What is expand (f(x^2-5+h)-x^2-5)/h ?

The solution to expand (f(x^2-5+h)-x^2-5)/h is (fx^2)/h-(5f)/h+f-(x^2)/h-5/h

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expand (f(x^2-5+h)-x^2-5)/h

expand

\frac{f ( x ^{2} - 5 + h ) - x ^{2} - 5}{h}

\frac{f x ^{2}}{h} - \frac{5 f}{h} + f - \frac{x ^{2}}{h} - \frac{5}{h}

Solution steps

\frac{f ( x ^{2} - 5 + h ) - x ^{2} - 5}{h}

Expand

f (x^{2} - 5 + h) - x^{2} - 5 : f x^{2} - 5 f + f h - x^{2} - 5

= \frac{f x ^{2} - 5 f + f h - x ^{2} - 5}{h}

Apply the fraction rule:

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

\frac{f x ^{2} - 5 f + f h - x ^{2} - 5}{h} = \frac{f x ^{2}}{h} - \frac{5 f}{h} + \frac{f h}{h} - \frac{x ^{2}}{h} - \frac{5}{h}

= \frac{f x ^{2}}{h} - \frac{5 f}{h} + \frac{f h}{h} - \frac{x ^{2}}{h} - \frac{5}{h}

Cancel

\frac{f h}{h} : f

= \frac{f x ^{2}}{h} - \frac{5 f}{h} + f - \frac{x ^{2}}{h} - \frac{5}{h}

ln (t) (t ln (t) - t)

\frac{1}{s ( s ^{s} + 2 s + 2 )}

\frac{2 x ^{2} + x + 8}{( 2 x - 1 ) ( x + 3 )}

\frac{( ( x - 4 ) ^{4} )}{4}

(2 + \frac{5}{8} x)^{8}

What is expand (f(x^2-5+h)-x^2-5)/h ?
The solution to expand (f(x^2-5+h)-x^2-5)/h is (fx^2)/h-(5f)/h+f-(x^2)/h-5/h