# Factorising Cubic Equations

1. Nov 25, 2008

### TFM

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:

$$-x^3 +10x^2 - 27x + 18$$

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:

$$abc + (acx - bcx + abx) - (x^2 + ax^2 - bx^2) - x^3$$

Any ideas for a better way to solve these?

Any help will be most appreciated,

TFM

2. Nov 25, 2008

### LogicalTime

3. Nov 25, 2008

### tiny-tim

Hi TFM!

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. Nov 25, 2008

### TFM

Lets see then, the Rational Root Theroem implies that, for this equation:

$$x = \pm\frac{1,-1}{1,18}$$

that gives:

$$\frac{1}{1}, \frac{-1}{1}, \frac{-1}{1}, \frac{1}{1}, \frac{-1}{1}, \frac{1}{1}, \frac{1}{18}, \frac{-1}{18}$$

so the possible answers seem to be 1, -1, -1/18 and 1/18

Does this look correct?

TFM

5. Nov 25, 2008

### TFM

Hi Tiny-Tim

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. Nov 25, 2008

### tiny-tim

Well they have to multiply to 18 (or is it -18? ),

so the third one is … ?

7. Nov 25, 2008

### HallsofIvy

Staff Emeritus
The "rational root theorem", that LogicalTime mentions, says that if m/n is a rational number root of $a_px^p+ a_{p-1}x^{p-1}+ \cdot\cdot\cdot+ a_0= 0$ then the numerator m must divide $a_0$ and and n must divide $a_p$.

In your example, $-x^3 +10x^2 - 27x + 18$, 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 $$-(1)^3 +10(1)^2 - 27(1) + 18= 0$$ so x-1 is a factor. Divide $$-x^3 +10x^2 - 27x + 18$$ by x-1 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 $x^2- 3x+ 4$ has no rational roots but we can solve $x^3- 5x+ 3= 0$ using the quadratic formula:
$$x= \frac{3\pm\sqrt{25-12}}{2}= \frac{3\pm\sqrt{13}}{2}$$
which tells us
$$x^3-5x+ 3= (x- \frac{3- \sqrt{13}}{2})(x-\frac{3+\sqrt{13}}{2})$$

For cubic polynomials you could use the cubic formula but that is much more complicated.

8. Nov 25, 2008

### LogicalTime

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. Nov 25, 2008

### TFM

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. Nov 25, 2008

### LogicalTime

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. Nov 26, 2008

### kenewbie

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.

$$(-x^3 +10x^2 - 27x + 18) : (x-1) = -x^2 + 9x + 2$$
so
$$-x^3 +10x^2 - 27x + 18 = (x-1)(-x^2 + 9x + 2)$$

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

Last edited: Nov 26, 2008
12. Nov 26, 2008

### Дьявол

-x3+x2+9x2-9x-18x+18=
=x2(1-x)-9x(1-x)+18(1-x)=
=(1-x)(x2-9x+18)=
=(1-x)(x2-3x-6x+18)=
=(1-x)[x(x-3)-6(x-3)]=
=(1-x)(x-3)(x-6)=
=-(x-1)(x-3)(x-6)

13. Nov 26, 2008

### icystrike

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

14. Nov 26, 2008

### HallsofIvy

Staff Emeritus
Have you bothered to read what everyone has been saying?

15. Jun 3, 2011

### vatsalsparsh

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 (x-a) where a is the factor we found earlier.Now write the cubic equation in the form of divisor*quotient(Note:divisor here is x-a and quotient is the quotient found by dividing the given polynomial by x-a).

Now the question is as easy as solving a quadratic equation.