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# Class 10 RD Sharma Solutions – Chapter 2 Polynomials – Exercise 2.3

• Last Updated : 20 May, 2021

### 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 â€“ 6x2 + 11x â€“ 6, g(x) = x2 + x + 1

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

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

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

Solution:

(i) Here we have to divide f(x) = x3 â€“ 6x2 + 11x â€“ 6 by g(x) = x2 + x + 1

So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 17x – 1

Quotient q(x) = x – 7

(ii) Here we have to divide f(x) = 10x4 + 17x3 â€“ 62x2 + 30x â€“ 105 by g(x) = 2x2 + 7x + 1

So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 53x – 1

Quotient q(x) = 5x2 – 9x – 2

(iii) Here we have to divide f(x) = 4x3 + 8x2 + 8x + 7 by g(x) =2x2 â€“ x + 1

So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 11x + 2

Quotient q(x) = 2x – 5

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

Here we have to divide f(x) =  15x3 â€“ 20x2 + 13x â€“ 12 by g(x) = x2 â€“ 2x + 2

So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 3x + 32

Quotient q(x) = 15x + 10

### Question 2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:

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

(ii) g(x) = x2 â€“ 3x + 1; f(x) = x5 â€“ 4x3 + x2 + 3x + 1

(iii) g(x) = 2x2 â€“ x + 3; f(x) = 6x5 âˆ’ x4 + 4x3 â€“ 5x2 â€“ x â€“ 15

Solution:

(i) Here, we have to check whether g(t) = t2 â€“ 3 is a factor of f(t) = 2t4 + 3t3 â€“ 2t2 â€“ 9t â€“ 12

So, by using division algorithm, we get

As the remainder left is 0.

Therefore,

g(t) = t2 â€“ 3 is a factor of f(t) = 2t4 + 3t3 â€“ 2t2 â€“ 9t â€“ 12

(ii) Here, we have to check whether g(x) = x2 â€“ 3x + 1 is a factor off(x) = x5 â€“ 4x3 + x2 + 3x + 1

So, by using division algorithm, we get

As the remainder left is 2.

Therefore,

g(x) = x2 â€“ 3x + 1 is not a factor of  f(x) = x5 â€“ 4x3 + x2 + 3x + 1

(iii) Here, we have to check whether g(x) = 2x2 â€“ x + 3 is a factor of f(x) = 6x5 âˆ’ x4 + 4x3 â€“ 5x2 â€“ x â€“ 15

So, by using division algorithm, we get

As the remainder left is 0.

Therefore,

g(x) = 2x2 â€“ x + 3 is a factor of f(x) = 6x5 âˆ’ x4 + 4x3 â€“ 5x2 â€“ x â€“ 15

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

Solution:

Given: f(x) = 2x4 + x3 â€“ 14x2 â€“ 19x â€“ 6

Here we have given the two zeroes of the polynomial that are -2 and -1,

Hence, its factors will be  (x + 2) and (x + 1)

Further,

(x + 2)(x + 1) = x2 + x + 2x + 2 = x2 + 3x + 2

So, by using division algorithm, we get

f(x) = 2x4 + x3 â€“ 14x2 â€“ 19x â€“ 6 = (2x2 â€“ 5x â€“ 3)(x2 + 3x + 2)

= (2x + 1)(x â€“ 3)(x + 2)(x + 1)

Hence, the factors of f(x) = 2x4 + x3 â€“ 14x2 â€“ 19x â€“ 6 are (2x + 1), (x â€“ 3), (x + 2), (x + 1)

Therefore, the zeroes of the polynomial are -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:

We have been given the zero of the polynomial f(x) = x3 + 13x2 + 32x + 20 is -2.

Hence, its factor is (x + 2).

So, by using division algorithm, we get

Thus,

f(x) = x3 + 13x2 + 32x + 20

= (x2 + 11x + 10)(x + 2)

= (x2 + 10x + x + 10)(x + 2)

= (x + 10)(x + 1)(x + 2)

Hence, the factors of f(x) = x3 + 13x2 + 32x + 20 are (x + 10), (x + 1), (x + 2)

Thus, the zeroes of the polynomial are -1, -10, -2.

### Question 5. Obtain all zeroes of the polynomial f(x) = x4 â€“ 3x3 â€“ x2 + 9x â€“ 6, if the two of its zeroes are -âˆš3 and âˆš3.

Solution:

Here, we are given two zeros of the polynomial f(x) = x4 â€“ 3x3 â€“ x2 + 9x â€“ 6 that are -âˆš3 and âˆš3.

Thus, the factors are (x + âˆš3â€‹)(x âˆ’ âˆš3â€‹) â‡’ x2 – 3.

So, by using division algorithm, we get

Hence,

f(x) = x4 â€“ 3x2 â€“ x2 + 9x â€“ 6 = (x2 â€“ 3)(x2 â€“ 3x + 2)

(x + âˆš3)(x â€“ âˆš3)(x2 â€“ 2x â€“ 2 + 2)

= (x + âˆš3)(x â€“ âˆš3)(x â€“ 1)(x â€“ 2)

Thus, the factors of f(x) = x4 â€“ 3x3 â€“ x2 + 9x â€“ 6 are (x + âˆš3)(x â€“ âˆš3)(x â€“ 1)(x â€“ 2).

Therefore, the zeroes of the polynomial are -âˆš3, âˆš3, 1, 2.

### Question 6. Obtain all zeroes of the polynomial f(x) = 2x4 â€“ 2x3 â€“ 7x2 + x â€“ 1, if the two of its zeroes are -âˆš3/2 and âˆš3/2.

Solution:

Here, we are given two zeros of the polynomial f(x) = 2x4 â€“ 2x3 â€“ 7x2 + x â€“ 1 that are -âˆš3/2 and âˆš3/2.

Thus, the factors are  â‡’ x2 – 3/2.

So, by using division algorithm, we get

Hence,

Factors of f(x) = 2x4 â€“ 2x3 â€“ 7x2 + x â€“ 1 are .

Thus, the zeroes of the polynomial are -1, 2, -âˆš3/2 and âˆš3/2.

### Question 7. Find all the zeroes of the polynomial x4 + x3 â€“ 34x2 â€“ 4x + 120, if the two of its zeroes are 2 and â€“ 2.

Solution:

Here, we are given two zeros of the polynomial x4 + x3 â€“ 34x2 â€“ 4x + 120 that are 2 and -2.

Thus, the factors are (x + 2)(x – 2)â‡’ x2 – 4.

So, by using division algorithm, we get

Hence,

x4 + x3 â€“ 34x2 â€“ 4x + 120 = (x2 â€“ 4)(x2 + x â€“ 30)

= (x â€“ 2)(x + 2)(x2 + 6x â€“ 5x â€“ 30)

=  (x â€“ 2)(x + 2)(x + 6)(x â€“ 5)

So, the factors of x4 + x3 â€“ 34x2 â€“ 4x + 120 are (x â€“ 2), (x + 2), (x + 6), (x â€“ 5)

Thus, the zeroes of the polynomial = x = 2, â€“ 2, â€“ 6, 5

### Question 8. Find all the zeroes of the polynomial 2x4 + 7x3 â€“ 19x2 â€“ 14x + 30, if the two of its zeroes are âˆš2 and -âˆš2.

Solution:

Here, we are given two zeros of the polynomial 2x4 + 7x3 â€“ 19x2 â€“ 14x + 302 that are âˆš2 and -âˆš2.

Thus, the factors are (x + âˆš2)(x – âˆš2) â‡’ x2 – 2.

So, by using division algorithm, we get

Hence,

2x4 + 7x3 â€“ 19x2 â€“ 14x + 30 = (x2 â€“ 2)(2x2 + 7x â€“ 15)

= (2x2 + 10x â€“ 3x â€“ 15)(x + âˆš2)(x â€“ âˆš2)

= (2x â€“ 3)(x + 5)(x + âˆš2)(x â€“ âˆš2)

So, the factors of 2x4 + 7x3 â€“ 19x2 â€“ 14x + 30 are (2x â€“ 3), (x + 5), (x + âˆš2), (x â€“ âˆš2)

Thus, the zeroes of the polynomial is âˆš2, -âˆš2, -5, 3/2.

### Question 9. Find all the zeroes of the polynomial f(x) = 2x3 + x2 â€“ 6x â€“ 3, if two of its zeroes are -âˆš3 and âˆš3.

Solution:

Here, we are given two zeros of the polynomial f(x) = 2x3 + x2 â€“ 6x â€“ 3 that are -âˆš3 and âˆš3.

Thus, the factors are (x + âˆš3)(x – âˆš3) â‡’ x2 – 3.

So, by using division algorithm, we get

Hence,

f(x) = 2x3 + x2 â€“ 6x â€“ 3

= (x2 â€“ 3)(2x + 1)

= (x + âˆš3)(x – âˆš3)(2x + 1)

Factors of f(x) = 2x3 + x2 â€“ 6x â€“ 3 are (x + âˆš3), (x – âˆš3), 2x + 1

Thus, the zeroes for the given polynomial are âˆš3, -âˆš3, -1/2

### Question 10. Find all the zeroes of the polynomial f(x) = x3 + 3x2 â€“ 2x â€“ 6, if the two of its zeroes are âˆš2 and -âˆš2.

Solution:

Here, we are given two zeros of the polynomial f(x) = x3 + 3x2 â€“ 2x â€“ 6 that are âˆš2 and -âˆš2.

Thus, the factors are (x + âˆš2)(x – âˆš2)â‡’ x2 – 2.

So, by using division algorithm, we get

Hence,

f(x) = x3 + 3x2 â€“ 2x â€“ 6

= (x2 â€“ 2)(x + 3)

= (x + âˆš2)(x – âˆš2)(x + 3)

Factors of f(x) = x3 + 3x2 â€“ 2x â€“ 6 are (x + âˆš2), (x – âˆš2), (x + 3)

Thus, the zeroes of the given polynomial is -âˆš2, âˆš2, and â€“ 3.

### Question 11. What must be added to the polynomial f(x) = x4 + 2x3 â€“ 2x2 + x âˆ’ 1 so that the resulting polynomial is exactly divisible by g(x) = x2 + 2x âˆ’ 3.

Solution:

Here we have to add to the polynomial f(x) = x4 + 2x3 â€“ 2x2 + x âˆ’ 1 so that the

resulting polynomial is exactly divisible by g(x) = x2 + 2x âˆ’ 3.

So, divide f(x) = x4 + 2x3 â€“ 2x2 + x âˆ’ 1 by g(x) = x2 + 2x âˆ’ 3 to get the answer.

As the remainder left is (x â€“ 2) to get the resulting polynomial exactly divisible by

g(x) = x2 + 2x âˆ’ 3 we must add (x – 2) to f(x) = x4 + 2x3 â€“ 2x2 + x âˆ’ 1.

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

Solution:

Here we have to subtract to the polynomial f(x) = x4 + 2x3 â€“ 13x2 â€“ 12x + 21

so that the resulting polynomial is exactly divisible by g(x) = x2 – 4x + 3.

So, divide f(x) = x4 + 2x3 â€“ 13x2 â€“ 12x + 21 by  g(x) = x2 – 4x + 3 to get the answer.

As the remainder left is (2x – 3) to get the resulting polynomial exactly divisible by

g(x) = x2 – 4x + 3 we must add (2x – 3) to f(x) = x4 + 2x3 â€“ 13x2 â€“ 12x + 21.

### Question 13. Given that âˆš2 is a zero of the cubic polynomial f(x) = 6x3 + âˆš2x2â€“ 10x â€“ 4âˆš2, find its other two zeroes.

Solution:

Here, we are given that âˆš2 is the zero of the cubic polynomial

f(x) = 6x3 + âˆš2x2â€“ 10x â€“ 4âˆš2, thus, factor of the polynomial is (x – âˆš2)

So, by using division algorithm, we get

Hence,

f(x) = 6x3 + âˆš2x2 â€“ 10x â€“ 4âˆš2

= (x – âˆš2)(6x2 + 7âˆš2x + 4)

= (x – âˆš2)(6x2 + 4âˆš2x + 3âˆš2x + 4)

= (x – âˆš2)(3x + 2âˆš2)(2x + âˆš2)

The factors of f(x) = 6x3 + âˆš2x2â€“ 10x â€“ 4âˆš2 are (x – âˆš2), (3x + 2âˆš2), (2x + âˆš2)

Therefore, the zeros of the polynomial are -2âˆš2/3, -âˆš2/2, âˆš2

### Question 14. Given that x – âˆš5 is a factor of the cubic polynomial x3 – 3âˆš5x2 + 13x – 3âˆš5, find all the zeroes of the polynomial.

Solution:

Here, we have x – âˆš5 as factor of the cubic polynomial x3 – 3âˆš5x2 + 13x – 3âˆš5

To find all the zeros of the polynomial, we have to divide the polynomial x3 – 3âˆš5x2 + 13x – 3âˆš5 by the factor x – âˆš5

Hence,

x3 – 3âˆš5x2 + 13x – 3âˆš5

= (x – âˆš5)(x2 – 2âˆš5 + 3)

= (x – âˆš5)(x – (âˆš5 + âˆš2))(x – (âˆš5 – âˆš2))

So, the factors of the cubic polynomial x3 – 3âˆš5x2 + 13x – 3âˆš5 are (x – âˆš5), (x – (âˆš5 + âˆš2)), (x – (âˆš5 – âˆš2))

Therefore, the zero of the polynomial are âˆš5, (âˆš5 – âˆš2), (âˆš5 + âˆš2)

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