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**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) = the remainder is equal to f(1/2)

Let , g(x) = 0

⇒ 2x – 1 = 0

⇒ 2x = 1

⇒ x = 1/2

Substitute the value of x in f(x)

Taking L.C.M

Therefore, the remainder is

**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) = - the remainder is equal to f(1/2)

Let , g(x) = 0

⇒ 1 – 2x = 0

⇒ -2x = -1

⇒ 2x = 1

⇒ x = 1/2

Substitute the value of x in f(x)

Taking L.C.M

Therefore, the remainder is

**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) = 24–3(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) **

**Sol :**

Here, f(x) = 9x^{3}–3x^{2}+x–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}+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

⇒

**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. 2R_{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)

⇒ R1 = 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

⇒

3.Given, 2R_{1}–R_{2} = 0

⇒ 2(64a + 45) – 108 – a = 0

⇒ 128a + 90 – 108 – a = 0

⇒ 127a – 18 = 0

⇒ a = 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. 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

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

By remainder theorem

⇒

⇒

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

⇒ 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 + 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 = -5/2

substitute the value of x in f(x)

Taking L.C.M

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