How to solve a rational equation

Question

anonymous · Answer

The following equation is made up and Need you to show me how to answer it so I can do it to my own problem

anonymous · Answer

$$f(x) = x^3 - 2x^2 + 8x^2 + x + 9$$

anonymous · Answer

@agreene

anonymous · Answer

http://www.quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp#c=solve_stepssolveequation&v1=f%28x%29%3Dx3%E2%88%922x2%2B8x2%2Bx%2B9&v2=x

phi · Answer

that is not an equation until you put in an equal sign
if you mean
$$  x^3 - 2x^2 + 8x^2 + x + 9 =0 $$

in general, it is a mess formula to solve for the roots of a cubic.
For a polynomial of degree 4 there is a *very* messy formula,
and for degree 5 and above, we use numerical techniques (no formula)

phi · Answer

*sometimes* we can factor the cubic, but not usually

phi · Answer

if you make up a cubic by multiplying out (for example) (x-1)(x-2)(x+1) =0 you can "solve" it by factoring back into this form. People use https://en.wikipedia.org/wiki/Descartes'_rule_of_signs and the rational root theorem https://en.wikipedia.org/wiki/Rational_root_theorem

phi · Answer

The rational root theorem gives *possible* roots. But the cubic does *not* have to have a rational root, in which case, we are out of luck. In the case of $$ x^3 - 2x^2 + 8x^2 + x + 9 =0 $$ there are no rational roots, so we have to use the *complicated* formula Here are the roots http://www.wolframalpha.com/input/?i=x%5E3+-+2x%5E2+%2B+8x%5E2+%2B+x+%2B+9+%3D0

anonymous · Answer

they didn't do that in the equation they gave me @phi

phi · Answer

what did they give you ?