Do you plug in g(x) in x2 and x?
you get f(g(x)) by replacing each x is f(x) by g(x)
* each x in
ok I have my work on my paper could you do it too so I can compare and see If I did it right?
ill medal and fan
3(5x^2 - 1)^2 - (5x^1 - 1) + 2
so far so good
what do you get for the next line?
75x^4-3-5x^2-1+2 that's what I get after I deal wit the parenthesis. Not sure if its correct though.
no you deal with the (5x^2 - 1)^2 first this is a binomial = (5x^2 - 1)(5x^1 - 1) first you multiply th e second parentheses vy the 5 x^2 then by the -1 so its 25x^" - 5x^2 - 5x^2 + 1 = 25x^2 - 10x^2 + 1
oh 'im making a lot of typos - its 3(25x^4 - 10x^2 + 1) - (5x^2 - 1) + 2 now be careful with - (5x^2 - 1) because you have to distribute the negative over the 5^2 - 1 and the -1 becomes + 1:- = 75x^4 - 30 x^2 + 3 - 5x^2 + 1 + 2 = 75x^4 - 35x^3 + 6
Is that just the first part of the equation or is that the answer after you combine like terms and stuff
- thats another typo!! its 35x^2 of course
- thats the answer
Oh I see what you did I multiplied wrong lol thanks for the help !
I'm really tired so i'd better call it a day on OS today.
when multiplying binomials its like this |dw:1437423013917:dw|
ya foil I just didn't do it for some reason too im tired aswell lol