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Factorizing Cubic Functions 
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#1
May1006, 03:20 AM

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


#2
May1006, 03:41 AM

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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. 


#3
May1006, 03:50 AM

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P: 1,131

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^3x^210x8[/tex] Thats one of the functions ive been given to practice. 


#4
May1006, 05:03 AM

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Factorizing Cubic Functions



#5
May1006, 05:13 AM

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How do i divide [tex]x^3x^210x8[/tex] by (x+1)?



#6
May1006, 05:34 AM

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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.
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. 


#7
May1006, 05:50 AM

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To dividea polynomial by another polynomial, we use Long division. You can view it here.
 There's another way that you can do it straight in the paper: Say, you want to divide x^{3}  x^{2}  10x  8 by (x + 1). We have: First, look at the coefficient of the highest order term, i.e x^{3}, and x, we have: x^{3} / x = x^{2}, so we write: (x^{3}  x^{2}  10x  8) / (x + 1) = x^{2} And we multiply x^{2} by 1 (bolded above) to obtain x^{2}, we have x^{2}, i.e, there's a 2x^{2} left right? We continue to divide 2x^{2} by x, we have 2x. We write (x^{3}  x^{2}  10x  8) / (x + 1) = x^{2}  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: (x^{3}  x^{2}  10x  8) / (x + 1) = x^{2}  2x  8 Can you get it? :) 


#8
May1006, 09:15 AM

P: 213

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)(ax^{2}+bx+c) = x^{3}x^{2}10x8 2) compare coefficients ie: for c; 1 multiplied by c will give us 8...c = 8 3) plug 8 in...(x+1)(ax^{2}+bx8) = x^{3}x^{2}10x8 4) now for x... 8x+bx = 10x...b = 2 5) plug 2 in...(x+1)(ax^{2} 2x8) = x^{3}x^{2}10x8 6) Now since x(ax^{2}) = x^{3}, a must surely = 1...as a check though...lets look at x^{2}... 2x^{2} +x^{2} = x^{2} which is what we wanted. With practice you just throw a and c in straight away and you only need to look for b 


#9
May1106, 03:42 AM

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P: 1,131

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



#10
May1106, 09:06 PM

P: 81

Something else that might help is the Rational Zeros Theorem. If you have any polynomial with integer coefficients, a_{n}x^{n}+a_{n1}x^{n1}+...+a_{1}x+a_{0}=0, all the rational zeros will be of the form p/q, where p is a factor of a_{0}, and q is a factor of a_{n}. So if p/q is a zero, there is a corresponding factor, (qxp).
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. 


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