Is there any way to factor this x^3 - 5x^2 - 8x + 4 ???

Discussion in 'General Math' started by Eddie, Apr 16, 2007.

1. EddieGuest

Hi,

I need to factor [ x^3 - 5x^2 - 8x + 4 ] in order to solve a
recurrence for an algorithms class, but I seem to be stuck. The
closest I've been able to get is (x - 1)(x - 2)(x - 2), which yields
[ x^3 - 5x^2 + 8x - 4 ] , but as you can see, the signs on the 3rd and
last term are off.

The particular method for solving the recurrence strictly calls for
factoring of this polynomial, but I'm starting to think this
polynomial can't be factored. Anyway, before I say this to my
professor, I want to make sure there is no other way to factor
this....

Ed

Eddie, Apr 16, 2007

2. FelicisGuest

Must it be factored into three monomials with integer coefficients?

Felicis, Apr 16, 2007

3. PubkeybreakerGuest

Hint: If the roots are r1, r2, and r3, what is their product? What
possible rational roots?
Hint: If the polynomial is reducible then it must have at least one
linear factor and correspondingly
at least one rational root.
Hint: Since the polynomial is monic, if it has a rational root, then
one can conclude
Hint: If you find one root, (say) r, then you can divide by (x-r)
I assume you know how to factor a quadratic.

Pubkeybreaker, Apr 16, 2007
4. EddieGuest

Hmm... I belive so because I need to plug-in both the roots and the
coefficients into the general solution for the recurrence I'm trying
to solve (which is of the form [ a_n = C_1r_1^n + C_2r_2^n +
C_3r_3^n ] (note I used the underscore to indicate a subscript).

Eddie, Apr 16, 2007
5. TempGuest

No "nice" roots. There are a number of online polynomial root
finders. Here's one:
<http://home.att.net/~srschmitt/script_cubic.html>

Temp, Apr 16, 2007
6. MateGuest

(x-14*COS(2*ATAN(SQRT(1518)/99)/3)/3-5/3)*
(x+14*SIN(2*ATAN(SQRT(1518)/99)/3+pi/6)/3-5/3)*
(x+14*COS(2*ACOT(-SQRT(1518)/99)/3)/3-5/3)

Mate

Mate, Apr 16, 2007
7. W. Dale HallGuest

Are you sure you have the signs correct? If this were

x^3 - 5 x^2 + 8 x - 4

then it would factor easily.

Dale

W. Dale Hall, Apr 17, 2007