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anonymous
 one year ago
Help?
Which of the following is an actual zero of f(x) = x3 + x2  4x  4? (4 points)
4
1
1
4
anonymous
 one year ago
Help? Which of the following is an actual zero of f(x) = x3 + x2  4x  4? (4 points) 4 1 1 4

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0use the remainder theorem if 4 is, then f(4) = 0 if 1 is, then f(1) = 0 if 1 is, then f(1) =0 and if 4 is instead, then f(4) =0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jdoe0001 im so confused. It has 4 in the equation... would it be D?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04 in the equation means nothing, in relation to a root the remainder theorem only pertains to the constant in the binomial

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so get the f(x) values, see which of the choices give you 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the one that gives 0, is the root

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so... what's 1 again? or...what's f(1) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf f(x)=x^3+x^24x4 \\ \quad \\ f({\color{brown}{ 1}})=({\color{brown}{ 1}})^3+({\color{brown}{ 1}})^24({\color{brown}{ 1}})4\implies f(1)\ne 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well. what did you get for f(1)? assume that you did it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf f(x)=x^3+x^24x4 \\ \quad \\ f({\color{brown}{ 1}})=({\color{brown}{ 1}})^3+({\color{brown}{ 1}})^24({\color{brown}{ 1}})4\implies f(1)=1+144 \\ \quad \\ f(1)=28\implies f(1)=6\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i guess i wrote it out wrong, you had it a different way
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