How would I factor this x^3 equation?

[MrCodeDude]

Determinant of a 3x3 matrix comes out to be:

3 - 7x + 5x^2 - x^3 = 0

I'm trying to find the Eigenvalues, which means factoring this out to solve for x.

Unfortuantly, I have forgotten how and the book does not explain how (I wouldn't expect a Linear Analysis book to cover it anyway).

Help please?

 

[PurdueRy]

-(x-3)(x-1)^2= 0

3 and 1 mult. 2

 

[MrCodeDude]

No, I know the answer. I can find that out using my calculator.

I need to know how to do it. How do I figure out to use 3 and 1 in long division?

 

[Howard]

http://www.purplemath.com/modules/synthdiv.htm

 

[MrCodeDude]

So, guess and check is the only method?

 

[JustAnAverageGuy]

This was from a different math thread, but hopefully it'll inspire you

Pic

-x^3 + 5x^2 - 7x + 3 = 0

You can do Ps and Qs to find all possible factors. (factors of constant (±1, ±3) divided by factors of the coefficient of largest exponent (±1)

± (1, 3)

All possible real x values are +1, -1, +3, -3

Trial and error via synthetic division.

After you get it down to quadratic, you could probably just factor it.

 

[PurdueRy]

You can try and use some logic...

3 - 7x + 5x^2 - x^3 = 0

you need -x...so x * X *-x or all negatives

you need a 3 so 1 * 1 * 3

(x-1)^2*(-x+3)

and its done

 

[chuckywang]

You just have to notice that x-1 is a factor. Long divide x-1 out of the equation, and factor the resulting quadratic.

 

[DrPizza]

What they said, particularly JustAnAverageGuy about the possible rational roots. If a polynomial factors over the rational numbers, then the roots must all be factors of the constant divided by factors of the leading coefficient. (plus or minus)

However, it's not a matter of "trial and error" - this is where graphing calculators ARE a useful tool in mathematics. Graph the function and look at the roots. Does it look like the graph is going through x=1? Then, use (x-1) and do either long division or synthetic division. (Or, of course, simply allow your graphing calculator to give you the roots.)

Keep in mind that after factoring out one of the roots of this equation, you could use the quadratic formula to find the other two roots (which would be one of the simple methods if the other two roots were irrational or complex) or you could use the process of completing the square to find the other two roots.

There is a formula for finding the roots of a polynomial of degree = 2 (quadratic formula), and for degree = 3 and degree = 4. I don't recall those formulas, but I'm sure they can be googled for.

edit: (without looking at a graphing calculator) Actually, I should have said "does it look like the function touches the x-axis at x=1? Since it touches it at that point without going through it, you end up with the same root twice. In this case, it goes through x=-3.