What is (2p-4)/(2p^3+4p^2)*(p^2+p-2)/(p^2-3p+2) ?

The solution to (2p-4)/(2p^3+4p^2)*(p^2+p-2)/(p^2-3p+2) is 1/(p^2)

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simplify (2p-4)/(2p^3+4p^2)*(p^2+p-2)/(p^2-3p+2)

simplify

\frac{2 p - 4}{2 p ^{3} + 4 p ^{2}} \cdot \frac{p ^{2} + p - 2}{p ^{2} - 3 p + 2}

\frac{1}{p ^{2}}

Solution steps

\frac{2 p - 4}{2 p ^{3} + 4 p ^{2}} \cdot \frac{p ^{2} + p - 2}{p ^{2} - 3 p + 2}

Cancel

\frac{2 p - 4}{2 p ^{3} + 4 p ^{2}} : \frac{p - 2}{p ^{2} ( p + 2 )}

= \frac{p - 2}{p ^{2} ( p + 2 )} \cdot \frac{p ^{2} + p - 2}{p ^{2} - 3 p + 2}

Cancel

\frac{p ^{2} + p - 2}{p ^{2} - 3 p + 2} : \frac{p + 2}{p - 2}

= \frac{p - 2}{p ^{2} ( p + 2 )} \cdot \frac{p + 2}{p - 2}

Cross-cancel common factor:

p - 2

= \frac{p + 2}{p ^{2} ( p + 2 )}

Cancel the common factor:

p + 2

= \frac{1}{p ^{2}}

2 y^{2} - 2 y - 4

- \frac{1}{4} \cdot (- \frac{1}{4})

(a + 2) (3 a^{2} + 12) (a - 2)

\frac{3}{8} - \frac{1}{4} + \frac{5}{6}

84 \div 12

What is (2p-4)/(2p^3+4p^2)*(p^2+p-2)/(p^2-3p+2) ?
The solution to (2p-4)/(2p^3+4p^2)*(p^2+p-2)/(p^2-3p+2) is 1/(p^2)