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anonymous
 4 years ago
((x^2+3x+4)2)
simplifies to 2x^26x8
anonymous
 4 years ago
((x^2+3x+4)2) simplifies to 2x^26x8

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0fog.. when f(x)=x2 g(x)=x^2+3x+4 =((x^2+3x+4)2) = 2x^26x8

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1f(x)=x2 f(g(x))=g(x)2 f(g(x))=x^2+3x+42 f(g(x))=x^2+3x+2

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1You're not multiplying, you're subtracting in this case

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh.. so I don't distribute the 2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so for fof.. would is be.. (x2)2=x4?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1no, you're not distributing since it's originally x2 and not 2x

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1yes, fof is x4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no the roots is X=1 or 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what about ... (x2)^2+3(x2)+4?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1(x2)^2+3(x2)+4 x^24x+4+3(x2)+4 x^24x+4+3x6+4 x^2x+2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the polynomial discriminant: =1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how'd you go from (x2)^2+3(x2)+4 to this.. x^24x+4+3(x2)+4 ?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1(x2)^2 (x2)(x2) x(x2)  2(x2) x^2  2x  2x + 4 x^2  4x + 4

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1So (x2)^2 is the same as x^2  4x + 4

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1which makes (x2)^2+3(x2)+4 the same as x^24x+4+3(x2)+4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea, yeah, I wasn't thinking.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is 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
 4 years ago
Best ResponseYou've already chosen the best response.0x^4+3x^2+16+3x^2+9x+12+4..

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1no, you're supposed to have cubed terms in there

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1keep 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
 4 years ago
Best ResponseYou've already chosen the best response.0(x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1that 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
 4 years ago
Best ResponseYou've already chosen the best response.0x^4+6x^3+20x^2+33x+24?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1x^4+6x^3+20x^2+33x+32 is the correct final answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my goodness! finally lol! Now, one final question... How do I find the domain of those... :/

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1The domain of any polynomial is the set of all real numbers.

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1since you can plug in any real number into a polynomial function and get a real number out.

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1If 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
 4 years ago
Best ResponseYou've already chosen the best response.1So the domain of each composite function is the set of all real numbers.

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1Notice 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
 4 years ago
Best ResponseYou've already chosen the best response.0so the domain for all four polynomials is (inf,inf)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0phew.. 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
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, any number will do for the input of a polynomial. So the domain in interval notation for any polynomial is (inf,inf)

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1You'll only run into domain restrictions if you deal with division of variables or having variables in square roots (or logs)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, I gotta remember that :) thank you!

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1you're welcome
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