What is expand ((x+3)(2x-1))/(P(x)) ?

The solution to expand ((x+3)(2x-1))/(P(x)) is (2x)/P+5/P-3/(xP)

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expand ((x+3)(2x-1))/(P(x))

expand

\frac{( x + 3 ) ( 2 x - 1 )}{P ( x )}

\frac{2 x}{P} + \frac{5}{P} - \frac{3}{x P}

Solution steps

\frac{( x + 3 ) ( 2 x - 1 )}{P ( x )}

Remove parentheses:

(a) = a

= \frac{( x + 3 ) ( 2 x - 1 )}{P x}

Expand

(x + 3) (2 x - 1) : 2 x^{2} + 5 x - 3

= \frac{2 x ^{2} + 5 x - 3}{x P}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 x ^{2} + 5 x - 3}{P x} = \frac{2 x ^{2}}{P x} + \frac{5 x}{P x} - \frac{3}{P x}

= \frac{2 x ^{2}}{x P} + \frac{5 x}{x P} - \frac{3}{x P}

Cancel

\frac{2 x ^{2}}{P x} : \frac{2 x}{P}

= \frac{2 x}{P} + \frac{5 x}{x P} - \frac{3}{x P}

Cancel

\frac{5 x}{P x} : \frac{5}{P}

= \frac{2 x}{P} + \frac{5}{P} - \frac{3}{x P}

(2 + x cos (y)) d

(e^{6 (3 + 5 t)}) (6 t^{5}) + (t^{6} e^{6 (3 + 5 t)}) (5)

\frac{n \cdot ( n + 1 )}{2}

(e^{i θ} + e^{- i θ})^{4}

\frac{x}{( x + 1 ) ( x - 3 )}

What is expand ((x+3)(2x-1))/(P(x)) ?
The solution to expand ((x+3)(2x-1))/(P(x)) is (2x)/P+5/P-3/(xP)