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(2x³+4x²-x+8) / (x² + 3x + 2)

(2x³+4x²-x+8) / (x² + 3x + 2)
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(2x³+4x²-x+8) / (x² + 3x + 2)

User Megabeets
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1 Answer

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(2x^3+4x^2 -x+8)/(x^2 +3x+2)\\\\\\=\frac{2x^3+6x^2+4x-2x^2-6x-4+x+12}{}\\\\=(2x(x^2 +3x+2)-2(x^2 +3x+2)+x+12)/(x^2 +3x+2)\\\\\\=(2x(x^2+3x+2))/(x^2 +3x +2 )- (2(x^2 +3x +2))/(x^2 +3x +2 ) + (x+12)/(x^2 +3x+2)\\\\\\=2x-2 + (x+2)/(x^2 +3x +2)\\\\\\=2x-2 + (x+12)/(x^2 +2x +x +2)\\\\\\=2x-2 + (x+12)/(x(x+2) + x+2)\\\\\\=2x-2 + (x+12)/((x+1)(x+2))\\

User Wyattisimo
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