Using division algorithm find the quotient and remainder of the following:

I) x³-6x²+11x-6 by x²+x+1

### Asked by renuneeraj2005 | 2nd Jun, 2020, 11:00: AM

###
We have

f(x) = x^{3 }- 6x^{2} + 11x - 6 and g(x) = x^{2} + x + 1

Clearly, degree of f(x) = 3 and degree of g(x) = 2.

Therefore, the degree of quotient is q(x) = 3 - 2 = 1 and the degree of remainder is r(x) less than 2.

Let quotient q(x) = ax + b and remainder r(x) = cx +d.

Using division algorithm, we have

f(x) = g(x) × q(x) + r(x)

Comparing the coefficient of same powers of x on both sides, we get

a = 1 [Comparing the coefficient of x^{3}]

a + b = -6 [Comparing the coefficient of x^{2}]

a + b + c = 11 [Comparing the coefficient of x]

b + d = -6 [Comparing the constant terms ]

Solving the above equations, we get the following values:

a = 1, b = -7, c = 17, and d = 1

∴ Quotient is q (x) = x - 7 and remainder is r(x) = 17x + 1

###

(i) We have:

f(x) = x^{3 }- 6x^{2} + 11x - 6 and g(x) = x + 1

Clearly, degree of f(x) = 3 and degree of g(x) = 1. Therefore, the degree of quotient is q(x) = 3 - 1 = 2 and the degree of remainder is r(x) = 0

Let quotient q(x) = ax^{2} + bx + c and remainder r(x) = k.

Using division algorithm, we have

f(x) = g(x) × q(x) + r(x)

_{}

Comparing the coefficient of same powers of x on both sides, we get

a = 1 [Comparing the coefficient of x^{3}]

a + b = -6 [Comparing the coefficient of x^{2}]

b + c = -11 [Comparing the coefficient of x]

c + k = -6 [Comparing the constant terms ]

Solving the above equations, we get the following values:

a = 1, b = -7, c = 18, and k = -24

∴ Quotient is q (x) = x^{2} - 7x + 18 and remainder is r(x) = -24.

(i) We have:

f(x) = x^{3 }- 6x^{2} + 11x - 6 and g(x) = x + 1

Clearly, degree of f(x) = 3 and degree of g(x) = 1. Therefore, the degree of quotient is q(x) = 3 - 1 = 2 and the degree of remainder is r(x) = 0

Let quotient q(x) = ax^{2} + bx + c and remainder r(x) = k.

Using division algorithm, we have

f(x) = g(x) × q(x) + r(x)

Comparing the coefficient of same powers of x on both sides, we get

a = 1 [Comparing the coefficient of x^{3}]

a + b = -6 [Comparing the coefficient of x^{2}]

b + c = -11 [Comparing the coefficient of x]

c + k = -6 [Comparing the constant terms ]

Solving the above equations, we get the following values:

a = 1, b = -7, c = 18, and k = -24

∴ Quotient is q (x) = x^{2} - 7x + 18 and remainder is r(x) = -24.

We have

f(x) = x^{3 }- 6x^{2} + 11x - 6 and g(x) = x^{2} + x + 1

Clearly, degree of f(x) = 3 and degree of g(x) = 2.

Therefore, the degree of quotient is q(x) = 3 - 2 = 1 and the degree of remainder is r(x) less than 2.

Let quotient q(x) = ax + b and remainder r(x) = cx +d.

Using division algorithm, we have

f(x) = g(x) × q(x) + r(x)

Comparing the coefficient of same powers of x on both sides, we get

a = 1 [Comparing the coefficient of x^{3}]

a + b = -6 [Comparing the coefficient of x^{2}]

a + b + c = 11 [Comparing the coefficient of x]

b + d = -6 [Comparing the constant terms ]

Solving the above equations, we get the following values:

a = 1, b = -7, c = 17, and d = 1

∴ Quotient is q (x) = x - 7 and remainder is r(x) = 17x + 1

(i) We have:

f(x) = x^{3 }- 6x^{2} + 11x - 6 and g(x) = x + 1

Clearly, degree of f(x) = 3 and degree of g(x) = 1. Therefore, the degree of quotient is q(x) = 3 - 1 = 2 and the degree of remainder is r(x) = 0

Let quotient q(x) = ax^{2} + bx + c and remainder r(x) = k.

Using division algorithm, we have

f(x) = g(x) × q(x) + r(x)

_{}

Comparing the coefficient of same powers of x on both sides, we get

a = 1 [Comparing the coefficient of x^{3}]

a + b = -6 [Comparing the coefficient of x^{2}]

b + c = -11 [Comparing the coefficient of x]

c + k = -6 [Comparing the constant terms ]

Solving the above equations, we get the following values:

a = 1, b = -7, c = 18, and k = -24

∴ Quotient is q (x) = x^{2} - 7x + 18 and remainder is r(x) = -24.

(i) We have:

f(x) = x^{3 }- 6x^{2} + 11x - 6 and g(x) = x + 1

Clearly, degree of f(x) = 3 and degree of g(x) = 1. Therefore, the degree of quotient is q(x) = 3 - 1 = 2 and the degree of remainder is r(x) = 0

Let quotient q(x) = ax^{2} + bx + c and remainder r(x) = k.

Using division algorithm, we have

f(x) = g(x) × q(x) + r(x)

Comparing the coefficient of same powers of x on both sides, we get

a = 1 [Comparing the coefficient of x^{3}]

a + b = -6 [Comparing the coefficient of x^{2}]

b + c = -11 [Comparing the coefficient of x]

c + k = -6 [Comparing the constant terms ]

Solving the above equations, we get the following values:

a = 1, b = -7, c = 18, and k = -24

∴ Quotient is q (x) = x^{2} - 7x + 18 and remainder is r(x) = -24.

### Answered by Renu Varma | 3rd Jun, 2020, 10:39: AM

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