(x^4+7x^3+17x^2+20x)÷(x^2+4)
\(x^2+7x+13+\frac{-8x-52}{x^2+4}\)
is this what u are looking for?
Yes but how'd you get the answer, I got stuck in the beginning
\(\frac{x^4+7x^3+17x^2+20x}{x^2+4} \)
\(=x^2+\frac{7x^3+13x^2+20x}{x^2+4}\)
\(=x^2+7x+\frac{13x^2-8x}{x^2+4}\)
\(=x^2+7x+13+\frac{-8x-52}{x^2+4}\)
x^2 + 7x + 13
x^2 + 4 [ x^4 + 7x^3 + 17x^2 + 20x + 0 ]
x^4 + 4x^2
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7x^3 + 13x^2 + 20x
7x^3 + 28x
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13x^2 - 8x + 0
13x^2 + 52
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-8x - 52
x^2 + 7x + 13 R - (8x + 52) / (x^2 + 4)
+514 
