List the potential rational zeros of the following function. Please explain.
f(x)= -2x^3+4x^2-3x+8 List the potential rational zeros of the following function. Please explain.
f(x)= -2x^3+4x^2-3x+8 @Mathematics

Question

anonymous · Answer

Use the rational zeros theorem to find all the real zeros of the polynomial function. 

We can see that this function has a degree of 3, which means that the maximum number of zeros that the function can have is 3.

anonymous · Answer

Use, Descartes' Rule of Signs to determine the possible number of positive zeros and negative zeros.

$$f(x)= -2x^3+4x^2-3x+8$$ Here you see there is a change in variation 4 times, so we can expect either 4 positive real zeros or one positive real zero.

$$f(-x)= -2(-x)^3+4(-x)^2-3(-x)+8$$
$$f(−x)=2x^3 +4x^2 +3x + 8$$ Here there is isn't any change in variation.

anonymous · Answer

The rational zeros theorem states that if p/q, in lowest terms, is a rational zero of f, then p must factor of a(0), and q must be a factor of a(n).

Because f(x)=−2x^3 +4x^2 −3x+8 has integer coeeficiens, we can use the rational zeros theorem. List all the integers p that are factors of the constant term a(0) = 8 and integers q that are factors of the leading coefficient a(3) = -2.
$$p: +/- 1, +/- 2, +/- 4, +/- 8$$
$$q: +/- 1, +/- 1$$

Then, form all possible ratios p/q.

anonymous · Answer

whoops, q: +/- 1, +/- 2

anonymous · Answer

p/q: +/- 1, +/- 2, +/- 4,.....

anonymous · Answer

Now, you're going to stary by testing the potential rational zero, 1, by using substitution.

$$f(1)=−2(1)^3 +4(1)^2 −3(1)+8$$
$$f(1) = -2 + 4 - 4 + 8$$
$$= -2 + 8$$
So you're going to do this for the rest of the possibilities until you get it to 0.

then you'll use synthetic division and go from there.

anonymous · Answer

Are you understanding any of this?

anonymous · Answer

a little bit. lol

anonymous · Answer

so how many possibilities are there then? Im not really sure how to get them to zero..

anonymous · Answer

The degree of the function tells you there are three possibilities. 
Due to th echange in th evariation of f(x), there are 4 possible 0s, doesn't necessarily mean they are all going to work. Which is why you need to plug them in and work them out.

anonymous · Answer

have you tried using interactmath.com? it has a lot of textbooks on there that would be able to go through your problem step by step.

anonymous · Answer

I will look into that. THank you