² + 37x – 30, find them.
Solution: 
Let, α = a – d, β = a and γ = a + d be the zeroes of the polynomial.
f(x) = 2x³ – 15x² + 37x – 30

And, a (a² + d²) = 15

Question: 4
Find the condition that the zeroes of the polynomial f(x) = x³ + 3px² + 3qx + r may be in A.P.
Solution: 

f(x) = x³ + 3px² + 3qx + r
Let, a – d, a, a + d be the zeroes of the polynomial.
Then,
The sum of zeroes = – b/a
a + a – d + a + d = -3p 3a = -3p a = -p Since, a is the zero of the polynomial f(x),
Therefore, f(a) = 0
f(a) = a³ + 3pa² + 3qa + r = 0
Therefore, f(a) = 0f(a) = 0
⇒ a³ + 3pa² + 3qa + r = 0
= ⇒ (−p)³ + 3p(−p)² + 3q(−p) + r = 0
= − p³ + 3p³ – pq + r = 0
= 2p³ – pq + r = 0

Question: 5
If zeroes of the polynomial f(x) = ax³ + 3bx² + 3cx + d are tin A.P., prove that 2b³ – 3abc + a²d = 0.
Solution:

f(x) = x³ + 3px² + 3qx + r
Let, a – d, a, a + d be the zeroes of the polynomial.
Then,
The sum of zeroes = - b/a
a + a – d + a + d = – 3b/a

Since, f(a) = 0
⇒ a(a²) + 3b(a)² + 3c(a) + d = 0
⇒ a(a²) + 3b(a)² + 3c(a) + d = 0 

Question: 6
If the zeroes of the polynomial f(x) = x³ – 12x² + 39x + k are in A.P., find the value of k.
Solution: 

f(x) = x³ – 12x² + 39x + k
Let, a-d, a, a + d be the zeroes of the polynomial f(x).
The sum of the zeroes = 12
3a = 12
a = 4
Now,
f(a) = 0
f(a) = a³ – 12a² + 39a + k f(4) = 43 – 12(4)² + 39(4) + k = 0
f(4) = 43 –12(4)² + 39(4) + k = 0
64 – 192 + 156 + k = 0
k = – 28

                                                        Exercise 2.3

Question: 1
Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:
(i) f(x) = x³ – 6x² + 11x – 6, g(x) = x² + x + 1
(ii) f(x) = 10x⁴ + 17x³ – 62x² + 30x – 105, g(x) = 2x² + 7x + 1
(iii) f(x) = 4x³ + 8x² + 8x + 7, g(x) = 2x² – x + 1
(iv) f(x) = 15x³ – 20x² + 13x – 12, g(x) = x² – 2x + 2
Solution: 
(i) f(x) = x³ – 6x² + 11x – 6 and g(x) = x² + x + 1

(ii) f(x) = 10x⁴ + 17x³ – 62x² + 30x – 105, g(x) = 2x² + 7x + 1

(iii) f(x) = 4x3 + 8x2 + 8x + 7, g(x) = 2x2 – x + 1

(iv) f(x) = 15x³ – 20x² + 13x – 12, g(x) = x² – 2x + 2

Question: 2
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
(i) g(t) = t² – 3; f(t) = 2t⁴ + 3t³ – 2t² – 9t – 12
(ii) g(x) = x² – 3x + 1; f(x) = x⁵ – 4x³ + x² + 3x + 1
(iii) g(x) = 2x² – x + 3; f(x) = 6x⁵ − x⁴ + 4x³ – 5x² – x – 15
Solution:

(i) g(t) = t² – 3; f(t) = 2t⁴ + 3t³ – 2t² – 9t – 12

g(t) = t² – 3g(t) = t² – 3 f(t) = 2t⁴ + 3t³ – 2t² – 9t Therefore, g(t) is the factor of f(t).
(ii) g(x) = x² – 3x + 1; f(x) = x⁵ – 4x³ + x² + 3x + 1

g(x) = x² – 3x + 1 g(x) = x² – 3x + 1 f(x) = x⁵ – 4x³ + x² + 3x + 1.Therefore, g(x) is not the factor of f(x).
(iii) g(x) = 2x² – x + 3; f(x) = 6x⁵ − x4 + 4x³ – 5x² – x – 15

g(x) = 2x2 – x + 3 g(x) = 2x2 – x + 3 f(x) = 6x5 − x4 + 4x3 – 5x2 – x – 15

Question: 3
Obtain all zeroes of the polynomial f(x) = f(x) = 2x⁴ + x³ – 14x² – 19x – 6, if two of its zeroes are -2 and -1.
Solution:

f(x) = 2x⁴ + x³ – 14x² – 19x – 6
If the two zeroes of the polynomial are -2 and -1, then its factors are (x + 2) and (x + 1)
(x + 2)(x + 1) = x² + x + 2x + 2 = x² + 3x + 2

f(x) = 2x⁴ + x³ – 14x² – 19x – 6 = (2x² – 5x – 3)(x² + 3x + 2)
= (2x + 1)(x – 3)(x + 2)(x + 1)
Therefore, zeroes of the polynomial = - 1/2, 3, -2 , -1

Question: 4
Obtain all zeroes of f(x) = x³ + 13x² + 32x + 20, if one of its zeroes is -2.
Solution:

f(x) = x³ + 13x² + 32x + 20
Since, the zero of the polynomial is -2 so, it means its factor is (x + 2).

So, f(x) = x³ + 13x² + 32x + 20 =  (x² + 11x + 10)(x + 2)
= (x² + 10x + x + 10)(x + 2)

= (x + 10)(x + 1)(x + 2)
Therefore, the zeroes of the polynomial are – 1, – 10, – 2.

Question: 5
Obtain all zeroes of the polynomial f(x) = x⁴ – 3x³ – x² + 9x – 6, if the two of its zeroes are – √3 and √3.
Solution:

f(x) = x⁴ – 3x³ – x² + 9x – 6 Since, two of the zeroes of polynomial are -√3 and √3 so,(x + √3)(x – √3) = x²– 3x² – 3

So, f(x) = x⁴ – 3x² – x² + 9x – 6 = (x² – 3)(x² – 3x + 2)
= (x + √3)(x – √3)x²– 2x – 2 + 2)

= (x + √3)(x – √3)(x – 1)(x – 2)
Therefore, the zeroes of the polynomial are - √3, √3, 1, 2.

Question: 6
Obtain all zeroes of the polynomial f(x) = 2x⁴ – 2x³ – 7x² + x – 1, if the two of its zeroes are – √(3/2) and √(3/2).
Solution:

f(x) = 2x⁴ – 2x³ – 7x² + x – 1 Since, - √(3/2)  and √(3/2) are the zeroes of the polynomial, so the factors are


So, f(x) = 2x⁴– 2x³–7x²+ x – 1

Therefore, the zeroes of the polynomial = x = -1, 2, -√(3/2) and √(3/2).

Question: 7
Find all the zeroes of the polynomial x⁴ + x³ – 34x² – 4x + 120, if the two of its zeroes are 2 and – 2.
Solution:

x⁴ + x³ – 34x² – 4x + 120 Since, the two zeroes of the polynomial given is 2 and – 2 So, factors are (x + 2)(x – 2) =  x² + 2x – 2x – 4 = x² – 4x² – 4
    
So, x⁴ + x³ – 34x² – 4x + 120 = (x² – 4)(x² + x – 30)
= (x – 2)(x + 2)(x² + 6x – 5x – 30)
=  (x – 2)(x + 2)(x + 6)(x – 5)
Therefore, the zeroes of the polynomial = x = 2, – 2, – 6, 5

Question: 8 
Find all the zeroes of the polynomial 2x⁴ + 7x³ – 19x² – 14x + 30, if the two of its zeroes are √2 and - √2.
Solution:

2x⁴ + 7x³ – 19x² – 14x + 302 Since, √2 and-√2 are the zeroes of the polynomial given. So, factors are (x + √2) and (x - √2)
= x²+√2x – √2x – 2 = x² – 2

So, 2x⁴ + 7x³ – 19x² – 14x + 30 = (x2 – 2)(2x² + 7x – 15)
= (2x²+ 10x – 3x – 15)(x + √2)(x – √2)
= (2x – 3)(x + 5)(x + √2)(x – √2)
Therefore, the zeroes of the polynomial is √2,- √2,-5, 3/2.

Question: 9
Find all the zeroes of the polynomial f(x) = 2x³ + x² – 6x – 3, if two of its zeroes are – √3 and √3.
Solution: 

f(x) = 2x³ + x² – 6x – 3 Since, -√3 and √3 are the zeroes of the given polynomial So, factors are (x - √3) and (x + √3)
= (x²– √3x + √3x – 3) = (x² – 3)

So, f(x) = 2x³ + x2 – 6x – 3 = (x² – 3)(2x + 1)
= (x – √3)(x + √3)(2x + 1)
Therefore, set of zeroes for the given polynomial = √3,– √3, –1/2

Question: 10
Find all the zeroes of the polynomial f(x) = x³ + 3x² – 2x – 6, if the two of its zeroes are √2 and - √2.
Solution:

f(x) = x³ + 3x² – 2x – 6 Since, √2 and -√2 are the two zeroes of the given polynomial. So, factors are (x + √2) and (x - √2)
= x²+ √2x – √2x – 2 = x² – 2

By division algorithm, we have:
f(x) = x3 + 3x2 – 2x – 6 = (x2 – 2)(x + 3)
= (x – √2)(x + √2)(x + 3)
Therefore, the zeroes of the given polynomial is - √2, √2 and – 3.

Question: 11
What must be added to the polynomial f(x) = x⁴ + 2x³ – 2x² + x − 1 so that the resulting polynomial is exactly divisible by g(x) = x² + 2x − 3.
Solution:

f(x) = x⁴ + 2x³ – 2x² + x − 1

We must add (x – 2) in order to get the resulting polynomial exactly divisible by g(x) = x² + 2x − 3.

Question: 12
What must be subtracted from the polynomial f(x) = x⁴ + 2x³ – 13x² –12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x² – 4x + 3.
Solution:

f(x) = x⁴ + 2x³ – 13x² – 12x + 21

We must subtract (2x – 3) in order to get the resulting polynomial exactly divisible by g(x) = x² – 4x + 3.

Question: 13
Given that √2 is a zero of the cubic polynomial f(x) = 6x³+ √2x²– 10x – 4√2, find its other two zeroes.
Solution:

f(x) = 6x³+√2x² – 10x – 4√2 Since, √2 is a zero of the cubic polynomial So, factor is (x–√2)

Question: 14
Given that x – √5 is a factor of the cubic polynomial  find all the zeroes of the polynomial.
Solution: 

In the question, it’s given that x – √5 is a factor of the cubic polynomial.


So, the zeroes of the polynomial