# Factorizing Cubic Functions

by danago
Tags: cubic, factorizing, functions
 PF Patron P: 1,132 Hi. Im a bit lost as to how to factorize a cubic function. Would anyone be able to help me out? Thanks, Dan.
 HW Helper Sci Advisor P: 11,717 Depends on its shape. Post the entire problem. In the general case, you'd have to use Cardano's formulae, but i'm sure that it might not be the case. Daniel.
 PF Patron P: 1,132 Well i havnt been given en exact question, but from what ive been told, the functions that i will have to factorize will be specially made to be simple. Things like: $$y=x^3-x^2-10x-8$$ Thats one of the functions ive been given to practice.
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P: 2,634

## Factorizing Cubic Functions

 Quote by danago Well i havnt been given en exact question, but from what ive been told, the functions that i will have to factorize will be specially made to be simple. Things like: $$y=x^3-x^2-10x-8$$ Thats one of the functions ive been given to practice.
y has an obvious zero at x = -1, meaning that (x+1) is a factor of y. (Factor Theorem). Use long division to divide y by (x+1) to get a quadratic, which you should know how to factor.
 PF Patron P: 1,132 How do i divide $$x^3-x^2-10x-8$$ by (x+1)?
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P: 2,634
 Quote by danago How do i divide $$x^3-x^2-10x-8$$ by (x+1)?
There's a long hand old fashioned way, which I dislike, and a fast shorthand way which I use. Note that the fast way only works for linear factors with the x-coefficient reduced to one.

You can google for the long way. The short way :

Array the coefficients of the expression out in decreasing order. Put zeros where the term is missing. Then draw lines like this and put -1 at the right hand side. That's the zero of the divisor (x + 1 = 0 implies x = -1).

The algorithm for division is simplicity itself. Bring the first number (the x^3 coefficient) down. Multiply by the number on the right hand side (-1) and write that under the second number (x^2 coeff). *ADD* the corresponding top number (coefficient) to the bottom number to get a new number. In traditional long division, you subtract, but here you add. Work from left to right till you're left with a zero under the constant term. You will always get a zero when the divisor is a factor of the orig. expression. If you get a nonzero number at the end, then the divisor is not a factor and the number you ended up with is in fact the remainder from the division.

x^3             x^2             x          const

1                -1           -10         -8        | -1
-1             2          8        |
____________________________________________________|
|
1                -2            -8          0

x^2              x          const       Rem
The final answer is obtained by just putting those three bottom-most numbers (don't take the zero, which is the remainder) and using them as coefficients in an expression of degree one less than the orig. expression. In this case, the quotient has degree 2, so the answer is (x^2 - 2x - 8)

So (x^3 - x^2 - 10x - 8)/(x+1) = (x^2 - 2x - 8)

This way works a lot faster than the conventional way because you're working only with numbers (easier for the brain than expressions in x) and you're adding (mentally easier than subtraction). But I think before you use this method practise with the traditional long hand method to learn it well. Then use this for routine purposes (though I don't know if your teacher will be OK with it).

BTW, I just found out that this is called Ruffini's synthetic division method. I've been using it for ages, yet never knew it had a special name!