write the binomial expansion of (4x-3)^5 i really and lost on tis topic please explain cleary thanks
general form is \[(a+b)^n=\sum_{k=0}^n\dbinom{n}{k}a^kb^{n-k}\]
here \(n=5, a=4x,b=-3\) so your but before we compute, lets find the fifth line of pascal's triangle. do you know how to do that?
i do no the triangle
the last line of the triangle pictured on the right is what you need http://en.wikipedia.org/wiki/Pascal%27s_triangle
i do so the coefficients are 1,5,10,10,5,1
the last line is 1 5 10 10 5 1 so your expansion is going to look like \[(4x)^5+5\times (4x)^4\times (-3)^1+10\times (4x)^3\times (-3)^2\] \[+10\times (4x)^2\times (-3)^3+5\times (4x)^1\times (-3)^4+(-3)^5\] yes and now you have a bunch of arithmetic to do to clean this up
wait wheres the variable y
There is no variable y in your question...
no my teacher told me that u change (4x-3)^5 to (x+y)^5 to make it easier then
u just plug it in
but see thats were i get lost
Hmm... in that case x=4x, y=-3 That's what satellite73 has plugged in :|
Here it goes... \[(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 +5xy^4 + y^5\] Put, x=4x and y=-3 into above expression. \[(4x-3)^5 = (4x)^5 + 5(4x)^4(-3) + 10(4x)^3(-3)^2 \]\[+ 10(4x)^2(-3)^3 +5(4x)(-3)^4 + (-3)^5\]
okay so then do a add all of the 3s with the exponets
Then you should simplify the terms... for example, (4x)^5 = 1024x^5 , 5(4x)^4(-3) = -3840x^4 and so on..
ohhhh okay so i can combine the coefficent with "Y"
What do you mean?
so let me solve it and ill type the answer and u can tell me if my work is rite k just hold on a sec.
iwhat i mean is i can muitply 4 and ^5
Hmm... 4^5 = 4x4x4x4x4 ...
1024x^5-768x^4+576x^3-432x^2+324x-243
is that rite
The first term and the last terms are correct. The rest are not
okay i see wat i did ring i didnt mulptiy the coffients with it lol
i got it now thanks
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