Question

Multiplying...

Answer 1

what

Answer 2

Wait a moment please.

Answer 3

k

Answer 4

\[(4\sqrt{2m^2 + 4m})(4\sqrt{3m^2}-3\sqrt{5m^2})\]

Answer 5

how about to convert radical to an exponent form first :=))

Answer 6

\[4\sqrt{2m^2}(4\sqrt{3m^2})+4\sqrt{2m^2}(-3\sqrt{5m^2})+4m(4\sqrt{3m^2})+4m(-3\sqrt{5m^2})\]

Answer 7

If you have an easier way, please tell me.

Answer 8

ye i would to convert radical to an exponent form first

\[\huge\rm \sqrt[\color{Red}{n}]{x^\color{blue}{m}}=x^\frac{\color{blue}{ m }}{ \color{ReD}{n} }\]index becomes denominator of fraction

Answer 9

like*

Answer 10

aww i see wait is it \[\large\rm (4\sqrt{2m^2} +\sqrt{ 4m})(4\sqrt{3m^2}-3\sqrt{5m^2})\]
like this ?

Answer 11

yeah

Answer 12

alright then.
you can multiply the numbers under the square root
like \[\large\rm 4\sqrt{2m^2}*4\sqrt{3m^2} \rightarrow 16 \sqrt{6m^4}\]

Answer 13

yes i got that

Answer 14

that's the easy i can see
instead writing a*b just directly right ab

Answer 15

way*

Answer 16

so after i used foil i got:
\[16\sqrt{6m^4} - 12\sqrt{10m^4} + 16m \sqrt{3m^2}-12m \sqrt{5m^2}\]

Answer 17

hmm isn't it\[\large\rm (4\sqrt{2m^2} +\color{Red}{\sqrt{ 4m}})(4\sqrt{3m^2}-3\sqrt{5m^2})\]
so it should be sqrt{4m} times 4sqrt3m^2}

Answer 18

\[16\sqrt{6m^4} - 12\sqrt{10m^4} + \color{Red}{16}m \sqrt{3m^2}-12m \sqrt{5m^2}\]
how did you get 16 here

Answer 19

no it's only 4m without the radical

Answer 20

\(\color{#0cbb34}{\text{Originally Posted by}}\) @calculusxy
\[(4\sqrt{2m^2 + 4m})(4\sqrt{3m^2}-3\sqrt{5m^2})\]
\(\color{#0cbb34}{\text{End of Quote}}\)


THIS IS NOT CORRECT! I JUST NOTICED!

Answer 21

yea so that's why i asked
is it like this \[\large\rm (4\sqrt{2m^2} +\sqrt{ 4m})(4\sqrt{3m^2}-3\sqrt{5m^2})\]

Answer 22

\[\large\rm (4\sqrt{2m^2} +4m)(4\sqrt{3m^2}-3\sqrt{5m^2})\]

Answer 23

\[\rm 16\sqrt{6m^4} - 12\sqrt{10m^4} + \color{Red}{16}m \sqrt{3m^2}-12m \sqrt{5m^2}\]
this is correct :=)

Answer 24

\[16\sqrt{6m^4} - 12\sqrt{10m^4} + 16m \sqrt{3m^2}-12m \sqrt{5m^2}\]
\[16m^2\sqrt{6} - 12m^2\sqrt{10} +16m^2\sqrt{3}-12m^2\sqrt{5}\]

Answer 25

\[16\sqrt{6m^4} - 12\sqrt{10m^4} + 16m \sqrt{3m^2}-12m \sqrt{5m^2}\]
\[16m^2\sqrt{6} - 12m^2\sqrt{10} +16m^2\sqrt{3}-12m^2\sqrt{5}\]
\[16\sqrt{6}-12\sqrt{10}+16\sqrt{3}-12\sqrt{5}\]

Answer 26

\[16\sqrt{6}-12\sqrt{10}+16\sqrt{3}-12\sqrt{5}\] would that be correct?

Answer 27

what happened to m^2?
remember we should keep the GCF outside the parentheses
ab+ac= a(b+C)

Answer 28

and what about common number ?? :=))

Answer 29

well the answer key says that that is the correct answer

Answer 30

well there is GCF(greatest common factor )
you took out the m^2 from each term
\[m^2(16\sqrt{6}-12\sqrt{10}+16\sqrt{3}-12\sqrt{5})\]
and 4 is also the common factor

Answer 31

ok thanks :)

Answer 32

np :=))