goobi
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This question is from Ex. 3.2 of Terry Lee's textbook.
Factorise
as real factors, given that it has a repeated zero.
The suggested solution is:
![](https://latex.codecogs.com/png.latex?\bg_white f(x)=x^5+5x^4+6x^3-2x^2-7x-3)
![](https://latex.codecogs.com/png.latex?\bg_white f'(x)=5x^4+20%20x^3+18%20x^2-4%20x-7)
![](https://latex.codecogs.com/png.latex?\bg_white f''(x)%20=%2020%20x^3+60%20x^2+36%20x-4)
![](https://latex.codecogs.com/png.latex?\bg_white f'''(x)%20=%2060x^2+120%20x%20+36)
![](https://latex.codecogs.com/png.latex?\bg_white f'''(x)%20=%2012%20(5%20x^2+10%20x+3))
As f'''(x) has no real linear factors, we must find the multiple root from f''(x).
(^This is the part that I don't understand. When we factorise f'''(x), we get
which are actually real and linear factors despite not being rational.)
![](https://latex.codecogs.com/png.latex?\bg_white f''(-1)=0)
![](https://latex.codecogs.com/png.latex?\bg_white f(-1)=0)
Therefore, (x+1) is the triple root.
![](https://latex.codecogs.com/png.latex?\bg_white \therefore%20f(x)=x^5+5x^4+6x^3-2x^2-7x-3)
![](https://latex.codecogs.com/png.latex?\bg_white =%20(x+1)^3(x^2+2x-3))
![](https://latex.codecogs.com/png.latex?\bg_white =%20(x+1)^3(x+3)(x-1))
The solution is undoubtedly correct, but I still don't understand the bold part.
Any help would be greatly appreciated![Smile :) :)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Factorise
The suggested solution is:
As f'''(x) has no real linear factors, we must find the multiple root from f''(x).
(^This is the part that I don't understand. When we factorise f'''(x), we get
Therefore, (x+1) is the triple root.
The solution is undoubtedly correct, but I still don't understand the bold part.
Any help would be greatly appreciated
Last edited: