We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
82
1
avatar

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

 Jul 31, 2019
 #1
avatar+23082 
+2

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

 

laugh

 Aug 1, 2019

16 Online Users

avatar