Distributive property.
\[3xy(2x^2+xy)\]
I’m 16
almost 17, remember?
Thanks for reminding me. Have fun distributing.
Gonna call the cops on me? I dunno where to start, it seems weird.
Even if I wanted to call the cops (which I don't) but if I did, how would I know what address to give them? Utimately, you are your parents responsibility. I am merely guidance and support. Of course understand that even God did not make a rule that said you must be over a certain age in order to drink certain drinks. That's strictly a manmade rule.
Touché, how do I start?
Well, for starters the distributive Property still applies. You start by identifying your a, b, and c.
Distributive Propery is \(a(b + c) = ab + ac\)
a=3xy b=2x^2 c=xy
The expression on the outside of the parentheses is always \(a\). The first term in parentheses is always b. The last term in parentheses is always c. Yes exactly
Now you may proceed from there with distribution.
\[6x^3+3x^2y^2\]
You have at least one variable missing from your final result amongst other possible errors.
I don’t... um... see one...
sigh
This is why you should post all your steps. Mind doing so?
I uh... did it mentally..
Yes and it is good practice to do for yourself, but when you are presenting work to others, you should show your steps explicitly.
The result you have on paper does not match what you posted here. Do you see that?
Observe the difference between what you have on paper and what you posted as your initial answer.
A left the y out of \[6x^3y\]
Yes. It is otherwise correct.
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