what is the remainder when $(x^{3}+1)$ is divided by $(x^{2}-x + 1)$?\n$x + 1$\n$x$\n$2$\n$0$
what is the remainder when $(x^{3}+1)$ is divided by $(x^{2}-x + 1)$?\n$x + 1$\n$x$\n$2$\n$0$
Answer
Answer:
D. 0
Explanation:
Step1: Use polynomial long - division
We perform long - division of (x^{3}+1) by (x^{2}-x + 1). First, divide the leading term of the dividend (x^{3}) by the leading term of the divisor (x^{2}), we get (x). Multiply (x^{2}-x + 1) by (x): (x(x^{2}-x + 1)=x^{3}-x^{2}+x). Subtract this from the dividend: ((x^{3}+1)-(x^{3}-x^{2}+x)=x^{2}-x + 1).
Step2: Divide the new dividend by the divisor
Divide the leading term of (x^{2}-x + 1) (which is (x^{2})) by the leading term of (x^{2}-x + 1) (which is also (x^{2})), we get (1). Multiply (x^{2}-x + 1) by (1): (1\times(x^{2}-x + 1)=x^{2}-x + 1). Subtract this from (x^{2}-x + 1): ((x^{2}-x + 1)-(x^{2}-x + 1)=0). So the remainder is (0).