Taking factors out of a radical (extra practice)

Text

Take out of the radical sign every power you can:

• (a) $\sqrt[5]{2^{12}\cdot 3^{24}}$
• (b) $\sqrt[3]{12^7\cdot 3^{18}}$
• (c) $\sqrt[3]{4^{55}\cdot 5^{13}\cdot 3^{19}}$

Solution

First factor into primes ($12=2^2\cdot 3$, $4=2^2$), then extract the quotient of each exponent divided by the index.

(a) $12=5\cdot 2+2$, $24=5\cdot 4+4$: $\sqrt[5]{2^{12}\cdot 3^{24}}=2^2\cdot 3^4\sqrt[5]{2^2\cdot 3^4}=324\sqrt[5]{324}$.

(b) $12^7=(2^2\cdot 3)^7=2^{14}\cdot 3^{7}$, hence $2^{14}\cdot 3^{7+18}=2^{14}\cdot 3^{25}$. With $14=3\cdot 4+2$, $25=3\cdot 8+1$: $\sqrt[3]{2^{14}\cdot 3^{25}}=2^4\cdot 3^8\sqrt[3]{2^2\cdot 3}=104976\sqrt[3]{12}$.

(c) $4^{55}=2^{110}$, so $\sqrt[3]{2^{110}\cdot 5^{13}\cdot 3^{19}}$. With $110=3\cdot 36+2$, $13=3\cdot 4+1$, $19=3\cdot 6+1$: $2^{36}\cdot 5^{4}\cdot 3^{6}\sqrt[3]{2^{2}\cdot 5\cdot 3}=2^{36}\cdot 5^{4}\cdot 3^{6}\sqrt[3]{60}$.

$\boxed{\;(a)\;324\sqrt[5]{324}\qquad (b)\;104976\sqrt[3]{12}\qquad (c)\;2^{36}\cdot 5^{4}\cdot 3^{6}\sqrt[3]{60}\;}$