What is (z^2)/(z+1)*(z^2+3z+2)/(z^2+3z) ?

The solution to (z^2)/(z+1)*(z^2+3z+2)/(z^2+3z) is (z(z+2))/(z+3)

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simplify (z^2)/(z+1)*(z^2+3z+2)/(z^2+3z)

simplify

\frac{z ^{2}}{z + 1} \cdot \frac{z ^{2} + 3 z + 2}{z ^{2} + 3 z}

\frac{z ( z + 2 )}{z + 3}

Solution steps

\frac{z ^{2}}{z + 1} \cdot \frac{z ^{2} + 3 z + 2}{z ^{2} + 3 z}

Apply the fraction rule:

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

= \frac{z ^{2} ( z ^{2} + 3 z + 2 )}{( z + 1 ) ( z ^{2} + 3 z )}

Factor

z^{2} + 3 z + 2 : (z + 1) (z + 2)

= \frac{z ^{2} ( z + 1 ) ( z + 2 )}{( z + 1 ) ( z ^{2} + 3 z )}

Cancel the common factor:

z + 1

= \frac{z ^{2} ( z + 2 )}{z ^{2} + 3 z}

Factor

z^{2} + 3 z : z (z + 3)

= \frac{z ^{2} ( z + 2 )}{z ( z + 3 )}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

z^{2} = zz

= \frac{zz ( z + 2 )}{z ( z + 3 )}

Cancel the common factor:

z

= \frac{z ( z + 2 )}{z + 3}

\frac{2}{x + 1} - \frac{4}{x - 1}

(\frac{2}{3})^{- 2} (\frac{2}{3})^{3}

\frac{1}{\frac{1}{3}}

8^{2} - 6^{2}

\frac{- 6 ^{- 2} x ^{0} z ^{- 6}}{2 y ^{- 4}}

What is (z^2)/(z+1)*(z^2+3z+2)/(z^2+3z) ?
The solution to (z^2)/(z+1)*(z^2+3z+2)/(z^2+3z) is (z(z+2))/(z+3)