((x^2+3x+4)-2)
simplifies to -2x^2-6x-8

- anonymous

((x^2+3x+4)-2)
simplifies to -2x^2-6x-8

- katieb

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- anonymous

fog..
when f(x)=x-2
g(x)=x^2+3x+4
=((x^2+3x+4)-2) = -2x^2-6x-8

- jim_thompson5910

f(x)=x-2
f(g(x))=g(x)-2
f(g(x))=x^2+3x+4-2
f(g(x))=x^2+3x+2

- jim_thompson5910

You're not multiplying, you're subtracting in this case

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## More answers

- anonymous

oh.. so I don't distribute the -2?

- anonymous

so for fof..
would is be.. (x-2)-2=x-4?

- jim_thompson5910

no, you're not distributing since it's originally x-2 and not -2x

- jim_thompson5910

yes, fof is x-4

- anonymous

no the roots is X=-1 or -2

- anonymous

what about ... (x-2)^2+3(x-2)+4?

- jim_thompson5910

(x-2)^2+3(x-2)+4
x^2-4x+4+3(x-2)+4
x^2-4x+4+3x-6+4
x^2-x+2

- anonymous

the polynomial discriminant: =1

- anonymous

:)

- anonymous

hope that helps

- anonymous

how'd you go from (x-2)^2+3(x-2)+4
to this..
x^2-4x+4+3(x-2)+4
?

- jim_thompson5910

(x-2)^2
(x-2)(x-2)
x(x-2) - 2(x-2)
x^2 - 2x - 2x + 4
x^2 - 4x + 4

- jim_thompson5910

So (x-2)^2 is the same as x^2 - 4x + 4

- jim_thompson5910

which makes
(x-2)^2+3(x-2)+4
the same as
x^2-4x+4+3(x-2)+4

- anonymous

yea, yeah, I wasn't thinking.

- jim_thompson5910

that's ok

- anonymous

lol :)

- anonymous

I'm gonna try to do this one by myself.. and i'll report back what answer I get lol..
(x^2+3x+4)^2+3(x^2+3x+4)+4...

- jim_thompson5910

alright

- anonymous

is it simplified to..
x^4+3x^2+8+3x^2+9x+12+4?
I know that not all the way simplified.. I just wanna know if im sorta on the right track?

- anonymous

x^4+3x^2+16+3x^2+9x+12+4..

- jim_thompson5910

no, you're supposed to have cubed terms in there

- jim_thompson5910

keep in mind that
(x^2+3x+4)^2
(x^2+3x+4)(x^2+3x+4)
x^2(x^2+3x+4)+3x(x^2+3x+4)+4(x^2+3x+4)

- anonymous

(x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?

- jim_thompson5910

that is correct, so
(x^2+3x+4)^2+3(x^2+3x+4)+4
becomes
x^4+6x^3+17x^2+24x+16+3(x^2+3x+4)+4

- anonymous

x^4+6x^3+20x^2+33x+24?

- anonymous

+32*

- jim_thompson5910

good catch

- jim_thompson5910

x^4+6x^3+20x^2+33x+32 is the correct final answer

- anonymous

my goodness! finally lol! Now, one final question... How do I find the domain of those... :/

- jim_thompson5910

The domain of any polynomial is the set of all real numbers.

- jim_thompson5910

since you can plug in any real number into a polynomial function and get a real number out.

- jim_thompson5910

If f(x) and g(x) are both polynomial functions, then the following are also polynomial functions
f(g(x)), g(f(x)), f(f(x)), and g(g(x))
or any combination of composition you can think of are all polynomial functions

- jim_thompson5910

So the domain of each composite function is the set of all real numbers.

- jim_thompson5910

Notice nowhere do we have to worry about a) dividing by zero or b) taking the square root of negative numbers. So there are no restrictions on x.

- anonymous

so the domain for all four polynomials is (-inf,inf)?

- jim_thompson5910

exactly

- anonymous

phew.. I thought it was gonna b longer then that. lol! Thank you so much. so for any polynomial.. the domain will be (-inf,inf) because there are no restrictions?

- jim_thompson5910

Yes, any number will do for the input of a polynomial. So the domain in interval notation for any polynomial is (-inf,inf)

- jim_thompson5910

You'll only run into domain restrictions if you deal with division of variables or having variables in square roots (or logs)

- anonymous

okay, I gotta remember that :) thank you!

- jim_thompson5910

you're welcome

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