Factorizing Cubic Functions

danago

Hi. Im a bit lost as to how to factorize a cubic function. Would anyone be able to help me out?

Thanks,
Dan.

dextercioby

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.

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:

[tex]y=x^3-x^2-10x-8[/tex]

Thats one of the functions ive been given to practice.

Curious3141

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.

danago

How do i divide [tex]x^3-x^2-10x-8[/tex] by (x+1)?

Curious3141

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.

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!

http://en.wikipedia.org/wiki/Synthetic_division --> I just added this link as a tutorial.

VietDao29

To dividea polynomial by another polynomial, we use Long division. You can view it here.
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There's another way that you can do it straight in the paper:
Say, you want to divide x³ - x² - 10x - 8 by (x + 1). We have:
First, look at the coefficient of the highest order term, i.e x³, and x, we have: x³ / x = x², so we write:
(x³ - x² - 10x - 8) / (x + 1) = x²
And we multiply x² by 1 (bolded above) to obtain x², we have -x², i.e, there's a -2x² left right? We continue to divide -2x² by x, we have -2x. We write
(x³ - x² - 10x - 8) / (x + 1) = x² - 2x. Now we again multiply -2x by 1 to get -2x, we have -10x, that means we are left with -8x. We continue to divide -9x by x to obtain -8, multiply -8 by 1 to get -8, and that's done.
So we have:
(x³ - x² - 10x - 8) / (x + 1) = x² - 2x - 8
Can you get it? :)

GregA

I only do algebraic long division only when I absolutely have to...(wastes too much paper)
The way I tend to factorise these types of expressions is as follows:

1) write down what I know so far... (x+1)(ax²+bx+c) = x³-x²-10x-8
2) compare co-efficients ie: for c; 1 multiplied by c will give us -8...c = -8
3) plug -8 in...(x+1)(ax²+bx-8) = x³-x²-10x-8
4) now for x... -8x+bx = -10x...b = -2
5) plug -2 in...(x+1)(ax² -2x-8) = x³-x²-10x-8
6) Now since x(ax²) = x³, a must surely = 1...as a check though...lets look at x²... -2x² +x² = -x² which is what we wanted.

With practice you just throw a and c in straight away and you only need to look for b

danago

Thanks for all the replies. Helped me alot, especially that link you posted VietDao29.

Nimz

Something else that might help is the Rational Zeros Theorem. If you have any polynomial with integer coefficients, a_nxⁿ+a_n-1x^n-1+...+a₁x+a₀=0, all the rational zeros will be of the form p/q, where p is a factor of a₀, and q is a factor of a_n. So if p/q is a zero, there is a corresponding factor, (qx-p).

In your example, the candidates for rational zeros are 8/1, 4/1, 2/1, 1/1, -1/1, -2/1, -4/1, and -8/1 (Of course, you can write them as integers instead of fractions). From here, you can use synthetic division (or long division) on the candidate zeros until you find one that leaves you with no remainder.