## musicalrose 3 years ago ( (1/7)x^2 - (2/5)) * ( (1/2)x + (1/2) )

1. apoorvk

multiply each term of the anyone binomial into the other. Like this: (a+b)(c+d) = a(c+d) + b(c+d)

2. musicalrose

so..[ (1/7)x^2 * ((1/2)x^2) + (1/2)) ] + [ ((2/5) * ((1/2)x^2) + (1/2)) ] ?

3. math456

yup thts right..

4. apoorvk

good except that the (2/5) has a 'minus' sign with it.

5. math456

$\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{3}+\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{2}-\left(\begin{matrix}2 \\ 10\end{matrix}\right)x-\left(\begin{matrix}2 \\ 10\end{matrix}\right)$

6. apoorvk

@math456 oop - that notation is actually used for 'combination' or 'C' :P. Other wise it's good.

7. musicalrose

crap im so lost.

8. math456

ohh yea i know i couldnt find the division sign thts y i use this..

9. apoorvk

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.

10. math456

@musicalrose i will make it clear.. u use the same concept..

11. math456

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

12. musicalrose

yes itd be written like that

13. musicalrose

.07x^3 + 1/2 for the first part?

14. math456

(1/14)x3+(1/14)x2−(2/10)x−(2/10) <-- like this

15. musicalrose

oooh because you made like denominators?

16. math456

yup... u'll get 0.07 for 1/14

17. musicalrose

so shouldnt the 1/14x2 actually be 7/14 since it was 1/2x2? or am I not getting it lol

18. math456

|dw:1339011637725:dw|

19. musicalrose

I think I get it now..

20. math456

kool..

21. musicalrose

thank you tons for spending so much time helping me :)

22. math456

no problem..!