Factoring Polynomials

lexieaj

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Sep 18, 2009
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I have a couple of problems, I am trying to completely factor a couple of polynomials down to factors with real coefficient, however I get to a point where I have to use the quadratic formula but cannot get the equation into quadratic form.
P(x)=x^4-7x^3+14x^2-3x-9
I have managed to find the possible zeroes: {1,3,9} (positive and negative). I got: P(x)= (x-3)(x^3-4x^2+2x+3).

I can't seem to further factor it, and am unable to put it into the quadratic formula. Please help! I have more problems like this and don't know how to proceed.

Alexis
 
lexieaj said:
I have a couple of problems, I am trying to completely factor a couple of polynomials down to factors with real coefficient, however I get to a point where I have to use the quadratic formula but cannot get the equation into quadratic form.
P(x)=x^4-7x^3+14x^2-3x-9
I have managed to find the possible rational zeroes: {1,3,9} (positive and negative). I got: P(x)= (x-3)(x^3-4x^2+2x+3).

"3" is a root again for your function (x^3-4x^2+2x+3)

I can't seem to further factor it, and am unable to put it into the quadratic formula. Please help! I have more problems like this and don't know how to proceed.

Alexis
 
P(x) = x47x3+14x23x9 = (x3)2(x2x1), x2x1 is prime.\displaystyle P(x) \ = \ x^{4}-7x^{3}+14x^{2}-3x-9 \ = \ (x-3)^{2}(x^{2}-x-1), \ x^{2}-x-1 \ is \ prime.

Solve f(x) = 0 = x2x1, x = 1±52\displaystyle Solve \ f(x) \ = \ 0 \ = \ x^{2}-x-1, \ x \ = \ \frac{1\pm\sqrt5}{2}

Ergo, factors of P(x) = (x3)(x3)[x(1+52)][x(152)]\displaystyle Ergo, \ factors \ of \ P(x) \ = \ (x-3)(x-3)\bigg[x-\bigg(\frac{1+\sqrt5}{2}\bigg)\bigg]\bigg[x-\bigg(\frac{1-\sqrt5}{2}\bigg)\bigg]

Hence, P(x) is completely factor for all real numbers.\displaystyle Hence, \ P(x) \ is \ completely \ factor \ for \ all \ real \ numbers.
 
Did you notice that the quadratic x2x1\displaystyle x^{2}-x-1 is directly related to the Golden Ratio?.

τ=5+12,   τ=(51)2\displaystyle {\tau}=\frac{\sqrt{5}+1}{2}, \;\ {\tau}=\frac{-(\sqrt{5}-1)}{2}

Cool.
 
galactus, so are the Fibonacci numbers (related to the Golden Mean), how?\displaystyle galactus, \ so \ are \ the \ Fibonacci \ numbers \ (related \ to \ the \ Golden \ Mean), \ how?
 
limnFn+1Fn=L\displaystyle \lim_{n\to \infty}\frac{F_{n+1}}{F_{n}}=L

Form the ratio rn=Fn+1Fn\displaystyle r_{n}=\frac{F_{n+1}}{F_{n}}

Right here we could go on about a nested sequence of closed and bounded intervals, but I won't do that. I am not going to be that anal.

Start with the recursive formula Fn+2=Fn+1+Fn\displaystyle F_{n+2}=F_{n+1}+F_{n} and divide through by Fn+1\displaystyle F_{n+1}

We get: rn+2=1+1rn\displaystyle r_{n+2}=1+\frac{1}{r_{n}}

Let n\displaystyle n\to {\infty} and we get L=1+1L\displaystyle L=1+\frac{1}{L}

Therefore, we have L2L1=0\displaystyle L^{2}-L-1=0

Whose roots are.........drum roll................The Golden Ratio. :D
 
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