I seem to be having a bit of a brain-lock....?

Question

f(x)=x^4 -3x^3+2

I must factor and/or use the quadratic formula to find all zeros of the give function.

Obviously I can't use the quadratic formula right off the bat being it's not a quadratic equation.....I was thinking of using the rational root theorem and synthetic division to get it to degree 2, but there must be an easier way....any help would be great...

I have the correct equation.

I've graphed it and it appears to have 2 real zeros..... 1 & approx. 3

I need to have the work though and I'm just running in circles on my paper

Answer 1

By inspection (i.e., by squinting at the equation a little and seeing if there are any obvious roots), one root is x=1. That means you can divide the original fourth order polynomial by (x-1) and solve the resulting cubic equation for the remaining roots:

(x^4 -3x^3 +2)/(x-1) = x^3 - 2x^2 - 2x - 2

You can use the methods described at the first source (or cheat and use the on-line calculator at the second source) to show that there is only one additional real root at:

[u^(2/3) + 2u^(1/3) + 10]/[3u^(1/3)], where

u = (53 + 1809^(1/2))

The other two roots are imaginary.

Are you *sure* the original problem isn't supposed to be:

f(x) = x^4 - 3x^2 + 2?

In which case, this factors nicely into:

f(x) = (x^2 -2)(x^2-1), which has 4 real roots:

x = +sqrt(2), -sqrt(2), 1, and -1

Answer 2

i remember doing this question 6-7 years back. it does have a very specific 2 step solution.
but at the moment, what i can think of is:-
x^4 - x^3 -2X^3+2 =0
x^3 (x-1) -2(x^3 -1) =0
x^3 (x-1) - 2 [ (x-1) (x^2 +1 +x)] = 0
(x-1) [x^3 - 2x^2 - 2 - 2x] = 0
thus, x=1 and
[x^3 - 2x^2 - 2x - 2] = 0
here onwards you can divide the second equation by simple division (hint try dividing the equation by(x^2-1))

Answer 3

the cheater way is to graph it... Im thinking its going to have 3 zeros... If it was DiffEq Id use the Anhillator method

Answer 4

I don't believe this equation has any zeros. Are you sure you have the correct equation?

Answer 5

go on www.ask.com and type in the question