Factoring Polynomials

[lexieaj]

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

 

[Deleted member 4993]

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

 

[BigGlenntheHeavy]

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

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

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

\(\displaystyle Hence, \ P(x) \ is \ completely \ factor \ for \ all \ real \ numbers.\)

 

[galactus]

Did you notice that the quadratic \(\displaystyle x^{2}-x-1\) is directly related to the Golden Ratio?.

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

Cool.

 

[BigGlenntheHeavy]

\(\displaystyle galactus, \ so \ are \ the \ Fibonacci \ numbers \ (related \ to \ the \ Golden \ Mean), \ how?\)

 

[galactus]

\(\displaystyle \lim_{n\to \infty}\frac{F_{n+1}}{F_{n}}=L\)

Form the ratio \(\displaystyle r_{n}=\frac{F_{n+1}}{F_{n}}\)

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 \(\displaystyle F_{n+2}=F_{n+1}+F_{n}\) and divide through by \(\displaystyle F_{n+1}\)

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

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

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

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