Step by step solution to "The intial fraction is:3x3−x+4x3+2x2+6xHow should to make the..."

misangela4gi

Answered question

2022-04-22

The intial fraction is:

\frac{3 x^{3} - x + 4}{x^{3} + 2 x^{2} + 6 x}

How should to make the decomposition for the denominator?
Thanks!

elvis0217t2x

Beginner2022-04-23Added 13 answers

First divide the numerator by the denominator!

\frac{3 x^{3} - x + 4}{x^{3} + 2 x^{2} + 6 x} = 3 + \frac{- 6 x^{2} - 19 x + 4}{x^{3} + 2 x^{2} + 6 x}

Then note that the denominator factors, at best, to:

x^{3} + 2 x^{2} + 6 x = x (x^{2} + 2 x + 3)

3 + \frac{- 6 x^{2} - 19 x + 4}{x^{3} + 2 x^{2} + 6 x} = 3 + \frac{A}{x} + \frac{B x + C}{(x^{2} + 2 x + 3)}

Note that since the numerator and denominator of the fraction have the same order so we can divide them to find out:

\frac{3 x^{3} - x + 4}{x^{3} + 2 x^{2} + 6 x} = 3 + \frac{- 6 x^{2} - 19 x + 4}{x^{3} + 2 x^{2} + 6 x}

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