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Find the remainder when the polynomial x^1000 is divided by the polynomial (x^2+1)(x+1).

Jul 31, 2019

#1
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Find the remainder when the polynomial $$\large{x^{1000}}$$ is divided by the polynomial $$(x^2+1)(x+1)$$

$$\begin{array}{|lrcll|} \hline &\mathbf{ x^{1000} }&=& \mathbf{(x^2+1)(x+1)q(x) +\underbrace{ax^2+bx+c}_{=r(x)}} \\ \hline x=-1:& (-1)^{1000} &=& \left((-1)^2+1\right)(-1+1)q(x) + a(-1)^2+b(-1)+c \\ & 1 &=& 0 + a-b+c \\ &\mathbf{ a-b+c} &=& \mathbf{1} \qquad (1) \\ \hline x=i:& (i)^{1000} &=& (i^2+1)(i+1)q(x) + ai^2+bi+c \quad | \quad \boxed{i^2=-1\\i^{1000}=\left(i^2\right)^{500}=\left(-1\right)^{500}=1} \\ & 1 &=& 0 -a+bi+c \\ &\mathbf{-a+bi+c} &=& \mathbf{1} \qquad (2) \\ \hline x=-i:& (-i)^{1000} &=& \left((-i)^2+1\right)(-i+1)q(x) + a(-i)^2+b(-i)+c \quad | \quad \boxed{(-i)^2=-1\\ (-i)^{1000}=1} \\ & 1 &=& 0 -a-bi+c \\ &\mathbf{-a-bi+c} &=& \mathbf{1} \qquad (3) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline &\mathbf{ a-b+c} &=& \mathbf{1} \qquad (1) \\ &\mathbf{-a+bi+c} &=& \mathbf{1} \qquad (2) \\ &\mathbf{-a-bi+c} &=& \mathbf{1} \qquad (3) \\ \hline (2)+(3): & -a+bi+c+-a-bi+c &=& 1+1 \\ &-2a+2c&=& 2 \quad &| \quad :2 \\ &- a+ c&=& 1 \\ & \mathbf{c} &=& \mathbf{1+a} \qquad (4) \\ \hline (1) : & a-b+c &=& 1 \quad &| \quad \boxed{c = 1+a} \\ & a-b+1+a &=& 1 \\ & 2a-b &=& 0 \\ & \mathbf{b} &=& \mathbf{2a} \qquad (5) \\ \hline (2) : & -a+bi+c &=& 1 \quad &| \quad \boxed{b = 2a,\ c = 1+a}\\ & -a+2ai+1+a &=& 1 \\ & 2ai &=& 0 \quad &| \quad :2i \\ &\mathbf{a} &=& \mathbf{0} \\ & b &=& 2a \\ & b &=& 2*0 \\ &\mathbf{b} &=& \mathbf{0} \\ & c&=& 1+a \\ & c &=& 1+0 \\ &\mathbf{c} &=& \mathbf{1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline r(x) &=& ax^2+bx+c \\ r(x) &=& 0*x^2+0*x +1 \\ \mathbf{r(x)} &=& \mathbf{1} \\ \hline \end{array}$$

The remainder is 1

Aug 1, 2019