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rach_ell
 one year 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?
rach_ell
 one year 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
 one year 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.

rach_ell
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Do you know synthetic division?

rach_ell
 one year ago
Best ResponseYou've already chosen the best response.0We haven't learned that yet

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

myininaya
 one year 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
 one year 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
 one year 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.

rach_ell
 one year ago
Best ResponseYou've already chosen the best response.0Thank you, I will try that!
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