monroe17
((x^2+3x+4)-2)
simplifies to -2x^2-6x-8
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monroe17
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fog..
when f(x)=x-2
g(x)=x^2+3x+4
=((x^2+3x+4)-2) = -2x^2-6x-8
jim_thompson5910
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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
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You're not multiplying, you're subtracting in this case
monroe17
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oh.. so I don't distribute the -2?
monroe17
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so for fof..
would is be.. (x-2)-2=x-4?
jim_thompson5910
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no, you're not distributing since it's originally x-2 and not -2x
jim_thompson5910
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yes, fof is x-4
85295james
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no the roots is X=-1 or -2
monroe17
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what about ... (x-2)^2+3(x-2)+4?
jim_thompson5910
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(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
85295james
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the polynomial discriminant: =1
85295james
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:)
85295james
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hope that helps
monroe17
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how'd you go from (x-2)^2+3(x-2)+4
to this..
x^2-4x+4+3(x-2)+4
?
jim_thompson5910
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(x-2)^2
(x-2)(x-2)
x(x-2) - 2(x-2)
x^2 - 2x - 2x + 4
x^2 - 4x + 4
jim_thompson5910
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So (x-2)^2 is the same as x^2 - 4x + 4
jim_thompson5910
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which makes
(x-2)^2+3(x-2)+4
the same as
x^2-4x+4+3(x-2)+4
monroe17
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yea, yeah, I wasn't thinking.
jim_thompson5910
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that's ok
monroe17
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lol :)
monroe17
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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
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alright
monroe17
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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?
monroe17
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x^4+3x^2+16+3x^2+9x+12+4..
jim_thompson5910
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no, you're supposed to have cubed terms in there
jim_thompson5910
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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)
monroe17
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(x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?
jim_thompson5910
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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
monroe17
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x^4+6x^3+20x^2+33x+24?
monroe17
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+32*
jim_thompson5910
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good catch
jim_thompson5910
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x^4+6x^3+20x^2+33x+32 is the correct final answer
monroe17
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my goodness! finally lol! Now, one final question... How do I find the domain of those... :/
jim_thompson5910
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The domain of any polynomial is the set of all real numbers.
jim_thompson5910
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since you can plug in any real number into a polynomial function and get a real number out.
jim_thompson5910
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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
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So the domain of each composite function is the set of all real numbers.
jim_thompson5910
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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.
monroe17
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so the domain for all four polynomials is (-inf,inf)?
jim_thompson5910
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exactly
monroe17
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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
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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
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You'll only run into domain restrictions if you deal with division of variables or having variables in square roots (or logs)
monroe17
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okay, I gotta remember that :) thank you!
jim_thompson5910
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you're welcome