anonymous
  • anonymous
((x^2+3x+4)-2) simplifies to -2x^2-6x-8
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
fog.. when f(x)=x-2 g(x)=x^2+3x+4 =((x^2+3x+4)-2) = -2x^2-6x-8
jim_thompson5910
  • 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
  • jim_thompson5910
You're not multiplying, you're subtracting in this case

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anonymous
  • anonymous
oh.. so I don't distribute the -2?
anonymous
  • anonymous
so for fof.. would is be.. (x-2)-2=x-4?
jim_thompson5910
  • jim_thompson5910
no, you're not distributing since it's originally x-2 and not -2x
jim_thompson5910
  • jim_thompson5910
yes, fof is x-4
anonymous
  • anonymous
no the roots is X=-1 or -2
anonymous
  • anonymous
what about ... (x-2)^2+3(x-2)+4?
jim_thompson5910
  • 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
  • anonymous
the polynomial discriminant: =1
anonymous
  • anonymous
:)
anonymous
  • anonymous
hope that helps
anonymous
  • 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
  • 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
  • jim_thompson5910
So (x-2)^2 is the same as x^2 - 4x + 4
jim_thompson5910
  • jim_thompson5910
which makes (x-2)^2+3(x-2)+4 the same as x^2-4x+4+3(x-2)+4
anonymous
  • anonymous
yea, yeah, I wasn't thinking.
jim_thompson5910
  • jim_thompson5910
that's ok
anonymous
  • anonymous
lol :)
anonymous
  • 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
  • jim_thompson5910
alright
anonymous
  • 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
  • anonymous
x^4+3x^2+16+3x^2+9x+12+4..
jim_thompson5910
  • jim_thompson5910
no, you're supposed to have cubed terms in there
jim_thompson5910
  • 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
  • anonymous
(x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?
jim_thompson5910
  • 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
  • anonymous
x^4+6x^3+20x^2+33x+24?
anonymous
  • anonymous
+32*
jim_thompson5910
  • jim_thompson5910
good catch
jim_thompson5910
  • jim_thompson5910
x^4+6x^3+20x^2+33x+32 is the correct final answer
anonymous
  • anonymous
my goodness! finally lol! Now, one final question... How do I find the domain of those... :/
jim_thompson5910
  • jim_thompson5910
The domain of any polynomial is the set of all real numbers.
jim_thompson5910
  • jim_thompson5910
since you can plug in any real number into a polynomial function and get a real number out.
jim_thompson5910
  • 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
  • jim_thompson5910
So the domain of each composite function is the set of all real numbers.
jim_thompson5910
  • 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
  • anonymous
so the domain for all four polynomials is (-inf,inf)?
jim_thompson5910
  • jim_thompson5910
exactly
anonymous
  • 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
  • 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
  • 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
  • anonymous
okay, I gotta remember that :) thank you!
jim_thompson5910
  • jim_thompson5910
you're welcome

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