Question

determine the quotient and remainder when the polynomial p(x) =x³ + 1 is divided by x + 1 by long divison method​

Answer 1

Answer:

remainder is 0

and quotient is x^-x+1

Answer 2

$\begin{array}{clc} \sf{x + 1)} & \sf{x^3 \: + \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1} & \sf{(x^2 - x + 1} \\ & \sf{x^3 + x^2} & \\ & \sf{(-)} \:\:\sf{( - )}&\\ &\dfrac{\qquad\qquad\qquad\qquad\qquad}{}& \\ & \sf{\:\:\:\:\:\:\:\:-x^2 } & \\ & \sf{\:\:\:\:\:\:\:\:-x^2 - x}& \\ & \sf{\:\:\:\:\:\:\:\:(+) \:\:\:\:( + )} & \\ & \dfrac{\qquad\qquad\qquad\qquad\qquad}{} & \\ &\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:x + 1} & \\ & \sf{ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: x + 1}& \\ & \sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(-)\:\:( - )} & \\ & \dfrac{\qquad\qquad\qquad\qquad\qquad}{}& \\ & \sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:0}&\end{array}$

Here the

• Reaminder is 0
• Quotient is x² - x + 1

EXPLANATION:-

First take the dividend first term x³ divide by divisor first term that is x

x³/x = x²

So, quotient first term will be x²

Now , multiply x² with divisor

x(x²+1) = x³ + x² Now subtract them that gives you -x²

Now repeat the above like

Divide -x² /x = -x

Now quotient 2nd term will be -x

Multiply -x with x+ 1 that gives you -x² -x

Now subtract both them that gives you x+1

Now quotient third term will be + 1

x/x = 1 now multiply 1 with x+ 1 gives you x+ 1 both subtract them then remainder is 0

So, the quotient is x² - x + 1

VERIFICATION

Dividend = (Quotient)(Divisor) + Remainder

Take R.H.S

= (x²-x+1) (x + 1) + 0

= x²-x +1(x) + x²-x + 1(1) + 0

= x³ -x² + x + x² - x + 1

R.H.S = x³ + 1

L.H.S = x³+ 1

Hence ,

L.H. S= R.H.S

Verified!