### Home > PC > Chapter 7 > Lesson 7.2.3 > Problem7-62

7-62.

Complete the following division problem. Express any remainder as a fraction.

$\frac { x ^ { 5 } - 1 } { x - 1 }$

Choose one of the methods in the box below.

Using Long division:

$\require{enclose} \begin{array}{rll} x^3-4x^2-8x+2\ \\[-3pt] x-2 \enclose{longdiv}{x^4-6x^3+0x^2+18x-1}\kern-.2ex \\[-3pt] \underline{x^4-2x^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phantom{00}} && \\[-3pt] -4x^3+0x^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phantom{0} \\[-3pt] \underline{\phantom{0}-4x^3+8x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phantom{0}} \\[-3pt] \phantom{0}-8x^2+18x\ \ \ \ \ \ \ \ \\[-3pt] \underline{\phantom{0}-8x^2+16x \ \ \ \ \ \ \phantom{0}} \\[-3pt] {\phantom{0}2x-1 \ \ \phantom{0}} \\[-3pt] \underline{\phantom{0}2x-4 \ \ \phantom{0}} \\[-3pt] {\phantom{0}3 \ \ \phantom{0}} \\[-3pt] \text{ remainder } \uparrow \ \ \ \end{array}$

Final answer:  $x^3-4x^2-8x+2+\frac{3}{x-2}$

Using Generic Rectangles:

 2 by 5 generic rectangle top edge first$x^3$ top edge second$-4x^2$ top edge third$-8x$ top edge fourth$+2$ top edge fifthRemainder left edge top $x$ interior top first$x^4$ interior top second$-4x^3$ interior top third$-8x^2$ interior top fourth$+2x$ interior top fifth$3$ left edge bottom$-2$ interior bottom first$-2x^3$ interior bottom second$+8x^2$ interior bottom third$+16x$ interior bottom fourth$-4$ interior bottom fifth blank $x^4$Below generic rectangle $-6x^3$ $+0x^2$ $+18x$ $-1$ Answer: $x^3-4x^2-8x+2+\frac{3}{x-2}$