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@_alex_urena_ Have you graphed the function yet? How many zeroes do you see?
yeah i graphed it but what do you mean by zeros?
A zero in a polynomial (or any other) equation is where the graph intersects the x-axis. In general, a polynomial of 6th degree (your case) has a maximum of 6 zeroes.
yeah one of them is in between 0 and 1 and the other is on 2
Good, so there are two (real) zeroes.
Have you learned Newton's method, or the bisection method, or anything similar to those?
uhm i dont think so. never heard those names
Can you do trial and error?
You're trying to solve for the zeroes of f(x)=3x6−5x5−4x3+x2+x+1 which means that where f(x) crosses the x axis. You know that there's one between 0 and 1. so you can try with f(0)=1, f(1)=-3. One is positive and the other is negative right?
yeah but idk where you got three from
f(1) means put x=1 in f(x), or f(1)=3(1)^6−5(1)^5−4(1)^3+(1)^2+(1)+1=-3
ok i got that
That means the curve has to cross the x-axis (y=0) somewhere between 0 and 1.
So you would try f(1/2), (between 0 and 1) to see if it is positive or negative. Use your calculator to find f(1/2)=1.14. So the solution f(x)=0 is between 1/2 and 1 because f(1/2)>0, and f(1)<0.
So we have narrowed down the gap by a factor of two. If you now repeat the same "trial" and "error", you can narrow down the zero to a much more accurate value.
Your next try should be (0.5+1)/2=0.75. So try f(0.75) and see what you have. The key is to narrow down the gap. If f(0.75) is positive, your next try is x=(0.75+1)/2=0.875, and if f(0.75) is negative, your next try is x=(0.5+0.75)/2=0.625.