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Factorising Cubic Equations 
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#1
Nov2508, 07:45 AM

P: 1,031

1. The problem statement, all variables and given/known data
Factroise equation 2. Relevant equations 3. The attempt at a solution Can anyone help shopw me the best way to factorise cubic equations? I am doing a question on matrices, and have a 3x3 matrix, which ends up giving me a cubic equation: [tex] x^3 +10x^2  27x + 18 [/tex] It should be in the from (x + a)(x + b)(x + c) but when I multiply these out, I get a very confusing function that doesn't seem to help: [tex] abc + (acx  bcx + abx)  (x^2 + ax^2  bx^2)  x^3 [/tex] Any ideas for a better way to solve these? Any help will be most appreciated, TFM 


#2
Nov2508, 07:49 AM

P: 115

you can easily find out all rational roots using this theorem:
http://en.wikipedia.org/wiki/Rational_root_theorem all you have to do is find one root, factor it out and then use the quadratic equation 


#3
Nov2508, 08:02 AM

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Essentially, if they're whole numbers, you know they must be factors of 18, so try putting x = ±1, then ±2, then ±3 … and seeing whether it comes to zero. (I can spot one answer immediately!) 


#4
Nov2508, 08:03 AM

P: 1,031

Factorising Cubic Equations
Lets see then, the Rational Root Theroem implies that, for this equation:
[tex] x = \pm\frac{1,1}{1,18} [/tex] that gives: [tex] \frac{1}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{18}, \frac{1}{18} [/tex] so the possible answers seem to be 1, 1, 1/18 and 1/18 Does this look correct? TFM 


#5
Nov2508, 08:11 AM

P: 1,031

Hi TinyTim
I had one value to be (x  1), which I got by inserting x = 1, Have now constructed a formula in a spreadsheet which now gives me x = 1 and 3, so I have 2 factors... TFM 


#6
Nov2508, 08:16 AM

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so the third one is … ? 


#7
Nov2508, 08:20 AM

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The "rational root theorem", that LogicalTime mentions, says that if m/n is a rational number root of [itex]a_px^p+ a_{p1}x^{p1}+ \cdot\cdot\cdot+ a_0= 0[/itex] then the numerator m must divide [itex]a_0[/itex] and and n must divide [itex]a_p[/itex].
In your example, [itex] x^3 +10x^2  27x + 18 [/itex], any rational root m/n must have n dividing 1 (and so is either 1 or 1) and m dividing 18: m/n must be one of 1, 1, 2, 2, 3, 3, 6, 6, 18, or 18. It is easy to calculate that [tex](1)^3 +10(1)^2  27(1) + 18= 0[/tex] so x1 is a factor. Divide [tex]x^3 +10x^2  27x + 18[/tex] by x1 to get the other, quadratic, factor. In this particular case you can factor that into linear factors. However, you should understand that most cubics, like most quadratics, cannot be factored into factors with only integer or rational number coefficients. You can factor a quadratic polynomial without rational zeros by using the quadratic formula: for example [itex]x^2 3x+ 4[/itex] has no rational roots but we can solve [itex]x^3 5x+ 3= 0[/itex] using the quadratic formula: [tex]x= \frac{3\pm\sqrt{2512}}{2}= \frac{3\pm\sqrt{13}}{2}[/tex] which tells us [tex]x^35x+ 3= (x \frac{3 \sqrt{13}}{2})(x\frac{3+\sqrt{13}}{2})[/tex] For cubic polynomials you could use the cubic formula but that is much more complicated. 


#8
Nov2508, 08:20 AM

P: 115

the last coefficient (18) corresponds to p, so 18 should in the numerator. I believe you need to look at all the factors of 18.



#9
Nov2508, 08:27 AM

P: 1,031

well, since they must add to 18, that must mean that the third value must be 6, giving:
(x  1)(x  3)(x + 6) I see what you mean about requiring all the factors, now for the Rational Root Theorem. Wikipedia has a bad example, which makes it look like you only need the 1,1 and the original values. slightly confusing, but I understand now. Thanks, TFM 


#10
Nov2508, 08:32 AM

P: 115

I think it's
(x  1)(x  3)(x  6) once you have all the roots in there you multiply by the leading coefficient and that reconstructs your function 


#11
Nov2608, 01:20 AM

P: 236

Once you have located a zero for the polynomial p() (x = 1 is easily spotted in this case) it can always be cleanly divided by that factor.
[tex] (x^3 +10x^2  27x + 18) : (x1) = x^2 + 9x + 2 [/tex] so [tex] x^3 +10x^2  27x + 18 = (x1)(x^2 + 9x + 2) [/tex] Since your original polynomial was a third degree, you now have a first and second degree part which you can further factor using the quadratic formula. k 


#12
Nov2608, 06:10 AM

P: 365

x^{3}+x^{2}+9x^{2}9x18x+18=
=x^{2}(1x)9x(1x)+18(1x)= =(1x)(x^{2}9x+18)= =(1x)(x^{2}3x6x+18)= =(1x)[x(x3)6(x3)]= =(1x)(x3)(x6)= =(x1)(x3)(x6) 


#13
Nov2608, 11:25 AM

P: 436

i wondering if there is other theorem that can help us in factorising?



#14
Nov2608, 02:04 PM

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Have you bothered to read what everyone has been saying?



#15
Jun311, 08:24 AM

P: 1

The simple method I know of is take the factors of the constant.You just need to find one that satisfies the cubic equation.Like in this case take "1".
Now divide the cubic equation by (xa) where a is the factor we found earlier.Now write the cubic equation in the form of divisor*quotient(Note:divisor here is xa and quotient is the quotient found by dividing the given polynomial by xa). Now the question is as easy as solving a quadratic equation. 


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