How to factor equations with an exponent of 3

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In summary, there are various methods for factoring a cubic equation, including using the factor theorem, finding rational roots using the rational root theorem, and using trial and error with small integers. However, these methods may not always work and there is a complicated formula for factoring cubics that is not recommended to derive on your own.
  • #1
Cafka
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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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  • #2
  • #3
That's impossible

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


Daniel.
 
  • #4
Practice.

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

(At least polynomials whose coefficients are integers)
 
  • #5
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.
 
  • #6
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
 
  • #7
[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 until you get a good enough approximation of the root.
 
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  • #8
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.
 
  • #9
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 ©)
 
  • #10
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.
 
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  • #12
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.
 
  • #13
^^ that's 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)
 
  • #14
Cafka said:
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.
 
  • #15
don't you have to look at -1, -2, and -3 also?
 
  • #16
Good point- thanks. -1, -2, and -3 are also factors of 6.
 
  • #17
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 don't 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 don't want to see the formula,also a poly greater than 4 can't be resolved, that was proven by abel in his impossibility theorem.see ya
 
  • #18
abia ubong said:
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 don't 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 don't want to see the formula,also a poly greater than 4 can't 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.
 

1. What is the general method for factoring equations with an exponent of 3?

The general method for factoring equations with an exponent of 3 is to use the difference of cubes formula, which states that a3 - b3 can be factored as (a - b)(a2 + ab + b2). This can be applied to any equation with an exponent of 3, as long as the terms are in the form of a3 and b3.

2. Can equations with an exponent of 3 be factored using the sum of cubes formula?

No, equations with an exponent of 3 cannot be factored using the sum of cubes formula. This formula only applies to equations in the form of a3 + b3, and does not work for equations with a negative sign between the terms.

3. Are there any other methods for factoring equations with an exponent of 3?

Yes, there are other methods for factoring equations with an exponent of 3. One method is to use the factoring by grouping technique, where the equation is broken down into smaller groups and common factors are identified and factored out. Another method is to use trial and error by testing different combinations of integers until the equation can be factored.

4. Can equations with an exponent of 3 be factored if there are variables in addition to numbers?

Yes, equations with an exponent of 3 can still be factored if there are variables in addition to numbers. The same methods, such as the difference of cubes formula and factoring by grouping, can be applied as long as the terms are in the form of a3 and b3.

5. How can factoring equations with an exponent of 3 be helpful in science?

Factoring equations with an exponent of 3 can be helpful in science because it allows us to simplify complex equations and make them easier to solve. This can be especially useful in fields such as physics and chemistry, where equations with exponents of 3 are common. Factoring can also help us identify key factors and relationships in an equation, making it easier to understand and apply in real-world situations.

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