How do i factor cubic trinomials?

[supernova1203]

Homework Statement

x³-13x+12=0

The book shows the following solutions, i should add that i have very little practice with factoring cubic TRINOMIALS, you see i factor trinomials through an intuitive understanding, i tried staring at the problem a few hours yesterday, tried youtube/google for help, even tried khan academy, nothing, there is plenty of information out there for cubic polynomials but none for trinomials, so i am posting here in hopes that someone will show me step by step how/what the book did step by step to solve the problem.

Anyway here is what the book says:

x³-13x+12=0

(x-3)(x²+3x-4)=0

(x-3)(x+4)(x-1)=0

so the factors are

x=3, x=-4, x=1

Thanks in advance

 

[SteamKing]

Factoring quadratic expressions is pretty straightforward. I am not aware of any particular procedure for factoring cubic expressions, except by taking a trial monomial and dividing it into the cubic to see if the quotient has no remainder. One other tip would be to graph the cubic and see what the roots are.

 

[Borek]

You can also try to use rational root theorem. In this particular case it would work perfectly.

 

[supernova1203]

Factoring quadratic expressions is pretty straightforward. I am not aware of any particular procedure for factoring cubic expressions, except by taking a trial monomial and dividing it into the cubic to see if the quotient has no remainder. One other tip would be to graph the cubic and see what the roots are.

you mean, trial and error?(i guess factor 'theorm'?) where after trial and error you see what turns the equation into a zero, and use that monomial to divide the equation, and via long division you find the other factor(and you then have 2 factors)If so...i thought you could only use long division for polynomials and not cubic trinomials...was i wrong?so i guess that begs the question, can you use long division on trinomials?

 

[Saitama]

you mean, trial and error?(i guess factor 'theorm'?) where after trial and error you see what turns the equation into a zero, and use that monomial to divide the equation, and via long division you find the other factor(and you then have 2 factors)


If so...i thought you could only use long division for polynomials and not cubic trinomials...was i wrong?

You should really pay attention to Borek's post, check out the rational root theorem, its helpful for guessing the roots.

 

[Borek]

i thought you could only use long division for polynomials and not cubic trinomials...

Are you suggesting that cubic trinomial is not a polynomial?

 

[supernova1203]

Are you suggesting that cubic trinomial is not a polynomial?

well..given the fact that they have 2 separate distinct terms to define them(trinomial and polynomial), and they don't really use it all under the umbrella term ' polynomial' i didnt really know that they are the same thing, at least until now#edit

do quadratics fall under polynomials as well??

 

[eumyang]

well..given the fact that they have 2 separate distinct terms to define them(trinomial and polynomial), and they don't really use it all under the umbrella term ' polynomial'

Who is "they", if I may ask?

do quadratics fall under polynomials as well??

Indeed they do.

 

[epenguin]

The general method for factoring a cubic (guaranteed) is in all moderately advanced algebra books.
For what the the level is, you are into complex numbers.
In fact it was through trying to solve the cubic that complex numbers were discovered - an oft-told tale.
You solve quadratics by expressing them as the difference of two squares, and cubics by the difference of two cubes.
But that is hard enough to do that you avoid it when possible.
As it is in this case and as you will look for when coefficients are whole numbers.
There are well known relationships between coefficients and roots. E.g. If the highest term has coefficient 1 (and if it isn't you can always reduce an equation to such form by dividing by the leading coefficient) then in terms of the roots α, β, γ, the polynomial is
(x - α)(x - β)(x - γ)
and you see the coefficient of x² is -(α + β + γ).
But in your example that is 0.
So you look for combinations of factors of 12, viz. +/- 1,2,3,4,6,12 that add up to 0 and see if any work, i.e. also fit the remaining coefficient.

 

[HallsofIvy]

well..given the fact that they have 2 separate distinct terms to define them(trinomial and polynomial), and they don't really use it all under the umbrella term ' polynomial' i didnt really know that they are the same thing, at least until now

Who is "they", if I may ask?

do quadratics fall under polynomials as well??

Indeed they do.

As do linear polynomials, y= ax+ b.

What definition of "polynomial" are you using? These questions are rather peculiar.