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January 31, 2015

January 31, 2015

Posted by **Anonymous** on Friday, March 8, 2013 at 8:27am.

- algebra -
**Steve**, Friday, March 8, 2013 at 11:27amIf we let

f(x) = (k1 x-r1)(k2 x-r2)...(kn x-rn)

then (r1/k1)(r2/k2)...(rn/kn) = 1024/1

Now, 1024 = 2^10, so all the k's are 1, and all the r's multiplied together are 2^10

the possible distinct roots are

2,4,8,16

so, f(x) = (x-2)(x-4)(x-8)(x-16) + 1024

has 4 values of x such that f(x) = 1024.

I'd say 3 is the max d such that there are n>d places where f(x) = 1024

If I'm way off base here, let me know. I'd be interested in how it's supposed to be done. What are you studying in the class now?

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