Would anyone be able to show me how to do this...
x is so small that in terms of x^3 and higher can be ignored and
(2-x)(3+x)^4 = a +bx+cx^2
find the values of the constants a, b, c
x Turn on thread page Beta
2nd Binomial Expansion PLEASE HELP!!! watch
- Thread Starter
- 26-09-2010 20:11
- 26-09-2010 21:40
Well, can you expand (3 + x)^4? What do you get?
Then multiply it by (2 - x), but don't do it completely. Since you only need up to x^2, the 2 from (2-x) only needs to multiply up to quadratic terms in (3 + x)^4 and the -x in (2-x) only needs to multiply up to x terms in (3 + x)^4.
This will sound confusing, but just try doing it.
- 26-09-2010 23:13
using this http://upload.wikimedia.org/math/0/5...71d5cdae8a.png
expand (3 + x)^4
multiply (2 - x) by expanded (3 + x)^4.
then i got