How do you divide $(3x^4 + 5x^2 + 12x + 3)/(x+3)$ ?

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Answer 1 · 2017-06-06T13:48:22

$3x^3-9x^2+32x-84+255/(x+3)$

Note that $0x^3$ is a place keeper and has no value. I use this to make formatting easier.

$" "3x^4+0x^3+5x^2+12x+3$
$color(magenta)(+3x^3)(x+3)-> ul(3x^4+9x^3 " "larr" Subtract")$

$" "0color(white)(.)-9x^3+5x^2+12x+3$
$color(magenta)(-9x^2)(x+3)->" " ul(-9x^3-27x^2larr" Subtract")$

$" "0color(white)(.)+32x^2+12x+3$
$color(magenta)(+32x)(x+3)-> " "ul(32x^2+96x larr" Subtract")$

$" "0" "-84x+3$
$color(magenta)(-84)(x+3)->" " ul(-84x-252larr" Subtract")$
$" "color(magenta)(+255larr" Remainder")$

$color(white)(.)$

$color(magenta)(3x^3-9x^2+32x-84+255/(x+3))$

How do you divide (3x^4 + 5x^2 + 12x + 3)/(x+3)(3x^4 + 5x^2 + 12x + 3)/(x+3)?