Question

f(x)=x^2+2x-1
g(x)=2x-4

solve

2f(x)-3g(x)

Answer 1

\[2(\overbrace{x^2+2x-1}^{f(x)})-3(\overbrace{2x-4}^{g(x)})\] is a start, then a raft of algebra

Answer 2

@satellite73 2(x^2+4x-2) -(6x+12)?

Answer 3

the second part is right, but you should also distribute the 2 for the first parentheses
then combine like terms

Answer 4

oh wait, there is a mistake in the last part too

Answer 5

oh this stuff is so confusing lol

Answer 6

lets go slow then, because it should not be that hard

Answer 7

lets tackle
\[2(x^2+2x-1)\]first
distribute the 2 and remove the parentheses
what do you get?

Answer 8

2x^2+4x-2

Answer 9

good, now lets put that aside and distribute the \(-3\)in
\[-3(2x-4)\]

Answer 10

-6x+12

Answer 11

ok so next step is to combine like terms in
\[2x^2+4x-2-6x+12\]

Answer 12

2x^2-2x+10

Answer 13

looks good to me
not that hard, is it?

Answer 14

thats it? haha thanks

Answer 15

yeah, that's it, but as judge judy says "the questions get harder"
yw