RD Sharma Solutions Class 9 Factorization Of Polynomials Exercise 6.3

RD Sharma Class 9 Solutions Chapter 6 Ex 6.3 Free Download

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

= \(\frac{-125+150-50+8}{8}\)

= \(\frac{-27}{8}\)

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