Question

x⁶+1 ÷x+1 find the reminder
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Answer 1

Solution :

( x⁶ + 1)/( x + 1 )

_______

x + 1 ) x⁶ + 1 ( x⁵ - x⁴ + x³ - x² + x - 1

x⁶ + x⁵

(-) (-)

_______

- x⁵ + 1

- x⁵ - x⁴

(+) (+)

_______

x⁴ + 1

x⁴ + x³

(-) (-)

_______

- x³ + 1

- x³ - x²

(+) (+)

______

x² + 1

+ x² + x

(-) (- )

_______

- x + 1

- x - 1

_________

2

On dividing x⁶ + 1 by x + 1 , the quotient is x⁵ - x⁴ + x³ - x² + x - 1 and the remainder is 2.

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Additional Information -

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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Answer 2

Answer :

2

Note :

★ Remainder theorem : If a polynomial p(x) is divided by (x - c) , then the remainder obtained is given as R = p(c) .

★ Factor theorem :

If the remainder obtained on dividing a polynomial p(x) by (x - c) is zero , ie. if R = p(c) = 0 , then (x - c) is a factor of the polynomial p(x) .

If (x - c) is a factor of the polynomial p(x) , then the remainder obtained on dividing the polynomial p(x) by (x - c) is zero , ie. R = p(c) = 0 .

Solution :

Let p(x) = x⁶ + 1 .

Here ,

We need to find the remainder when p(x) is divided by (x + 1) .

If x + 1 = 0 , then x = -1 .

Now ,

The remainder obtained on dividing p(x) = x⁶ + 1 by (x + 1) will be given as ;

=> R = p(-1)

=> R = (-1)⁶ + 1

=> R = 1 + 1

=> R = 2

Hence ,

Remainder = 2 .