A solvable polynomial with no factors?

[Bill_Nye_Fan]

I seem to have encountered a situation in which I have a quartic which has solutions, but no factors.

The polynomial is: $x^4 - 8x^2 + 224x - 160 = 0$

I attempted to find the factors for this quartic in the following manor
$f(x) = x^4 - 8x^2 + 224x - 160$
$f(1) = (1)^4 - 8(1)^2 + 224(1) - 160$
$f(1) = 60$
$f(2) = (2)^4 - 8(2)^2 + 224(2) - 160$
$f(2) = 272$
$...$
$f(8) = (8)^4 - 8(8)^2 + 224(8) - 160$
$f(8) = 5216$

So basically, after I got 8 (negative versions included) I gave up on this method and decided to attempt to factorise it on my calculator. However, my calculator refuses to break down this equation into it's respective equations. However, when I solve the equation it gives me two, real answers: $x = -6.705505492, x = 0.7321472234$. Also, the calculator refuses to give these answers in standard form, only decimal form. Normally when I solve an equation in standard form, it gives me the answer in fractional, surd, or even trigonometrical form.

This is really confusing me, so if anyone could explain to me how it is possible for an equation to have real answers, yet it can't be factorised?

 

[symbolipoint]

First guess, is either the solutions are irrational, or they are complex with imaginary components, or both. If any real solutions, there are some numerical ways of approximately finding them.

 

[SteamKing]

Wolfram Alpha gives one positive real and one negative real root along with a pair of complex conjugate roots.

If you want WA to print the fully symbolic expression for the real or imaginary roots using surds and what not, press the button labelled [Exact Form] next to the real or complex roots. I did it, and I wished I hadn't dunnit.

 

[Bill_Nye_Fan]

Wolfram Alpha gives one positive real and one negative real root along with a pair of complex conjugate roots.

If you want WA to print the fully symbolic expression for the real or imaginary roots using surds and what not, press the button labelled [Exact Form] next to the real or complex roots. I did it, and I wished I hadn't dunnit.

Ah, thanks for that. I knew the answers had to be expressible in some sort of fractional form. Still, it's rather odd my calculator refused to express this specific answer in fractional + surd form... I've seen it spit out some pretty god damn outrageous things in the past (imagine the exact form of this, expect with sine, cosine and tangents thrown in... I think there may have even been a few logarithms too). I wonder why it didn't want to do it for this one? Oh well.

 

[Mentallic]

Ah, thanks for that. I knew the answers had to be expressible in some sort of fractional form.

Keep in mind that this only applies for quartics and below. 5th degree polynomials and above may have real solutions but in some cases cannot be expressible with any finite mix of fractions, surds, logs, trigs, powers, exponentials, etc.

 

[HallsofIvy]

Don't think that "has no easy factors" is the same as "has no factors"! The "fundamental theorem of algebra" says that every polynomial can be factored into linear factors with complex coefficients. If you don't want to deal with complex numbers, it is still true, and follows from the "fundamental theorem of algebra", that any polynomial, with real coefficients, can be factored into a product of linear and quadratic factors with real coefficients.

If it is true that x = -6.705505492 and x = 0.7321472234 are two zeros of the polynomial, then it follows that (x+ 6.705505492) and (x- 0.7321472234) are factors of that polynomial. If you divide the given polynomial by x+ 6.705505492 you should get a third degree polynomial as quotient and 0 remainder. And if you divide that third degree polynomial by x- 0.7321472234, you should get a quadratic polynomial as quotient and 0 remainder.

You can solve that quadratic using the quadratic formula to see if there are two more real roots, so two more linear factors, or if has non-real roots so it is "irreducible" over the real numbers.