Question

Solve using long division Please

Answer 1

1. $Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}$

$Steps:$

$\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}$

$\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}$

$\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}$

$\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}$

$\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}$

2. $Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}$

$Steps:$

$\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}$

$\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}$

$\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$