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( (1/7)x^2 - (2/5)) * ( (1/2)x + (1/2) )

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multiply each term of the anyone binomial into the other. Like this: (a+b)(c+d) = a(c+d) + b(c+d)
so..[ (1/7)x^2 * ((1/2)x^2) + (1/2)) ] + [ ((2/5) * ((1/2)x^2) + (1/2)) ] ?
yup thts right..

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Other answers:

good except that the (2/5) has a 'minus' sign with it.
\[\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{3}+\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{2}-\left(\begin{matrix}2 \\ 10\end{matrix}\right)x-\left(\begin{matrix}2 \\ 10\end{matrix}\right)\]
@math456 oop - that notation is actually used for 'combination' or 'C' :P. Other wise it's good.
crap im so lost.
ohh yea i know i couldnt find the division sign thts y i use this..
forget that comment, @musicalrose , you're almost there, just keep in mind that it's a "-2/5" over there, so you include the 'minus' sign as well while multiplying.
@musicalrose i will make it clear.. u use the same concept..
[ (1/7)x^2 *((1/2)x+(1/2) ] - (2/5*((1/2)x^2) + (1/2)) ] ?
yes itd be written like that
.07x^3 + 1/2 for the first part?
(1/14)x3+(1/14)x2−(2/10)x−(2/10) <-- like this
oooh because you made like denominators?
yup... u'll get 0.07 for 1/14
so shouldnt the 1/14x2 actually be 7/14 since it was 1/2x2? or am I not getting it lol
I think I get it now..
thank you tons for spending so much time helping me :)
no problem..!

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