Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 11 - Rational Expressions and Functions - 11-3 Dividing Polynomials - Got It? - Page 680: 3

Answer

a. $q^3+q^2 +2q+3$. b. $ h^2 -3h+5-\frac{3}{h+3}$.

Work Step by Step

a. The given expression is $(q^4+q^2+q-3)\div(q-1)$ Rewrite in standard form. $(q^4+0q^3+q^2+q-3)\div(q-1)$ $\begin{matrix} & q^3 & +q^2 &+2q ​&+3& & \leftarrow &Quotient\\ &-- &-- &--&--& \\ q-1) &q^4&+0q^3&+q^2&+q&-3 & \\ ​& q^4 & -q^3 & && & \leftarrow &q^3(q-1) \\ & -- & -- & & && \leftarrow &subtract \\ & 0 & +q^3 & +q^2 & & \\ & & q^3 & -q^2 & & & \leftarrow & q^2(q-1) \\ & & -- & -- & & &\leftarrow & subtract \\ & & 0&+2q^2 &+q & \\ ​& & & 2q^2& -2q && \leftarrow & 2q(q-1) \\ & & & -- & -- && \leftarrow & subtract \\ & & & 0 & +3q &-3& \leftarrow & Remainder \\ ​& & & & 3q& -3 & \leftarrow & 3(q-1) \\ & & && -- & -- & \leftarrow & subtract \\ & & & & 0 &0& \leftarrow & Remainder ​\end{matrix}$ The answer is $\Rightarrow Quotient + \frac{Remainder}{Divisor}$ $\Rightarrow q^3+q^2 +2q+3+\frac{0}{q-1}$ Simplify. $\Rightarrow q^3+q^2 +2q+3$. b. The given expression is $(h^3-4h+12)\div(h+3)$ Rewrite in standard form. $(h^3+0h^2-4h+12)\div(h+3)$ $\begin{matrix} & h^2 & -3h &+5 & & \leftarrow &Quotient\\ &-- &-- &--&& \\ h+3) &h^3&+0h^2&-4h&+12 & \\ ​& h^3 & +3h^2 & & & \leftarrow &h^2(h+3) \\ & -- & -- & & & \leftarrow &subtract \\ & 0 & -3h^2 & -4h & \\ & & -3h^2 & -9h & & \leftarrow & -3h(h+3) \\ & & -- & -- & &\leftarrow & subtract \\ & & 0&+5h &+12& \\ ​& & & 5h& +15 & \leftarrow & 5(h+3) \\ & & & -- & -- & \leftarrow & subtract \\ & & & 0 & -3& \leftarrow & Remainder ​ ​\end{matrix}$ The answer is $\Rightarrow Quotient + \frac{Remainder}{Divisor}$ $\Rightarrow h^2 -3h+5+\frac{-3}{h+3}$ Simplify. $\Rightarrow h^2 -3h+5-\frac{3}{h+3}$.
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