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How to factor equations with an exponent of 3

by Cafka
Tags: equations, exponent, factor
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Cafka
#1
Apr16-05, 03:44 PM
P: 6
I know there's a formula somewhere, but how do you factor an equation with an exponent of three.
In my solution's manual it says: x^3 - x^2 + 11x - 6 = (x-1)(x-2)(x-3)

And i'm just trying to figure out how they got that.

Thank you.
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z-component
#2
Apr16-05, 04:00 PM
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P: 483
You must use the Factor Theorem.
dextercioby
#3
Apr16-05, 04:01 PM
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That's impossible

[tex] x^{3}-x^{2}+11x-6\neq (x-1)(x-2)(x-3) [/tex]


Daniel.

Hurkyl
#4
Apr16-05, 04:02 PM
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How to factor equations with an exponent of 3

Practice.

I find that looking for the roots of the equation is often the easiest... you should know a method that lets you identify all of the rational numbers that could possibly be the solution of a polynomial.

(At least polynomials whose coefficients are integers)
dextercioby
#5
Apr16-05, 04:04 PM
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You misstyped it

[tex] (x-1)(x-2)(x-3)=x^{3}-6x^{2}+11x-6 [/tex]

Look at the roots of the polyomial.U can see that 1 is a root.

Now u'll have to divide the ployonmial [itex] x^{3}-6x^{2}+11x-6 [/itex] through [itex] x-1 [/itex].


Daniel.
kleinwolf
#6
Apr17-05, 05:24 AM
P: 309
There is a way to factorize third degree polynomials by finding the roots :

Without restricting generality, a 3rd degree polynomial can always be put in the form [tex] p(x)=x^3+px+q [/tex]
[tex]
\textrm{let } x=r+s[/tex]
[tex]\Rightarrow p(x)=r^3+s^3+(3rs+p)(r+s)+q=0[/tex]

this can be solved by the trivial system of equations :

[tex]
r^3+s^3=-q[/tex]
[tex]3rs=-p
[/tex]

which can be transformed in a quadatic equation
Anzas
#7
Apr17-05, 09:26 AM
P: 88
[tex]y= \frac{2ax^3+bx^2-d}{3ax^2+2bx+c} [/tex]

this recrusive equation will converge to the real roots of any 3rd degree polynum ([tex]ax^3+bx^2+cx+d=0[/tex])
at first pick a random number x and solve the equation then use the result as x to solve it again and so on untill you get a good enough approximation of the root.
philosophking
#8
Apr17-05, 09:39 AM
P: 174
This is how I learned it. Say you have some cubic (let's pick an easy one):

x^3+x^2-x-1 = (x^3-x) + (x^2-1) = x(x^2-1) + (x^2-1), and you can factor out an x^2-1 here: (x^2-1)(x+1).

Obviously this does not work for all expressions that are possibly factorable, but it is a good way to make a quick check. If you find that you can immediately do it this way, then that's great. Otherwise, try one of the other ways listed above.
The Bob
#9
Apr17-05, 09:58 AM
P: 1,116
I was taught to find one root of a cubic (by trial and error) and then divide that root (i.e. x-1) into the cubic to find what remains. The quadratic that is left can be factorised or be applied to the quadratic equation to find the other two roots.

Basically it is the factor theorm.

The Bob (2004 )
HallsofIvy
#10
Apr17-05, 03:02 PM
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If by "factor" you mean "factor into terms with integer coefficients", the "rational root theorem" is useful: if x= m/n is a rational root of the polynomial axn+ bxn-1+ ...+ cx+ d= 0 (where all coefficients are integers) then the numerator m is a factor of the constant term d and the denominator n is a factor of the leaing coefficient a". Of course, if x= m/n is a root, then (x-m/n) is a factor and so is (nx- m).

In the example given x3 - 6x2+ 11x - 6 , we know that any rational root must have denominator that divides 1 (and so is an integer) and m (the integer root) must divide 6: it must be 1, 2, or 3. Of course, we still need to check to see IF each IS a root. 1- 6+ 11- 6= 12-12= 0, 8- 24+ 22- 6= 30-30= 0, and 27- 54+ 33- 6= 60-60= 0. Yes, each is a root and so x3 - 6x2+ 11x - 6 = (x-1)(x-2)(x-3).

In the example x3 - x2 + 11x - 6 , any rational (really integer) roots MUST divide 6 and so must be 1, 2, or 3. But 1- 1+ 11- 6 is NOT 0,
8- 4+ 22- 6 is NOT 0, 27- 9+ 33- 6 is NOT 0 so 1, 2, 3, are NOT roots. Since those are the only possible integer (or rational) roots, x3 - x2 + 11x - 6 cannot be factored using only integer (or rational) coefficients.

Of course, if you can find 3 real roots (perhaps by using the cubic formula kleinwolf referred to) you can factor using real (irrational) coefficients. It might happen that the cubic does not have 3 real roots (it must have at least one). In that case you could factor into a product of a linear term and a quadratic term with real coefficients or into linear terms using complex coefficients.
gregmead
#11
Apr18-05, 01:04 PM
P: 42
there is a formula for factorising cubics, but you sure as hell dont wanna see it.

http://www.math.vanderbilt.edu/~schectex/courses/cubic/
Cafka
#12
Apr18-05, 01:50 PM
P: 6
I'm trying to find the roots of an equation in the form of Ax^3 + Bx^2 + Cx + D

I've talked to other people at my school and they say to synthetically divide by a 1, 2, or 3 and see if it goes in evenly then factor the remaining equation.

Thanks for everyone's response.
gregmead
#13
Apr18-05, 01:55 PM
P: 42
^^ thats the way to do it (generally if you get this question in an exam or classwork then at least one of the roots will be a small integer - allowing you to use this method)
HallsofIvy
#14
Apr18-05, 02:48 PM
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Quote Quote by Cafka
I'm trying to find the roots of an equation in the form of Ax^3 + Bx^2 + Cx + D

I've talked to other people at my school and they say to synthetically divide by a 1, 2, or 3 and see if it goes in evenly then factor the remaining equation.

Thanks for everyone's response.
For this particular problem, divide by 1, 2, 3 and "see if it goes in evenly". The reason for choosing 1, 2, 3 is, as I said before because they divide the constant term D= 6.
JonF
#15
Apr18-05, 02:54 PM
P: 617
don't you have to look at -1, -2, and -3 also?
HallsofIvy
#16
Apr18-05, 03:08 PM
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Good point- thanks. -1, -2, and -3 are also factors of 6.
abia ubong
#17
Apr22-05, 06:26 AM
P: 70
all those that said use a trial and error method or substitute 1,2,3 and ten factorise are mistaken ,wat if i give a polynomial with highpower and large quoeficients that has neither 1,2,3,4,5or any samll number as its factor watdo u do,and for u cafka dont bother trying 2 derive it urself cos u might not get it 4 a long time ,it took many mathematicians 2 continue from where one stopped so u see .gregmead u are right i bet they dont wanna see the formula,also a poly greater than 4 cant be resolved, that was proven by abel in his impossibility theorem.see ya
HallsofIvy
#18
Apr22-05, 06:47 AM
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Quote Quote by abia ubong
all those that said use a trial and error method or substitute 1,2,3 and ten factorise are mistaken ,wat if i give a polynomial with highpower and large quoeficients that has neither 1,2,3,4,5or any samll number as its factor watdo u do,and for u cafka dont bother trying 2 derive it urself cos u might not get it 4 a long time ,it took many mathematicians 2 continue from where one stopped so u see .gregmead u are right i bet they dont wanna see the formula,also a poly greater than 4 cant be resolved, that was proven by abel in his impossibility theorem.see ya
What exactly is your point here? You seem to be saying, "What if we completely ignore the original question?" The original post asked about third degree polynomials and gave a specific example. Yes, I am aware that most polynomials cannot be factored except by actually find their zeroes- and for polynomials of degree greater than 4, that is impossible to do in terms of radicals- but that has nothing to do with the original question.


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