Question 1

Solve using long division
Please

Question 2

1.
$Solution,(2x^3+4x^2-5)/(x+3):\quad 2x^2-2x+6-(23)/(x+3)$

Steps:

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

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

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

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

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

2.
$Solution, (4x^3-2x^2-3)/(2x^2-1):\quad 2x-1+(2x-4)/(2x^2-1)$

Steps:

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

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

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

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

$\mathrm{-Austint1414}$

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