RD Sharma Solutions Class 10 Polynomials Exercise 2.3

RD Sharma Class 10 Solutions Chapter 2 Ex 2.3 PDF Free Download

Exercise – 2.3

 

Important Points to Note:

  • Division algorithms fall into two main categories: slow division and fast division.
  • Slow division algorithms produce one digit of the final quotient per iteration.
  • Fast division methods start with an approximation to the final quotient and produce twice as many digits of the final quotient on each iteration.
  • Discussion will refer to the form \(\frac{N}{D} = \left ( Q , R \right )\) where, N = Numerator (dividend) & D = Denominator (divisor) is the input, and Q = Quotient & R = Remainder is the output.

 

Q.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) \(f\left ( x \right ) = x ^{3} – 6x ^{2} + 11x – 6 , \; g\left ( x \right ) = x^{2} + x + 1\)

 

(ii) f(x) = \(10x^{4} + 17x^{3} – 62 x^{2} + 30x – 105 , \; g (x) = 2x^{2}\) + 7x + 1

 

(iii) \(f\left ( x \right ) = 4x ^{3} + 8 x^{2} + 8x + 7 , \; g\left ( x \right ) = 2x^{2} – x + 1\)

 

(iv) \(f\left ( x \right ) = 15x ^{3} – 20 x^{2} + 13x – 12 , \; g\left ( x \right ) = x^{2} – 2x + 2\)

 

Solution:

(i) Given:

\(f\left ( x \right ) = x ^{3} – 6x ^{2} + 11x – 6 \) and \( g\left ( x \right ) = x^{2} + x + 1\)

 

1

 

(ii) Given:

\(f\left ( x \right ) = 10x ^{4} + 17x ^{3} – 62 x^{2} + 30x – 105 , \; g\left ( x \right ) = 2x^{2} + 7x + 1\)

\(f\left ( x \right ) = 10x ^{4} + 17x ^{3} – 62 x^{2} + 30x – 105\)

\( g\left ( x \right ) = 2x^{2} + 7x + 1\)

 

2

 

(iii) Given:

\(f\left ( x \right ) = 4x ^{3} + 8 x^{2} + 8x + 7 , \; g\left ( x \right ) = 2x^{2} – x + 1\)

 

3

 

\(f\left ( x \right ) = 4x ^{3} + 8 x^{2} + 8x + 7\)

\( g\left ( x \right ) = 2x^{2} – x + 1\)

 

(iv) Given:

\(f\left ( x \right ) = 15x ^{3} – 20 x^{2} + 13x – 12 , \; g\left ( x \right ) = x^{2} – 2x + 2\)

 

4

 

\(f\left ( x \right ) = 15x ^{3} – 20 x^{2} + 13x – 12\)

\( g\left ( x \right ) = x^{2} – 2x + 2\)

 

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

 

(i) \(g\left ( t \right ) = t^{2} – 3 \; ; f\left ( t \right ) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12\)

(ii) \(g\left ( x \right ) = x^{2} – 3x  + 1\; ; f\left ( x \right ) = x^{5} –  4x^{3} + x^{2} + 3x + 1\)

(iii) \(g\left ( x \right ) = 2 x^{2} – x  + 3\; ; f\left ( x \right ) = 6x^{5}- x^{4} +  4x^{3} – 5x^{2} – x  – 15\)

 

Solution:

(i) \(g\left ( t \right ) = t^{2} – 3 \; ; f\left ( t \right ) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12\)

 

5

 

\(g\left ( t \right ) = t^{2} – 3\)

\( f\left ( t \right ) = 2t^{4} + 3t^{3} – 2t^{2} – 9t\)

Therefore, g(t) is the factor of f(t).

 

(ii) \(g\left ( x \right ) = x^{2} – 3x  + 1\; ; f\left ( x \right ) = x^{5} –  4x^{3} + x^{2} + 3x + 1\)

 

6

 

\(g\left ( x \right ) = x^{2} – 3x  + 1\)

\( f\left ( x \right ) = x^{5} –  4x^{3} + x^{2} + 3x + 1\)

Therefore, g(x) is not the factor of f(x).

 

(iii) \(g\left ( x \right ) = 2 x^{2} – x  + 3\; ; f\left ( x \right ) = 6x^{5}- x^{4} +  4x^{3} – 5x^{2} – x  – 15\)

 

7

 

\(g\left ( x \right ) = 2 x^{2} – x  + 3\)

\( f\left ( x \right ) = 6x^{5}- x^{4} +  4x^{3} – 5x^{2} – x  – 15\)

 

Q.3: Obtain all zeroes of the polynomial f(x) = \(f\left ( x \right ) = 2x^{4} + x^{3} – 14x^{2} – 19x – 6\), if two of its zeroes are -2 and -1.

 

Solution:

\(f\left ( x \right ) = 2x^{4} + x^{3} – 14x^{2} – 19x – 6\)

If the two zeroes of the polynomial are -2 and -1, then its factors are ( x + 2 ) and ( x + 1 )

\(\left ( x + 2 \right ) \left ( x + 1 \right ) = x^{2} + x + 2x + 2 = x ^{2} + 3x + 2\)

 

8

 

\(f\left ( x \right ) = 2x^{4} + x^{3} – 14x^{2} – 19x – 6\) = \(\left ( 2x^{2} – 5x – 3 \right ) \left ( x^{2} + 3x + 2 \right )\)

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

Therefore, zeroes of the polynomial = \(\frac{-1}{2}\) , 3, -2 , -1

 

Q-4: Obtain all zeroes of f(x) = \(f\left ( x \right ) = x^{3} + 13x^{2} + 32x + 20\), if one of its zeroes is -2.

Solution:

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

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

 

9

 

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

= \(\left (x^{2} + 10x + x + 10 \right ) \left ( x + 2 \right )\)

= \(\left ( x + 10 \right ) \left ( x + 1 \right ) \left ( x + 2 \right )\)

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

 

Q-5: Obtain all zeroes of the polynomial \(f\left ( x \right ) = x^{4} – 3x^{3} – x^{2} + 9x – 6\), if the two of its zeroes are \(-\sqrt{3} \; and \; \sqrt{3}\).

Solution:

\(f\left ( x \right ) = x^{4} – 3x^{3} – x^{2} + 9x – 6\)

Since, two of the zeroes of polynomial are \(-\sqrt{3} \; and \; \sqrt{3}\) so, \(\left ( x + \sqrt{3} \right ) \left ( x – \sqrt{3} \right )\) = \(x^{2} – 3\)

 

10

 

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

\(\left ( x + \sqrt{3} \right ) \left ( x – \sqrt{3} \right )\) \(\left ( x^{2} – 2x – 2 + 2 \right )\)

= \(\left ( x + \sqrt{3} \right ) \left ( x – \sqrt{3} \right )\) \(\left ( x – 1 \right ) \left ( x – 2 \right )\)

Therefore, the zeroes of the polynomial are \(-\sqrt{3} ,  \; \sqrt{3}\) , 1, 2.

 

Q-6: Obtain all zeroes of the polynomial \(f\left ( x \right ) = 2x^{4} – 2x^{3}  – 7x^{2}  + x – 1\), if the two of its zeroes are \(- \sqrt{\frac{3}{2}} \; and \; \sqrt{\frac{3}{2}}\).

 

Solution:

\(f\left ( x \right ) = 2x^{4} – 2x^{3}  – 7x^{2}  + x – 1\)

Since, \(- \sqrt{\frac{3}{2}} \; and \; \sqrt{\frac{3}{2}}\) are the zeroes of the polynomial, so the factors are

\(\left ( x – \sqrt{\frac{3}{2}} \right ) \; and \; \left ( x + \sqrt{\frac{3}{2}} \right )\)

 

11

 

So, \(f\left ( x \right ) = 2x^{4} – 2x^{3}  – 7x^{2}  + x – 1\) = \(\left ( x – \sqrt{\frac{3}{2}} \right ) \; \left ( x + \sqrt{\frac{3}{2}} \right )\) \(\left ( 2x^{2} – 2x – 4 \right )\)

\(\left ( x – \sqrt{\frac{3}{2}} \right ) \; \left ( x + \sqrt{\frac{3}{2}} \right )\) \(\left ( 2x^{2} – 4x + 2x – 4 \right )\)

= \(\left ( x – \sqrt{\frac{3}{2}} \right ) \; \left ( x + \sqrt{\frac{3}{2}} \right )\) \(\left ( x + 2 \right ) \left ( x – 2 \right )\)

Therefore, the zeroes of the polynomial = x = -1, 2, \(- \sqrt{\frac{3}{2}} \; and \; \sqrt{\frac{3}{2}}\).

 

Q.7: Find all the zeroes of the polynomial \(x^{4} + x^{3} – 34 x ^{2} – 4x + 120\), if the two of its zeroes are 2 and -2.

 

Solution:

\(x^{4} + x^{3} – 34 x ^{2} – 4x + 120\)

Since, the two zeroes of the polynomial given is 2 and -2

So, factors are ( x + 2 )( x – 2 ) = \(x^{2} + 2x – 2x – 4\) = \(x^{2} – 4\)

 

12

 

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

=\(\left ( x – 2 \right ) \left ( x + 2 \right ) \left ( x^{2} + 6x – 5x – 30 \right )\)

= \(\left ( x – 2 \right ) \left ( x + 2 \right ) \left ( x + 6 \right ) \left ( x – 5 \right )\)

Therefore, the zeroes of the polynomial = x = 2, -2, -6, 5

 

Q-8: Find all the zeroes of the polynomial \( 2x^{4} + 7x^{3} – 19 x^{2}\) – 14x + 30, if the two of its zeroes are \(\sqrt{2} \; \; and \; -\sqrt{2}\).

 

Solution:

\( 2x^{4} + 7x^{3} – 19 x ^{2} – 14x + 30\)

Since, \(\sqrt{2} \; \; and \; -\sqrt{2}\) are the zeroes of the polynomial given.

So, factors are \(\left (x + \sqrt{2} \right ) \; \; and \; \left (x -\sqrt{2} \right )\) = \(x^{2} + \sqrt{2}x – \sqrt{2}x – 2\) = \(x^{2} – 2\)

 

13

 

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

= \(\left ( 2x^{2} + 10x – 3x – 15 \right ) \left ( x + \sqrt{2} \right ) \left ( x – \sqrt{2} \right )\)

= \(\left ( 2x – 3 \right ) \left ( x + 5 \right ) \left ( x + \sqrt{2} \right ) \left ( x – \sqrt{2} \right )\)

Therefore, the zeroes of the polynomial is \(\sqrt{2} , -\sqrt{2} , -5, \frac{3}{2}\).

 

Q-9: Find all the zeroes of the polynomial \(f\left ( x \right ) = 2x^{3} + x^{2} – 6x – 3\) , if two of its zeroes are \(-\sqrt{3} \; and \; \sqrt{3}\).

 

Solution:

\(f\left ( x \right ) = 2x^{3} + x^{2} – 6x – 3\)

Since, \(-\sqrt{3} \; and \; \sqrt{3}\) are the zeroes of the given polynomial

So, factors are \(\left (x -\sqrt{3} \right ) \; and \; \left (x + \sqrt{3} \right )\) = \(\left ( x ^{2} – \sqrt{3}x + \sqrt{3}x – 3 \right )\) = \(\left ( x ^{2} – 3 \right )\)

 

14

 

So, \(f\left ( x \right ) = 2x^{3} + x^{2} – 6x – 3\) = \(\left ( x ^{2} – 3 \right ) \left ( 2x + 1 \right )\)

= \(\left ( x – \sqrt{3} \right ) \left ( x + \sqrt{3} \right ) \left ( 2x + 1 \right )\)

Therefore, set of zeroes for the given polynomial = \(\sqrt{3} , – \sqrt{3} , \frac{-1}{2}\)

 

Q-10: Find all the zeroes of the polynomial \(f\left ( x \right ) = x^{3} + 3x^{2} – 2x – 6\), if the two of its zeroes are \(\sqrt{2} \; \; and \; -\sqrt{2}\).

 

Solution:

\(f\left ( x \right ) = x^{3} + 3x^{2} – 2x – 6\)

Since, \(\sqrt{2} \; \; and \; -\sqrt{2}\) are the two zeroes of the given  polynomial.

So, factors are \(\left ( x + \sqrt{2} \right ) \; \; and \; \left (x -\sqrt{2} \right )\) = \(x^{2} + \sqrt{2}x – \sqrt{2}x – 2\) = \(x^{2} – 2\)

 

15

 

By division algorithm, we have:

\(f\left ( x \right ) = x^{3} + 3x^{2} – 2x – 6\) = \(\left (x^{2} – 2 \right ) \left ( x + 3 \right )\)

= \(\left ( x – \sqrt{2} \right ) \left ( x + \sqrt{2} \right ) \left ( x + 3 \right )\)

Therefore, the zeroes of the given polynomial is \(-\sqrt{2}, \sqrt{2} \; and \; -3\).

 

Q-11: What must be added to the polynomial \(f\left ( x \right ) = x^{4} + 2x^{3} – 2x^{2} + x- 1\) so that the resulting polynomial is exactly divisible by \(g\left ( x \right ) = x^{2} + 2x- 3\).

 

Sol:

\(f\left ( x \right ) = x^{4} + 2x^{3} – 2x^{2} + x- 1\)

 

16

 

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

 

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

 

Sol:

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

 

17

 

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

 

Q-13: Given that \(\sqrt{2}\) is a zero of the cubic polynomial \(f\left ( x \right ) = 6x^{3} + \sqrt{2}x^{2} – 10x – 4\sqrt{2}\), find its other two zeroes.

 

Solution:

\(f\left ( x \right ) = 6x^{3} + \sqrt{2}x^{2} – 10x – 4\sqrt{2}\)

Since, \(\sqrt{2}\) is a zero of the cubic polynomial

So, factor is \(\left (x – \sqrt{2} \right )\)

 

18

 

Since, \(f\left ( x \right ) = 6x^{3} + \sqrt{2}x^{2} – 10x – 4\sqrt{2}\) = \(\left (x – \sqrt{2} \right ) \left ( 6x^{2} + 7\sqrt{2}x + 4 \right )\)

= \(\left (x – \sqrt{2} \right ) \left ( 6x^{2} + 4\sqrt{2}x + 3\sqrt{2}x + 4 \right )\)

= \(\left (x – \sqrt{2} \right ) \left ( 3x + 2\sqrt{2} \right ) \left ( 2x + \sqrt{2} \right )\)

So, the zeroes of the polynomial is \(-\frac{2\sqrt{2}}{3} , \; -\frac{\sqrt{2}}{2} , \; \sqrt{2}\)

 

Q-14: Given that \(x – \sqrt{5}\) is a factor of the cubic polynomial \(x^{3} – 3\sqrt{5}x^{2} + 13x – 3\sqrt{5}\), find all the zeroes of the polynomial.

 

Solution:

\(x^{3} – 3\sqrt{5}x^{2} + 13x – 3\sqrt{5}\)

In the question, it’s given that \(x – \sqrt{5}\) is a factor of the cubic polynomial.

 

19

 

Since, \(x^{3} – 3\sqrt{5}x^{2} + 13x – 3\sqrt{5}\) = \(\left ( x – \sqrt{5} \right ) \left ( x^{2} – 2\sqrt{5}x + 3 \right )\)

= \(\left ( x – \sqrt{5} \right ) \left ( x -\left ( \sqrt{5} + \sqrt{2} \right ) \right ) \left ( x – \left ( \sqrt{5} – \sqrt{2} \right ) \right )\)

So, the zeroes of the polynomial = \(\sqrt{5} , \; \left ( \sqrt{5} – \sqrt{2} \right ), \; \left ( \sqrt{5} + \sqrt{2} \right )\).

 

1 Comment

  1. Thanks for helping to solve sums

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