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anonymous
 3 years ago
The function f(x)= kx^3  8x^2 x + 3k + 1 has a zero when x=2. Determine the value of k. Graph f(x) and determine all the zeros. Then rewrite in factored form.
I found that k=3 so far, but I'm not sure how to graph it without making a table of values, or how to find the zeros. Any ideas?
anonymous
 3 years ago
The function f(x)= kx^3  8x^2 x + 3k + 1 has a zero when x=2. Determine the value of k. Graph f(x) and determine all the zeros. Then rewrite in factored form. I found that k=3 so far, but I'm not sure how to graph it without making a table of values, or how to find the zeros. Any ideas?

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myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1So f is 0 when x=2 So write this f(2)=0 except actually evaluate the function given at x=2 Replace all of those x's with 2 and then set that result to 0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did that already to find the k value (3), but I'm stuck on the graphing part. And also on how to find the zeros.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know synthetic division?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We haven't learned that yet

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know long division then?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Great. Since x=2 is a zero, then x2 is a factor of 3x^38x^2x+9+1

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1See what another factor is by seeing how many times x2 goes into 3x^38x^2x+10

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1I must leave but you should get a quotient and no remainder. You should be able to factor the quotient. You should be able to now find your 3 zeros. Test the intervals between and before and after each zero to see if the function is above or below the xaxis. Then connect the dots.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you, I will try that!
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