What is expand 1/((s^2+2s+2))(1/((s+1)^2)) ?

The solution to expand 1/((s^2+2s+2))(1/((s+1)^2)) is 1/(s^4+4s^3+7s^2+6s+2)

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expand 1/((s^2+2s+2))(1/((s+1)^2))

expand

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

\frac{1}{s ^{4} + 4 s ^{3} + 7 s ^{2} + 6 s + 2}

Solution steps

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

Remove parentheses:

(a) = a

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

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

= \frac{1}{s ^{2} + 2 s + 2} \cdot \frac{1}{s ^{2} + 2 s + 1}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

Multiply the numbers:

1 \cdot 1 = 1

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

Expand

(s^{2} + 2 s + 2) (s^{2} + 2 s + 1) : s^{4} + 4 s^{3} + 7 s^{2} + 6 s + 2

= \frac{1}{s ^{4} + 4 s ^{3} + 7 s ^{2} + 6 s + 2}

e^{2 x} + e^{- 2 x} (2 e^{2 x} + 2 e^{- 2 x})

\frac{n ^{2} ( n + 1 ) ^{2}}{4} + (n + 1)

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

(x - 2 + 8)^{2}

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

What is expand 1/((s^2+2s+2))(1/((s+1)^2)) ?
The solution to expand 1/((s^2+2s+2))(1/((s+1)^2)) is 1/(s^4+4s^3+7s^2+6s+2)