Ooops... I forgot the steps on how to do this. (2x+1)(2x^2+6x=1)
First multply all terms in the second bracket by 2x
Then mutiply all terms in the second bracket by 1
Then collect like terms
Just to be safe: I believe you meant to say (2x+1)(2x^2 + 6x) = 1 Otherwise, it won't make sense.
no it was supposed to be (2x+1)(2x^2+6x+1)
Ah ok. Good thing we corrected that. :P
yeah, it's just the way I typed it. I can't seem to figured it out for some reason
still stuck?
yeah, I've been working on assignment for this class the last two days, I think I burned my brain out a bit.
ok lets go slow
this isn't an assignment question though, it's just been bugging me
\[(2x+1)(2x^2+6x+1)\] is the thing you need to multiply
ok
if you like you can write one on top of the other and multiply just like you would a 3 digit number by a 2 digit number. 6 multiplications and then additions (combining like terms)
but that is too cumbersome to write here, so the idea is just to multiply everything in the second parentheses by 2x and then multiply everything in the second parentheses by 1
we start with \[2x(2x^2+6x+1)\] \[=2x\times 2x^2+2x\times 6x+2x\times 1\] and get \[4x^3+12x^2+2x\]
then we multiply by 1, which is a lot like doing nothing: \[1\times (2x^2+6x+1)=2x^2+6x+1\]
the exponents confuse me
so we have \[4x^3+12x^2+2x+2x^2+6x+1\] and now combine like terms to get \[4x^3+14x^2+8x+1\]
ah when you multiply add the exponents. just like \[10\times 100=1000\] \[x\times x^2=x^3\]
huh, ok, that makes sense
so when you multiply \[2x\times 2x^2\] you get \[2\times 2\times x^1\times x^2=4\times x^{1+2}=4x^3\]
and similarly \[2x\times 6x=2\times 6\times x^1\times x^1=12\times x^{1+1}=12x^2\]
got it?
yeah thanks, I was just working out similar problems to make sure i got it down
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