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

Question

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

anonymous · Answer

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

jim_thompson5910 · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

jim_thompson5910 · Answer

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

jim_thompson5910 · Answer

yes, fof is x-4

anonymous · Answer

no the roots is X=-1 or -2

anonymous · Answer

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

jim_thompson5910 · Answer

(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 · Answer

the polynomial discriminant: =1

anonymous · Answer

:)

anonymous · Answer

hope that helps

anonymous · Answer

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

jim_thompson5910 · Answer

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

jim_thompson5910 · Answer

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

jim_thompson5910 · Answer

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

anonymous · Answer

yea, yeah, I wasn't thinking.

jim_thompson5910 · Answer

that's ok

anonymous · Answer

lol :)

anonymous · Answer

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 · Answer

alright

anonymous · Answer

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 · Answer

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

jim_thompson5910 · Answer

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

jim_thompson5910 · Answer

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 · Answer

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

jim_thompson5910 · Answer

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 · Answer

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

anonymous · Answer

+32*

jim_thompson5910 · Answer

good catch

jim_thompson5910 · Answer

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

anonymous · Answer

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

jim_thompson5910 · Answer

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

jim_thompson5910 · Answer

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

jim_thompson5910 · Answer

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 · Answer

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

jim_thompson5910 · Answer

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 · Answer

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

jim_thompson5910 · Answer

exactly

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

okay, I gotta remember that :) thank you!

jim_thompson5910 · Answer

you're welcome