Simplify!

- anonymous

Simplify!

- schrodinger

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- anonymous

\[\frac {12x^{5}}{20x^{2}}\]

- anonymous

..help?

- phi

can you simplify just the numbers? the 12/20 part?

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## More answers

- anonymous

I don't know how..

- anonymous

3x^3/5

- anonymous

divide them by 4..3/5 is the answer..

- anonymous

I want to know how to do this.

- phi

can you find the factors of 12? like 3*4 for example
for 20, 4*5
so the numbers part is
\[ \frac{3\cdot 4}{4\cdot 5} \]
the 4's "cancel" (that means 4/4 =1)
you get 3/5

- anonymous

but the thing is. how would i know how to do that?

- phi

the x's are actually not too hard, if you remember this rule:
if dividing x's to different powers, subtract the powers.

- anonymous

\[(12/20) \div4=3/5\]

- phi

but the thing is. how would i know how to do that?
for 12/20 notice that both are even. so you divide both by 2.
6/10
they are still even, divide by 2 again
3/5

- anonymous

but what aout the powers?

- phi

there a simple tricks to know if a number is divisible by another:
a number is even 2
sum of digits divisible by 3 then the number is divisible by 3
both of the above, then divisible by 6
ends in 0 or 5, divisible by 5

- phi

powers are easier. subtract when dividing. add when multiplying
here is how to remember:
x^2 means x*x
x^3 means x*x*x
multiply: x^2 * x^3 = x*x * x*x*x count the x's It is x^5 (the sum of 2 and 3)

- anonymous

hm...

- anonymous

So, on my problem, we'd subtract?

- phi

for
\[ \frac{x^5}{x^2} = \frac{x*x*x*x*x}{x*x}\]
we can cancel 2 of the x's from the top and bottom to get
\[ \frac{x^5}{x^2} = \frac{x*x*x*\cancel{x}*\cancel{x}}{\cancel{x}*\cancel{x}}= x*x*x \]
and x*x*x is written as \(x^3 \) because people don't like writing out all those x's

- phi

notice that the rule is just a fast way to get to the answer 5-2= 3

- anonymous

so my answer would ACTUALLY be \[\frac {3}{5}x^{3}\]

- phi

yes.

- anonymous

Could you help with another?

- phi

sure

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