The document RD Sharma Solutions: Exercise 2.2 & 2.3 - Polynomials Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

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** 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) = 2x ^{3}+ x^{2}– 5x + 2; 1/2, 1, – 2**

(ii) g(x) = x^{3}– 4x^{2} + 5x – 2; 2, 1, 1

Solution:

**(i) f(x) = 2x**^{3} + x^{2} – 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) = x**^{3} – 4x^{2}** + 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.**

Any cubic polynomial is of the form ax

= x

= x

= k[x

k is any non-zero real numbers.

**Question: 3****If the zeroes of the polynomial f(x) = 2x ^{3} – 15x^{2} + 37x – 30, find them.**

Let, α = a – d, β = a and γ = a + d be the zeroes of the polynomial.

f(x) = 2x

And, a (a

**Question: 4****Find the condition that the zeroes of the polynomial f(x) = x ^{3} + 3px^{2} + 3qx + r may be in A.P.**

f(x) = x^{3} + 3px^{2} + 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) = a^{3} + 3pa^{2} + 3qa + r = 0

Therefore, f(a) = 0f(a) = 0

⇒ a^{3} + 3pa^{2} + 3qa + r = 0

= ⇒ (−p)^{3} + 3p(−p)^{2} + 3q(−p) + r = 0

= − p^{3} + 3p^{3} – pq + r = 0

= 2p^{3} – pq + r = 0

**Question: 5****If zeroes of the polynomial f(x) = ax ^{3} + 3bx^{2} + 3cx + d are tin A.P., prove that 2b^{3} – 3abc + a^{2}d = 0.**

f(x) = x^{3} + 3px^{2} + 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(a^{2}) + 3b(a)^{2} + 3c(a) + d = 0

⇒ a(a^{2}) + 3b(a)^{2} + 3c(a) + d = 0

**Question: 6****If the zeroes of the polynomial f(x) = x ^{3} – 12x^{2} + 39x + k are in A.P., find the value of k.**

f(x) = x^{3} – 12x^{2} + 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) = a^{3} – 12a^{2} + 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) = x ^{3} – 6x^{2} + 11x – 6, g(x) = x^{2} + x + 1(ii) f(x) = 10x^{4} + 17x^{3} – 62x^{2} + 30x – 105, g(x) = 2x^{2} + 7x + 1(iii) f(x) = 4x^{3} + 8x^{2} + 8x + 7, g(x) = 2x^{2} – x + 1(iv) f(x) = 15x^{3} – 20x^{2} + 13x – 12, g(x) = x^{2} – 2x + 2**

**Question: 2****Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:(i) g(t) = t ^{2} – 3; f(t) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12(ii) g(x) = x^{2} – 3x + 1; f(x) = x^{5} – 4x^{3} + x^{2} + 3x + 1(iii) g(x) = 2x^{2} – x + 3; f(x) = 6x^{5} − x^{4} + 4x^{3} – 5x^{2} – x – 15Solution:**

**(i) g(t) = t ^{2} – 3; f(t) = 2t^{4} + 3t^{3} – 2t^{2}**

g(t) = t

g(x) = x^{2} – 3x + 1 g(x) = x^{2} – 3x + 1 f(x) = x^{5} – 4x^{3} + x^{2} + 3x + 1.Therefore, g(x) is not the factor of f(x).**(iii) g(x) = 2x**^{2} – x + 3; f(x) = 6x^{5} − x4 + 4x^{3} – 5x^{2}** – 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) = 2x ^{4} + x^{3} – 14x^{2} – 19x – 6, if two of its zeroes are -2 and -1.**

f(x) = 2x^{4} + x^{3} – 14x^{2} – 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) = x^{2} + x + 2x + 2 = x^{2} + 3x + 2

f(x) = 2x^{4} + x^{3} – 14x^{2} – 19x – 6 = (2x^{2} – 5x – 3)(x^{2} + 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) = x ^{3} + 13x^{2} + 32x + 20, if one of its zeroes is -2.**

f(x) = x^{3} + 13x^{2} + 32x + 20

Since, the zero of the polynomial is -2 so, it means its factor is (x + 2).

So, f(x) = x^{3} + 13x^{2} + 32x + 20 = (x^{2} + 11x + 10)(x + 2)

= (x^{2} + 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) = x ^{4} – 3x^{3} – x^{2} + 9x – 6, if the two of its zeroes are – √3 and √3.**

f(x) = x^{4} – 3x^{3} – x^{2} + 9x – 6 Since, two of the zeroes of polynomial are -√3 and √3 so,(x + √3)(x – √3) = x^{2}– 3x^{2} – 3

So, f(x) = x^{4} – 3x^{2} – x^{2} + 9x – 6 = (x^{2} – 3)(x^{2} – 3x + 2)

= (x + √3)(x – √3)x^{2}– 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) = 2x ^{4} – 2x^{3} – 7x^{2} + x – 1, if the two of its zeroes are – √(3/2) and √(3/2).**

f(x) = 2x^{4} – 2x^{3} – 7x^{2} + x – 1 Since, - √(3/2) and √(3/2) are the zeroes of the polynomial, so the factors are

So, f(x) = 2x^{4}– 2x^{3}–7x^{2}+ 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 x ^{4} + x^{3}^{ }– 34x^{2} – 4x + 120, if the two of its zeroes are 2 and – 2.**

x^{4} + x^{3} – 34x^{2} – 4x + 120 Since, the two zeroes of the polynomial given is 2 and – 2 So, factors are (x + 2)(x – 2) = x^{2} + 2x – 2x – 4 = x^{2} – 4x^{2} – 4

So, x^{4} + x^{3} – 34x^{2} – 4x + 120 = (x^{2} – 4)(x^{2} + x – 30)

= (x – 2)(x + 2)(x^{2} + 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 2x ^{4} + 7x^{3} – 19x^{2} – 14x + 30, if the two of its zeroes are √2 and - √2.**

2x^{4} + 7x^{3} – 19x^{2} – 14x + 302 Since, √2 and-√2 are the zeroes of the polynomial given. So, factors are (x + √2) and (x - √2)

= x^{2}+√2x – √2x – 2 = x^{2} – 2

So, 2x^{4} + 7x^{3} – 19x^{2} – 14x + 30 = (x2 – 2)(2x^{2} + 7x – 15)

= (2x^{2}+ 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) = 2x ^{3} + x^{2} – 6x – 3, if two of its zeroes are – √3 and √3.**

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

= (x^{2}– √3x + √3x – 3) = (x^{2} – 3)

So, f(x) = 2x^{3} + x2 – 6x – 3 = (x^{2} – 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) = x ^{3} + 3x^{2} – 2x – 6, if the two of its zeroes are √2 and - √2.**

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

= x^{2}+ √2x – √2x – 2 = x^{2} – 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) = x ^{4} + 2x^{3} – 2x^{2} + x − 1 so that the resulting polynomial is exactly divisible by g(x) = x^{2} + 2x − 3.**

f(x) = x^{4} + 2x^{3} – 2x^{2} + x − 1

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

**Question: 12****What must be subtracted from the polynomial f(x) = x ^{4} + 2x^{3} – 13x^{2} –12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x^{2} – 4x + 3.**

f(x) = x^{4} + 2x^{3} – 13x^{2} – 12x + 21

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

**Question: 13****Given that √2 is a zero of the cubic polynomial f(x) = 6x ^{3}+ √2x^{2}– 10x – 4√2, find its other two zeroes.**

f(x) = 6x^{3}+√2x^{2} – 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.**

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

So, the zeroes of the polynomial

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