# RD Sharma Solutions Class 10 Polynomials Exercise 2.3

## RD Sharma Solutions Class 10 Chapter 2 Exercise 2.3

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

#### Exercise – 2.3

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

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

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

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

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

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

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

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

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

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

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

$g\left ( t \right ) = t^{2} – 3$

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

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

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

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

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

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

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

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

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

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 )

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

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

= ( 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 ).

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

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

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

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 ) = $x^{2} + 2x – 2x – 4$ = $x^{2} – 4$

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 $\left (x + \sqrt{2} \right ) \; \; and \; \left (x -\sqrt{2} \right )$ = $x^{2} + \sqrt{2}x – \sqrt{2}x – 2$ = $x^{2} – 2$

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 $\left (x -\sqrt{3} \right ) \; and \; \left (x + \sqrt{3} \right )$ = $\left ( x ^{2} – \sqrt{3}x + \sqrt{3}x – 3 \right )$ = $\left ( x ^{2} – 3 \right )$

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 $\left ( x + \sqrt{2} \right ) \; \; and \; \left (x -\sqrt{2} \right )$ = $x^{2} + \sqrt{2}x – \sqrt{2}x – 2$ = $x^{2} – 2$

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:

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

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:

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

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 $\left (x – \sqrt{2} \right )$

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.

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 = $\sqrt{5} , \; \left ( \sqrt{5} – \sqrt{2} \right ), \; \left ( \sqrt{5} + \sqrt{2} \right )$..

#### Practise This Question

For real numbers x and y, we write xRyxy+2 is an irrational number. Then the relation R is