How do you divide #[4x^2-3x+1]/(3+x)#?

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1 Answer
EZ as pi
Aug 9, 2016

#[4x^2-3x+1]/(x+3) = (4x-15) " remainder " 46#

Explanation:

#color(red)[4x^2-3x+1]/color(blue)(x+3)#

In each line, divide #x# into the term with the highest power of #x#
Write the answer at the top.

#4x^2 div x = 4x#
#" " 4x-15#
#" "x+3bar(|" "color(red)(4x^2 -3x" "+1))#
#color(brown)(4x(x-2)->)" "underline(4x^2+12x)" " larr" Subtract#
#" "0 " "-15x#
#color(brown)(15(x-2))->" "underline(15x-45) larr Subtract#
#color(brown)("Remainder") ->" "0+46#

#:. [4x^2-3x+1]/(x+3) = (4x-15) " remainder " 46#

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