Factorization of polynomials

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

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
15
Factoring degree 3 integer polynomials over the integers is easy -- if it factors, it must have a rational root.
 
Factoring degree 3 integer polynomials over the integers is easy -- if it polynomial divides (synthetic or long), it must be factorable.:wink:


:bugeye:
 
that polynomial doesnt factor that i can see
 

HallsofIvy

Science Advisor
Homework Helper
41,665
857
Did you try x+ 1 as a factor? (-1)^3- 3(-1)^2+ 4= 0 doesn't it?
 
360
0
Or x-2 as a factor?

23 - 3(2)2 + 4 = 8 - 12 + 4 = 0.
 

HallsofIvy

Science Advisor
Homework Helper
41,665
857
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
[tex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0[/tex]
with integer coefficients, then n must divide [itex]a_n[/itex] and m must divide [itex]a_0[/itex].
With your example cubic, the equation is [itex]x^3- 3x^2+ 4= 0[/itex], [itex]a_3= 1[/itex] and [itex]a_0= 4[/itex] 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:
[tex]x^3- 3x^2- 4= (x+ 1)(x- 2)(x- 2)[/tex]
 
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
[tex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0[/tex]
with integer coefficients, then n must divide [itex]a_n[/itex] and m must divide [itex]a_0[/itex].
With your example cubic, the equation is [itex]x^3- 3x^2+ 4= 0[/itex], [itex]a_3= 1[/itex] and [itex]a_0= 4[/itex] 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:
[tex]x^3- 3x^2- 4= (x+ 1)(x- 2)(x- 2)[/tex]
i see your ability to factor is past my ability lol, well im still learning, so its alright, nice post.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top