- #1

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In my solution's manual it says: x^3 - x^2 + 11x - 6 = (x-1)(x-2)(x-3)

And i'm just trying to figure out how they got that.

Thank you.

- Thread starter Cafka
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- #1

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In my solution's manual it says: x^3 - x^2 + 11x - 6 = (x-1)(x-2)(x-3)

And i'm just trying to figure out how they got that.

Thank you.

- #2

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You must use the Factor Theorem.

- #3

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That's impossible

[tex] x^{3}-x^{2}+11x-6\neq (x-1)(x-2)(x-3) [/tex]

Daniel.

[tex] x^{3}-x^{2}+11x-6\neq (x-1)(x-2)(x-3) [/tex]

Daniel.

- #4

Hurkyl

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I find that looking for the roots of the equation is often the easiest... you should know a method that lets you identify all of the rational numbers that could possibly be the solution of a polynomial.

(At least polynomials whose coefficients are integers)

- #5

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[tex] (x-1)(x-2)(x-3)=x^{3}-6x^{2}+11x-6 [/tex]

Look at the roots of the polyomial.U can see that 1 is a root.

Now u'll have to divide the ployonmial [itex] x^{3}-6x^{2}+11x-6 [/itex] through [itex] x-1 [/itex].

Daniel.

- #6

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Without restricting generality, a 3rd degree polynomial can always be put in the form [tex] p(x)=x^3+px+q [/tex]

[tex]

\textrm{let } x=r+s[/tex]

[tex]\Rightarrow p(x)=r^3+s^3+(3rs+p)(r+s)+q=0[/tex]

this can be solved by the trivial system of equations :

[tex]

r^3+s^3=-q[/tex]

[tex]3rs=-p

[/tex]

which can be transformed in a quadatic equation

- #7

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[tex]y= \frac{2ax^3+bx^2-d}{3ax^2+2bx+c} [/tex]

this recrusive equation will converge to the real roots of any 3rd degree polynum ([tex]ax^3+bx^2+cx+d=0[/tex])

at first pick a random number x and solve the equation then use the result as x to solve it again and so on untill you get a good enough approximation of the root.

this recrusive equation will converge to the real roots of any 3rd degree polynum ([tex]ax^3+bx^2+cx+d=0[/tex])

at first pick a random number x and solve the equation then use the result as x to solve it again and so on untill you get a good enough approximation of the root.

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- #8

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x^3+x^2-x-1 = (x^3-x) + (x^2-1) = x(x^2-1) + (x^2-1), and you can factor out an x^2-1 here: (x^2-1)(x+1).

Obviously this does not work for all expressions that are possibly factorable, but it is a good way to make a quick check. If you find that you can immediately do it this way, then that's great. Otherwise, try one of the other ways listed above.

- #9

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Basically it is the factor theorm.

The Bob (2004 ©)

- #10

HallsofIvy

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If by "factor" you mean "factor into terms with integer coefficients", the "rational root theorem" is useful: if x= m/n is a rational root of the polynomial ax^{n}+ bx^{n-1}+ ...+ cx+ d= 0 (where all coefficients are integers) then the numerator m is a factor of the constant term d and the denominator n is a factor of the leaing coefficient a". Of course, if x= m/n is a root, then (x-m/n) is a factor and so is (nx- m).

In the example given x^{3} - 6x^{2}+ 11x - 6 , we know that any rational root must have denominator that divides 1 (and so is an integer) and m (the integer root) must divide 6: it must be 1, 2, or 3. Of course, we still need to check to see IF each IS a root. 1- 6+ 11- 6= 12-12= 0, 8- 24+ 22- 6= 30-30= 0, and 27- 54+ 33- 6= 60-60= 0. Yes, each is a root and so x^{3} - 6x^{2}+ 11x - 6 = (x-1)(x-2)(x-3).

In the example x^{3} - x^{2} + 11x - 6 , any rational (really integer) roots MUST divide 6 and so must be 1, 2, or 3. But 1- 1+ 11- 6 is NOT 0,

8- 4+ 22- 6 is NOT 0, 27- 9+ 33- 6 is NOT 0 so 1, 2, 3, are NOT roots. Since those are the only possible integer (or rational) roots, x^{3} - x^{2} + 11x - 6 cannot be factored using only integer (or rational) coefficients.

Of course, if you can find 3 real roots (perhaps by using the cubic formula kleinwolf referred to) you can factor using real (irrational) coefficients. It might happen that the cubic does not have 3 real roots (it must have at least one). In that case you could factor into a product of a linear term and a quadratic term with real coefficients or into linear terms using complex coefficients.

In the example given x

In the example x

8- 4+ 22- 6 is NOT 0, 27- 9+ 33- 6 is NOT 0 so 1, 2, 3, are NOT roots. Since those are the only possible integer (or rational) roots, x

Of course, if you can find 3 real roots (perhaps by using the cubic formula kleinwolf referred to) you can factor using real (irrational) coefficients. It might happen that the cubic does not have 3 real roots (it must have at least one). In that case you could factor into a product of a linear term and a quadratic term with real coefficients or into linear terms using complex coefficients.

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- #11

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http://www.math.vanderbilt.edu/~schectex/courses/cubic/

- #12

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I've talked to other people at my school and they say to synthetically divide by a 1, 2, or 3 and see if it goes in evenly then factor the remaining equation.

Thanks for everyone's response.

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- #14

HallsofIvy

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For thisCafka said:

I've talked to other people at my school and they say to synthetically divide by a 1, 2, or 3 and see if it goes in evenly then factor the remaining equation.

Thanks for everyone's response.

- #15

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don't you have to look at -1, -2, and -3 also?

- #16

HallsofIvy

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Good point- thanks. -1, -2, and -3 are also factors of 6.

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- #18

HallsofIvy

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What exactly is your point here? You seem to be saying, "What if we completely ignore the original question?" The original post asked about third degree polynomials and gave a specific example. Yes, I am aware that most polynomials cannot be factored except by actually find their zeroes- and for polynomials of degree greater than 4, that is impossible to do in terms of radicals- but that has nothing to do with the original question.abia ubong said:

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