#### Question

Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in the following f(x) = 10x^{4} + 17x^{3} − 62x^{2} + 30x − 3, g(x) = 2x^{2} + 7x + 1

#### Solution

We have

f(x) = 10x^{4} + 17x^{3} − 62x^{2} + 30x − 3

g(x) = 2x^{2} + 7x + 1

Therefore, quotient q(x) is of degree 4 - 2 = 2 and remainder r(x) is of degree less than 2

Let g(x) = ax^{2} + bx + c and

r(x) = px + q

Using division algorithm, we have

f(x) = g(x) x q(x) + r(x)

1Ox^{4}+ 17x^{3} - 62x^{2} + 30x - 3 = (2x^{2} + 7x+ 1)(ax^{2} +bx+c)+ px+q

1Ox^{4}+ 17x^{3} - 62x^{2} + 30x - 3 = 2ax^{4} + 7ax^{3} + ax^{2} + 2bx^{3} + 7bx^{2} + bx + 2cx^{2} + 7xc + c + px + q

1Ox^{4}+ 17x^{3} - 62x^{2} + 30x - 3 = 2ax^{4} + 7ax^{3} + 2bx^{3} + ax^{2} + 7bx^{2} + 2cx^{2} + bx + 7xc + px + c + q

a + 1Ox^{4}+ 17x^{3} - 62x^{2} + 30x - 3 = 2ax^{4} + x^{3} (7a + 2b) + x^{2} (a + 7b + 2c) + x(b + 7c + p) + c + q

Equating the co-efficients of various powers x on both sides, we get

On equating the co-efficient of x^{4 }

2a = 10

`a=10/2`

a = 5

On equating the co-efficient of x^{3}

7a + 2b = 17

Substituting a = 5 we get

7 x 5 2b = 17

35 + 2b = 17

2b = 17 - 35

2b = -18

`b=(-18)/2`

b = -9

On equating the co-efficient of x^{2}

a + 7b + 2c = -62

Substituting a = 5 and b = -9, we get

-9 + 7 x -2 + p = 30

-9 - 14 + p = 30

-23 + p = 30

p = 30 + 23

p = 53

On equating constant term, we get

c + q = -3

Substituting c = -2, we get

-2 + q = -3

q = -3 + 2

q = -1

Therefore, quotient q(x) = ax^{2} + bx + c

= 5x^{2} - 9x - 2

Remainder r(x) = px + q

= 53x - 1

Hence, the quotient and remainder are q(x) = 5x^{2} - 9x - 2 and r(x) = 53x - 1