Question

Find the real and complex zeros of the following function. Please show your work.
f(x) = x^3-x^2+4x-4 @Calculus1

Answer 1

This will require a bit of factoring:\[x^3-x^2+4x-4=x^2(x-1)+4(x-1)=(x^2+4)(x-1)=0\]So the only real solution is given by:\[x-1=0\iff x=1\]and the other 2 complex solutions are:\[x^2+4=0\iff x^2=-4 \iff x=\pm2i\]

Answer 2

im lost. is this the answer?

Answer 3

where are you lost? whenever you have a polynomial and they ask for "roots" or "zeros" or "solutions" you should always think about setting the guy equal to 0, and factoring.

Answer 4

what do you mean

Answer 5

im referring to your question. it asks for you to find the real an complex zeros of the function. The first thing that pops in my head when i hear that is to do:\[x^3-x^2+4x-4=0\] and to try and factor.

Answer 6

ok. But I guess Im wondering how to factor it.

Answer 7

its a little rough most of the time. especially for anything higher than a quadratic. This is cubic >.< I got lucky by looking at the first two terms:\[x^3-x^2=x^2(x-1)\]and the last two terms:\[4x-4=4(x-1)\]They both have the x-1 in common, so i can factor that out of both:\[x^2(x-1)+4(x-1)=(x^2+4)(x-1)\]

Answer 8

ok so the answer would be (x-1)(x^2+4) ?? grr.

Answer 9

thats just the factored form, now you need to actual zeros. Thats why I set each one equal to 0 in my first post and solved for x.

Answer 10

the*, not to, my bad.

Answer 11

x=1,2i,-2i ??

Answer 12

yes, thats correct. :)

Answer 13

yeah! Thank you so much!! :)