# Math Help - [SOLVED] long division/ Remainder Theorem

1. ## [SOLVED] long division/ Remainder Theorem

Problem:

Use long division and the Remainder Theorem to evaluate , where

The quotient is
The remainder is
=

2. Originally Posted by lsnyder

Problem:

Use long division and the Remainder Theorem to evaluate , where

The quotient is
The remainder is
=
Divide $x^3- 5x^2+ 8x- 6$ by $x-2$

x divides into $x^3$ $x^2$ times and $x^3- 5x^2- x^2(x- 2)= x^3- x^3- 5x^2+ 2x^2= -3x^2$
so we now have $-3x^2+ 8x- 6$ left.

x divides into $-3x^2$ $-3x$ times and $-3x^2+ 8x- (-3x)(x-2)= -3x^+ 3x^2+ 8x- 6x= 2x$
so we now have $2x- 6$ left.

x divides into $2x- 6$ 2 times and $2x- 6- 2(x-2)= 2x- 2x- 6+ 4= -2$

The quotient is $x^2- 3x+ 2$ and the remainder is -2.

(Check my arithmetic.)