**1. Find the remainder (without division) on dividing f(x) by (x – 2) where**

**(i) f(x) = 5x**

^{2}– 7x + 4**Solutions:-**

We have given that:

Let us assume x – 2 = 0

Then, x = 2

Given, f(x) = 5x

^{2}– 7x + 4Now, substitute the value of x in f(x), we get

f(2)= (5 × 22) – (7 × 2) + 4

= (5 × 4) – 14 + 4

= 20 – 14 + 4

= 24 – 14

= 10

Therefore, the remainder is 10.

(ii) f(x) = 2x

^{3}– 7x^{2}+ 3**Solution:-**

Let us assume x – 2 = 0

Then, x = 2

We have given that:

Therefore, the remainder is -2

Hence, the given values x = -2 are not roots of the equation.

**3. Find the remainder (without division) on dividing f(x) by (2x + 1) where,**

**(i) f(x) = 4x**

^{2}+ 5x + 3**Solution:-**

Let us assume 2x + 1 = 0

Then, 2x = -1

x=-1/2

We have given that:

**4. (i) find the remainder (without division) when 2x**

^{3}– 3x^{2}+ 7x – 8 is divided by x – 1.**Solution:-**

Let us assume x – 1 = 0

Then, x = 1

We have given that:

f(x) = 2x

^{3}– 3x^{2}+ 7x – 8Now, substitute the value of x in f(x), we get

f(1) = (2 × 1

^{3}) – (3 × 1^{2}) + (7 × 1) – 8= 2 – 3 + 7 – 8

= 9 – 11

= – 2

Hence, the given values x = -2 are not roots of the equation.

**(ii) Find the remainder (without division) on dividing 3x**

^{2}+ 5x – 9 by (3x + 2).**Solution:-**

Let us assume 3x + 2 = 0

3x = -2

x = -2/3

Then, x = -2/3

We have given that:

f(x) = 3x2 + 5x – 9

Now, substitute the value of x in f(x), we get

f(1) = 1

^{3}+ (5 × 1^{2}) – (a × 1) + 6= 1 + 5 – a + 6

= 12 – a

From the question it is given that, remainder is 2a

So, 2a = 12 – a

2a + a = 12

3a = 12

a = 12/3

a = 4

Hence, the given values a=4 are not roots of the equation.

Hence, remainder is p – 6

From the question it is given that, remainder is – 2.

P – 6 = – 2

P = -2 + 6

P = 4

Therefore,4 is to be added.

Hence, the given values p=4 are not roots of the equation.

x = -7/2

We have given that:

f(x) = 2x

^{3}+ 5x^{2}– 11x – 14Now, substitute the value of x in f(x), we get

f(-7/2) = 2(-7/2)

^{3}+ 5(-7/2)^{2}+ 11(-7/2) – 14= 2(-343/8) + 5(49/4) + (-77/2) – 14

= -343/4 + 245/4 – 77/2 – 14

= (-343 + 245 + 154 – 56)/4

= -399 + 399/4

= 0

Therefore, (2x + 7) is a factor of 2x

^{3}+ 5x2 – 11x – 14Then, dividing f(x) by (2x + 1), we get

Then, remainder is 0

So, (44 – 4k)/9 = 0

44 – 4k = 0 × 9

44 = 4k

K = 44/4

K = 11

Hence, the value of k is 11.

2 – a + b = 5

– a + b = 5 – 2

– a + b = 3 … (ii)

Now, subtracting equation (ii) from equation (i) we get,

(a + b) – (- a + b) = 13 – 3

a + b + a – b = 10

2a = 10

a = 10/2

a = 5

To find out the value of b, substitute the value of a in equation (i) we get,

a + b = 13

5 + b = 13

b = 13 – 5

b = 8

Therefore, value of a = 5 and b = 8

Hence, the given values b=8 are not roots of the equation.

**12. When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is, divided by (x – 2), the remainder is 7. Find the remainder when it is divided by (x – 1) (x – 2).**

**Solution:-**

We have given that:

Polynomial f(x) is divided by (x – 1),

Remainder = 5

Let x – 1 be 0

x = 1

f(1) = 5

and the divided be (x – 2), remainder = 7

let us assume x – 2 = 0

x = 2

Therefore, f(2) = 7

/So, f(x) = (x – 1) (x – 2) q(x) + ax + b

Where, q(x) is the quotient and ax + b is remainder,

Now put x = 1, we get,

f(1) = (1 – 1)(1 – 2)q(1) + (a × 1) + b

a + b = 5 … (i)

x = 2,

f(2) = (2 – 1)(2 – 2)q(2) + (a × 2) + b

2a + b = 7 … (ii)

Now subtracting equation (i) from equation (ii) we get,

(2a + b) – (a + b) = 7 – 5

2a + b – a – b = 2

a = 2

To find out the value of b, substitute the value of a in equation (i) we get,

a + b = 5

2 + b = 5

b = 5 – 2

b = 3

Hence, the remainder ax+b=2x+3