The document RD Sharma Solutions Ex-6.3, Factorization Of Polynomials, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

All you need of Class 9 at this link: Class 9

**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 â€“ 8)**

**Q1. f(x) = x ^{3}+4x^{2}â€“3x+10 , g(x) = x + 4**

**Sol :**

Here, f(x) = x^{3}+4x^{2}â€“3x+10

g(x) = x + 4

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

Let , g(x) = 0

â‡’ x + 4 = 0

â‡’ x = -4

Substitute the value of x in f(x)

f(-4) = (âˆ’4)^{3}+4(âˆ’4)^{2}â€“3(âˆ’4)+10

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

= -64 + 64 + 12 + 10

= 12 + 10

= 22

Therefore, the remainder is 22

**Q2. f(x) = 4x ^{4}â€“3x^{3}â€“2x^{2}+xâ€“7 , g(x) = x â€“ 1**

Sol :

Here, f(x) = 4x^{4}â€“3x^{3}â€“2x^{2}+xâ€“7

g(x) = x â€“ 1

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

Let , g(x) = 0

â‡’ x â€“ 1 = 0

â‡’ x = 1

Substitute the value of x in f(x)

f(1) = 4(1)^{4}â€“3(1)^{3}â€“2(1)2+1â€“7

= 4 â€“ 3 â€“ 2 + 1 â€“ 7

= 5 â€“ 12

= -7

Therefore, the remainder is 7

**Q3. f(x) = 2x ^{4}â€“6x^{3}+2x^{2}â€“x+2 , g(x) = x + 2**

**Sol :**

Here, f(x) = 2x^{4}â€“6x^{3}+2x^{2}â€“x+2

g(x) = x + 2

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

Let , g(x) = 0

â‡’ x + 2 = 0

â‡’ x = -2

Substitute the value of x in f(x)

f(-2) = 2(âˆ’2)^{4}â€“6(âˆ’2)^{3}+2(âˆ’2)^{2}â€“(âˆ’2)+2

= (2 * 16) â€“ (6 * (-8)) + (2 * 4) + 2 + 2

= 32 + 48 + 8 + 2 + 2

= 92

Therefore, the remainder is 92

**Q4. f(x) = 4x ^{3}â€“12x^{2}+14xâ€“3 , g(x) = 2x â€“ 1**

**Sol:**

Here, f(x) = 4x^{3}â€“12x^{2}+14xâ€“3

g(x) = 2x â€“ 1

from, the remainder theorem when f(x) is divided by g(x) = the remainder is equal to f(1/2)

Let , g(x) = 0

â‡’ 2x â€“ 1 = 0

â‡’ 2x = 1

â‡’ x = 1/2

Substitute the value of x in f(x)

Taking L.C.M

Therefore, the remainder is

**Q5. f(x) = x ^{3}â€“6x^{2}+2xâ€“4 , g(x) = 1 â€“ 2x**

**Sol :**

Here, f(x) = x^{3}â€“6x^{2}+2xâ€“4

g(x) = 1 â€“ 2x

from, the remainder theorem when f(x) is divided by g(x) = - the remainder is equal to f(1/2)

Let , g(x) = 0

â‡’ 1 â€“ 2x = 0

â‡’ -2x = -1

â‡’ 2x = 1

â‡’ x = 1/2

Substitute the value of x in f(x)

Taking L.C.M

Therefore, the remainder is

**Q6. f(x) = x ^{4}â€“3x^{2}+4 , g(x) = x â€“ 2**

**Sol :**

Here, f(x) = x^{4}â€“3x^{2}+4

g(x) = x â€“ 2

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

let , g(x) = 0

â‡’ x â€“ 2 = 0

â‡’ x = 2

Substitute the value of x in f(x)

f(2) = 24â€“3(2)^{2}+4

= 16 â€“ (3* 4) + 4

= 16 â€“ 12 + 4

= 20 â€“ 12

= 8

Therefore, the remainder is 8

**Q7. f(x) = 9x ^{3}â€“3x^{2}+xâ€“5 , g(x) **

**Sol :**

Here, f(x) = 9x^{3}â€“3x^{2}+xâ€“5

g(x) =

from, the remainder theorem when f(x) is divided by g(x) = the remainder will be equal to f(1/2) substitute the value of x in f(x)

Therefore, the remainder is -3

**Q8. **

**Sol :**

from remainder theorem when f(x) is divided by g(x) the remainder is equal to substitute the value of x in f(x)

= 0

Therefore, the remainder is 0

**Q9. If the polynomial 2x ^{3}+ax^{2}+3xâ€“5 and x^{3}+x^{2}â€“4x+a leave the same remainder when divided by x â€“ 2 , Find the value of a**

**Sol :**

Given , the polymials are

f(x) = 2x^{3}+ax^{2}+3xâ€“5

p(x) = x^{3}+x^{2}â€“4x+a

The remainders are f(2) and p(2) when f(x) and p(x) are divided by x â€“ 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)â€“5

= (2 * 8) + a4 + 6 â€“ 5

= 16 + 4a + 1

= 4a + 17 â€”â€”- 1

for, p(2)

substitute (x = 2) in p(x)

p(2) = 2^{3}+2^{2}â€“4(2)+a

= 8 + 4 â€“ 8 + a

= 4 + a â€”â€”â€” 2

Since, f(2) = p(2)

Equate eqn 1 and 2

â‡’ 4a + 17 = 4 + a

â‡’ 4a â€“ a = 4 â€“ 17

â‡’ 3a = -13

â‡’

**Q10. If polynomials ax ^{3}+3x^{2}â€“3 and 2x^{3}â€“5x+a when divided by (x â€“ 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. 2R_{1}â€“R_{2} = 0

**Sol :**

Here, the polynomials are

f(x) = ax^{3}+3x^{2}â€“3

p(x) = 2x^{3}â€“5x+a

let,

R_{1} is the remainder when f(x) is divided by x â€“ 4

â‡’ R_{1} = f(4)

â‡’ R1 = a(4)^{3}+3(4)^{2}â€“3

= 64a + 48 â€“ 3

= 64a + 45 â€”â€”â€” 1

Now , let

R_{2} is the remainder when p(x) is divided by x â€“ 4

â‡’ R_{2} = p(4)

â‡’ R_{2} = 2(4)^{3}â€“5(4)+a

= 128 â€“ 20 + a

= 108 + a â€”â€”â€“ 2

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

â‡’ 64a + 45 = 108 + a

â‡’ 63a = 63

â‡’ a = 1

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

â‡’ 64a + 45 + 108 + a = 0

â‡’ 65a + 153 = 0

â‡’

3.Given, 2R_{1}â€“R_{2} = 0

â‡’ 2(64a + 45) â€“ 108 â€“ a = 0

â‡’ 128a + 90 â€“ 108 â€“ a = 0

â‡’ 127a â€“ 18 = 0

â‡’ a = 18/127

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

**Sol :**

Here , the polynomials are

f(x) = ax^{3}+3x^{2}â€“13

p(x) = 2x^{3}â€“5x+a

equate , x â€“ 2 = 0

x = 2

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

f(2) = (2)^{3}+3(2)^{2}â€“13

= 8a + 12 â€“ 13

= 8a â€“ 1 â€”â€”â€”- 1

p(2) = 2(2)^{3}â€“5(2)+a

= 16 â€“ 10 + a

= 6 + a â€”â€”â€”â€“ 2

f(2) = p(2)

â‡’ 8a â€“ 1 = 6 + a

â‡’ 8a â€“ a = 6 + 1

â‡’ 7a = 7

â‡’ 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. 3. x 4. x + Ï€ 5. 5 + 2x**

**Sol :**

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

by remainder theorem

1. â‡’ x + 1 = 0

â‡’ x = -1

substitute the value of x in f(x)

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

= -1 + 3 â€“ 3 + 1

= 0

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

By remainder theorem

â‡’

â‡’

substitute the value of x in f(x)

= 27/8

3. x

Sol :

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

by remainder theorem

â‡’ x = 0

substitute the value of x in f(x)

f(0) = 03+3(0)^{2}+3(0)+1

= 0 + 0 + 0 + 1

= 1

4. x + Ï€

Sol :

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

by remainder theorem

â‡’ x + Ï€ = 0

â‡’ x = -Ï€

Substitute the value of x in f(x)

f(-Ï€) = (âˆ’Ï€)^{3}+3(âˆ’Ï€)^{2}+3(âˆ’Ï€)+1

= â€“(Ï€)^{3}+3(Ï€)^{2}â€“3(Ï€)+1

5. 5 + 2x

Sol :

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

by remainder theorem

5 + 2x = 0

2x = -5

x = -5/2

substitute the value of x in f(x)

Taking L.C.M

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

- RD Sharma Solutions Ex-6.4, (Part -1), Factorization Of Polynomials, Class 9, Maths
- RD Sharma Solutions Ex-6.4, (Part -2), Factorization Of Polynomials, Class 9, Maths
- RD Sharma Solutions Ex-6.5, (Part -1), Factorization Of Polynomials, Class 9, Maths
- RD Sharma Solutions Ex-6.5, (Part -2), Factorization Of Polynomials, Class 9, Maths