Question

Radicals...

Answer 1

\[\large -6 \sqrt{160p^4}\]

Answer 2

steps:
\[-6 \times \sqrt{160} \times \sqrt{p^4}\]
\[-6 \times \sqrt{4 \times 40} \times p^2\]
\[-6 \times \sqrt{2 \times 2 \times 2 \times 2 \times 2} \times p^2 \rightarrow -6 \times 4\sqrt{2} \times p^2 \rightarrow -24\sqrt{2p^2}\]

Answer 3

\[\huge\rm \sqrt{p^4}\]
convert square root to an exponent form (1/2) \[\huge\rm (p^4)^{\frac{1}{2}}\]
equal to p^2 no square root we already cancel that out

Answer 4

in other words p^4 is same as p^2 times p^2 \[\huge\rm \sqrt{p^4} \rightarrow \sqrt{p^2 *p^2} \rightarrow p*p\]

Answer 5

so am i not right?

Answer 6

I dont understand your last line @calculusxy

Answer 7

how does the square root of 160 become the square root of 32?

Answer 8

\[-6 \times \sqrt{4 *40} \times p^2\] this is correct take it from here
factor 40
don't divide 4 into 2 times 2
4 is perfect square so just take the square root

Answer 9

sqrt 160 = 4 sqrt10

Answer 10

if you are dividing 160 by 2

make a group of 2