Rational Root Theorem for Factoring Polynomials

[Darkbalmunk]

Hi I was wondering since i have problems factoring any polynomial past 2nd degree i was wondering if anyone can show a way i can remember for finals ^_^.

IE. let's say we have a 3rd degree polynomial.
X^3 - 3X^2 +4
i tried looking it up but most don't show how they did the work so i can understand the in between or like sparknotes that polynomial doesn't fit their formula so I am kinda stumped for the upcoming test where we do this, especially when its used in det of matrices.

 

[Hurkyl]

Factoring degree 3 integer polynomials over the integers is easy -- if it factors, it must have a rational root.

 

[BloodyFrozen]

Factoring degree 3 integer polynomials over the integers is easy -- if it polynomial divides (synthetic or long), it must be factorable.

 

[craig0303]

that polynomial doesn't factor that i can see

 

[HallsofIvy]

Did you try x+ 1 as a factor? (-1)^3- 3(-1)^2+ 4= 0 doesn't it?

 

[spamiam]

Or x-2 as a factor?

2³ - 3(2)² + 4 = 8 - 12 + 4 = 0.

 

[HallsofIvy]

Factoring degree 3 integer polynomials over the integers is easy -- if it factors, it must have a rational root.

A polynomial has "x- a" as a factor if and only x= a makes the polynomial equal to 0 and the "rational root theorem" says that if m/n is a rational root of the equation
$$a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$$
with integer coefficients, then n must divide $a_n$ and m must divide $a_0$.
With your example cubic, the equation is $x^3- 3x^2+ 4= 0$, $a_3= 1$ and $a_0= 4$ so n must be +/- 1 and m must be one of +/- 1, +/- 2, +/- 4. The only possible rational roots are +/- 1, +/- 2, +/- 4 and it is not hard to try those and see that -1 is a root and that 2 is a double root:
$$x^3- 3x^2- 4= (x+ 1)(x- 2)(x- 2)$$

 

[craig0303]

A polynomial has "x- a" as a factor if and only x= a makes the polynomial equal to 0 and the "rational root theorem" says that if m/n is a rational root of the equation
$$a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$$
with integer coefficients, then n must divide $a_n$ and m must divide $a_0$.
With your example cubic, the equation is $x^3- 3x^2+ 4= 0$, $a_3= 1$ and $a_0= 4$ so n must be +/- 1 and m must be one of +/- 1, +/- 2, +/- 4. The only possible rational roots are +/- 1, +/- 2, +/- 4 and it is not hard to try those and see that -1 is a root and that 2 is a double root:
$$x^3- 3x^2- 4= (x+ 1)(x- 2)(x- 2)$$

i see your ability to factor is past my ability lol, well I am still learning, so its alright, nice post.