# RD Sharma Solutions Class 9 Factorization Of Polynomials Exercise 6.3

## RD Sharma Solutions Class 9 Chapter 6 Ex 6.3

In each of the following , using the remainder theorem, find the remainder when f(x) is divided by  g(x) and verify the by actual division : (1 – 8)

Q1. f(x) = $x^{3} + 4x^{2} – 3x + 10$ , g(x) = x + 4

Sol :

Here,  f(x) = $x^{3} + 4x^{2} – 3x + 10$

g(x) = x + 4

from, the remainder theorem when f(x) is divided by g(x) = x – (-4)  the remainder will be equal to f(-4)

Let , g(x) = 0

=> x + 4 = 0

=> x = -4

Substitute the value of x in f(x)

f(-4) = $(-4)^{3} + 4(-4)^{2} – 3(-4) + 10$

= -64 + ( 4 * 16) + 12 + 10

= -64 + 64 + 12 + 10

= 12 + 10

= 22

Therefore, the remainder is 22

Q2. f(x) = $4x^{4} – 3x^{3} – 2x^{2} + x – 7$ , g(x) = x – 1

Sol :

Here, f(x) = $4x^{4} – 3x^{3} – 2x^{2} + x – 7$

g(x) = x – 1

from, the remainder theorem when f(x) is divided by g(x) = x – (-1) the remainder will be equal to f(1)

Let , g(x) = 0

=> x – 1 = 0

=> x = 1

Substitute the value of x in f(x)

f(1) = $4(1)^{4} – 3(1)^{3} – 2(1)^{2} + 1 – 7$

= 4 – 3 – 2 + 1 – 7

= 5 – 12

= -7

Therefore, the remainder is 7

Q3. f(x) = $2x^{4} – 6x^{3} + 2x^{2} – x + 2$ , g(x) = x + 2

Sol :

Here, f(x) = $2x^{4} – 6x^{3} + 2x^{2} – x + 2$

g(x) = x + 2

from, the remainder theorem when f(x) is divided by g(x) = x – (-2) the remainder will be equal to f(-2)

Let , g(x) = 0

=> x + 2 = 0

=> x = -2

Substitute the value of x in f(x)

f(-2) = $2(-2)^{4} – 6(-2)^{3} + 2(-2)^{2} – (-2) + 2$

= (2 * 16) – (6 * (-8)) + (2 * 4) + 2 + 2

= 32 + 48 + 8 + 2 + 2

= 92

Therefore, the remainder is 92

Q4. f(x) = $4x^{3} – 12x^{2} + 14x – 3$ , g(x) = 2x – 1

Sol:

Here, f(x) = $4x^{3} – 12x^{2} + 14x – 3$

g(x) = 2x – 1

from, the remainder theorem when f(x) is divided by g(x) = 2(x – $\frac{1}{2}$ ) , the remainder is equal to f($\frac{1}{2}$)

Let , g(x) = 0

=>  2x – 1 = 0

=> 2x = 1

=> x = $\frac{1}{2}$

Substitute the value of x in f(x)

f($\frac{1}{2}$) = 4($\frac{1}{2}$)³ – 12($\frac{1}{2}$)² + 14($\frac{1}{2}$ – 3

= $4(\frac{1}{8}) – 12(\frac{1}{4}) + 4(\frac{1}{2}) – 3$

= $(\frac{1}{2})$ – 3 + 7 – 3

= $(\frac{1}{2})$ + 1

Taking L.C.M

= $(\frac{2 + 1}{2})$

= $(\frac{3}{2})$

Therefore, the remainder is $(\frac{3}{2})$

Q5. f(x) = $x^{3} – 6x^{2} + 2x – 4$ , g(x) = 1 – 2x

Sol :

Here, f(x) = $x^{3} – 6x^{2} + 2x – 4$

g(x) = 1 – 2x

from, the remainder theorem when f(x) is divided by g(x) = -2(x – $\frac{1}{2}$ ) , the remainder is equal to f($\frac{1}{2}$)

Let , g(x) = 0

=> 1 – 2x  = 0

=> -2x = -1

=> 2x = 1

=> x = $\frac{1}{2}$

Substitute the value of x in f(x)

f($\frac{1}{2}$) = ( $\frac{1}{2}$)³ – 6($\frac{1}{2}$)² + 2($\frac{1}{2}$) – 4

= $\frac{1}{8} – 8(\frac{1}{4}) + 2(\frac{1}{2}) – 4$

= $\frac{1}{8} – (\frac{1}{2}) + 1 – 4$

= $\frac{1}{8} – (\frac{1}{2}) – 3$

Taking L.C.M

= $\frac{1 – 4 + 8 – 32}{8}$

= $\frac{1 – 36}{8}$

= $\frac{1 – 36}{8}$

= $\frac{-35}{8}$

Therefore, the remainder is $\frac{-35}{8}$

Q6. f(x) = $x^{4} – 3x^{2} + 4$ , g(x) = x – 2

Sol :

Here, f(x) = $x^{4} – 3x^{2} + 4$

g(x) = x – 2

from, the remainder theorem when f(x) is divided by g(x) = x – 2 the remainder will be equal to f(2)

let , g(x) = 0

=> x – 2 = 0

=> x = 2

Substitute the value of x in f(x)

f(2) = $2^{4} – 3(2)^{2} + 4$

= 16 – (3* 4) + 4

= 16 – 12 + 4

= 20 – 12

= 8

Therefore, the remainder is 8

Q7. f(x) = $9x^{3} – 3x^{2} + x – 5$ , g(x) = $x – \frac{2}{3}$

Sol :

Here, f(x) = $9x^{3} – 3x^{2} + x – 5$

g(x) = $x – \frac{2}{3}$

from, the remainder theorem when f(x) is divided by g(x) = x – $\frac{2}{3}$ the remainder will be equal to f($\frac{2}{3}$)

substitute the value of x in f(x)

f($\frac{2}{3}$) = 9($\frac{2}{3}$) – 3$(\frac{2}{3}$)² + ($\frac{2}{3}$) – 5

= $9(\frac{8}{27}) – 3(\frac{4}{9}) + \frac{2}{3} – 5$

= $(\frac{8}{3}) – (\frac{4}{3}) + \frac{2}{3} – 5$

= $\frac{8 – 4 + 2 – 15}{3}$

= $\frac{10 – 19}{3}$

= $\frac{-9}{3}$

= -3

Therefore, the remainder is -3

Q8. f(x) = $3x^{4} + 2x^{3} – \frac{x^{3}}{3} – \frac{x}{9} + \frac{2}{27}$ , g(x) = $x + \frac{2}{3}$

Sol :

Here, f(x) =  $3x^{4} + 2x^{3} – \frac{x^{3}}{3} – \frac{x}{9} + \frac{2}{27}$

g(x) = $x + \frac{2}{3}$

from remainder theorem when f(x) is divided by g(x) = $x – (-\frac{2}{3}$) , the remainder is equal to f($-\frac{2}{3}$)

substitute the value of x in f(x)

f($-\frac{2}{3}$) = 3($-\frac{2}{3}$)4 + 2($-\frac{2}{3}$)³ – $\frac{(-\frac{2}{3})³}{3}$$\frac{(-\frac{2}{3})}{9} + \(\frac{2}{27}$$\frac{2}{27}\]”> = \(3(\frac{16}{81}) + 2(\frac{-8}{27}) – \frac{4}{(9 * 3)} – (\frac{-2}{(9 * 3)}) + \frac{2}{27}$

= $(\frac{16}{27}) – (\frac{16}{27}) – \frac{4}{27} + (\frac{2}{27}) + \frac{2}{27}$

= $(\frac{4}{27}) – (\frac{4}{27})$

= 0

Therefore, the remainder is 0

Q9. If the polynomial $2x^{3} + ax^{2} + 3x – 5$ and $x^{3} + x^{2} – 4x + a$ leave the same remainder when divided by x – 2 , Find the value of a

Sol :

Given , the polymials are

f(x) = $2x^{3} + ax^{2} + 3x – 5$

p(x) = $x^{3} + x^{2} – 4x + a$

The remainders are f(2) and p(2) when f(x) and p(x) are divided by x – 2

We know that,

f(2) = p(2)     (given in problem)

we need to  calculate f(2) and p(2)

for, f(2)

substitute (x = 2) in f(x)

f(2) =  $2(2)^{3} + a(2)^{2} + 3(2) – 5$

= (2 * 8) + a4 + 6 – 5

= 16 + 4a + 1

= 4a + 17 ——- 1

for, p(2)

substitute (x = 2) in p(x)

p(2) = $2^{3} + 2^{2} – 4(2) + a$

= 8 + 4 – 8 + a

= 4 + a  ——— 2

Since, f(2) = p(2)

Equate eqn 1 and 2

=> 4a + 17 = 4 + a

=> 4a – a = 4 – 17

=> 3a = -13

=> a = $\frac{-13}{3}$

The value of a = $\frac{-13}{3}$

Q10. If polynomials $ax^{3} + 3x^{2} – 3$ and $2x^{3} – 5x + a$ when divided by (x – 4) leave the remainders as $R_{1} and R_{2}$ respectively. Find the values of a in each of the following cases, if

1. $R_{1} = R_{2}$
2. $R_{1} + R_{2}$ = 0
3. 2$R_{1} – R_{2}$ = 0

Sol :

Here, the polynomials are

f(x) = $ax^{3} + 3x^{2} – 3$

p(x) = $2x^{3} – 5x + a$

let,

$R_{1}$ is the remainder when f(x) is divided by x – 4

=>  $R_{1}$ = f(4)

=> $R_{1}$ = $a(4)^{3} + 3(4)^{2} – 3$

= 64a + 48 – 3

= 64a + 45               ——— 1

Now , let

$R_{2}$ is the remainder when p(x) is divided by x – 4

=>  $R_{2}$ = p(4)

=> $R_{2}$ = $2(4)^{3} – 5(4) + a$

= 128 – 20 + a

= 108 + a    ——– 2

1. Given , $R_{1} = R_{2}$

=> 64a + 45 = 108 + a

=> 63a = 63

=> a= 1

2. Given, $R_{1} + R_{2}$ = 0

=> 64a + 45 + 108 + a = 0

=> 65a + 153 = 0

=>  a = $\frac{-153}{65}$

3.Given, 2$R_{1} – R_{2}$ = 0

=> 2(64a + 45) – 108 – a = 0

=> 128a + 90 – 108 – a = 0

=> 127a – 18 = 0

=>  a = $\frac{18}{127}$

Q11. If the polynomials $ax^{3} + 3x^{2} – 13$ and $2x^{3} – 5x + a$ when divided by (x – 2) leave the same remainder, Find the value of a

Sol :

Here , the polynomials are

f(x) =  $ax^{3} + 3x^{2} – 13$

p(x) = $2x^{3} – 5x + a$

equate , x – 2 = 0

x = 2

substitute the value of x in f(x) and p(x)

f(2) = $(2)^{3} + 3(2)^{2} – 13$

= 8a + 12 – 13

= 8a – 1            ———- 1

p(2) = $2(2)^{3} – 5(2) + a$

= 16 – 10 + a

= 6 + a               ———– 2

f(2) = p(2)

=> 8a – 1 = 6 + a

=> 8a – a = 6 + 1

=>    7a = 7

=>     a = 1

The value of a = 1

Q12. Find the remainder when $x^{3} + 3x^{3} + 3x + 1$ is divided by,

1. x + 1
2. x – $\frac{1}{2}$
3. x
4. x + π
5. 5 + 2x

Sol :

Here, f(x) =  $x^{3} + 3x^{2} + 3x + 1$

by remainder theorem

1. =>  x + 1 = 0

=> x = -1

substitute the value of x in f(x)

f(-1) = $(-1)^{3} + 3(-1)^{2} + 3(-1) + 1$

= -1  + 3 – 3 + 1

= 0

2. x – $\frac{1}{2}$

Sol :

Here, f(x) =  $x^{3} + 3x^{2} + 3x + 1$

By remainder theorem

=>  x – $\frac{1}{2}$ = 0

=> x =  $\frac{1}{2}$

substitute the value of x in f(x)

f($\frac{1}{2}$) =  ( $\frac{1}{2}$)³ + 3($\frac{1}{2}$)² + 3($\frac{1}{2}$) + 1

= $(\frac{1}{2})^{3} + 3(\frac{1}{2})^{2} + 3(\frac{1}{2}) + 1$

= $\frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$

= $\frac{1+6+12+8}{8}$

= $\frac{27}{8}$

3. x

Sol :

Here, f(x) =  $x^{3} + 3x^{2} + 3x + 1$

by remainder theorem

=> x = 0

substitute the value of x in f(x)

f(0) =  $0^{3} + 3(0)^{2} + 3(0) + 1$

= 0 + 0 + 0 + 1

= 1

4. x + π

Sol :

Here, f(x) =  $x^{3} + 3x^{2} + 3x + 1$

by remainder theorem

=> x + π = 0

=>    x =  -π

Substitute the value of x in f(x)

f(-π) =  $( -π)^{3} + 3(-π)^{2} + 3(-π) + 1$

= –$( π)^{3} + 3(π)^{2} – 3(π) + 1$

5. 5 + 2x

Sol :

Here, f(x) =  $x^{3} + 3x^{2} + 3x + 1$

by remainder theorem

5 + 2x = 0

2x = -5

x = $\frac{-5}{2}$

substitute the value of x in f(x)

f($\frac{-5}{2}$) =  ( $\frac{-5}{2}$)³ + 3($\frac{-5}{2}$)² + 3($\frac{-5}{2}$) + 1

= $\frac{-125}{8}+3(\frac{25}{4})+3(\frac{-5}{2})+1$

= $\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1$

Taking L.C.M

= $\frac{-125+150-50+8}{8}$

= $\frac{-27}{8}$

182.25 = ?