Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Appendix A - Review - A.3 Polynomials - A.3 Assess Your Understanding - Page A30: 74

Answer

Quotient $=x^2+x-1$ Remainder $=-2x+2$

Work Step by Step

The given expression is $(1-x^2+x^4)\div(x^2-x+1)$ Rewrite the expression as $(x^4+0x^3-x^2+0x+1)\div(x^2-x+1)$ Perform long division: $\begin{matrix} & x^2 & +x & -1 ​& & & \leftarrow &\text{Quotient}\\ &-- &-- &--&--&-- \\ x^2-x+1) &x^4&+0x^3&-x^2&+0x&+1 & \\ ​& x^4 &-x^3 &+x^2 & && \leftarrow &x^2(x^2-x+1) \\ & -- & -- & --& && \leftarrow &\text{subtract} \\ & 0 & x^3& -2x^2 &+0x & \\ & & x^3& -x^2 &x && \leftarrow & x(x^2-x+1) \\ & & --& -- & -- & & \leftarrow & \text{subtract} \\ & & & -x^2&-x &+1& \\ & & & -x^2 &+x &-1& \leftarrow & -1(x^2-x+1) \\ & & & -- & -- & -- & \leftarrow & \text{subtract}\\ & & & &-2x &+2& \leftarrow & \text{Remainder} ​\end{matrix}$ Checking: $\text{(Quotient)(divisor)+ Remainder}$ $=(x^2+x-1)(x^2-x+1)-2x+2$ $=x^4+x^3-x^2-x^3-x^2+x+x^2+x-1-2x+2$ $=x^4-x^2+1$ $=\text{ Dividend}$ Hence, the quotient is $x^2+x-1$ and the remainder is $-2x+2$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.