Question 314986
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More art than science to this.


There are 4 terms, so that suggests a two and two grouping.  Let's try the first two terms as a group.  From the first two terms we can factor out an *[tex \Large x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ -\ 6x^2\ +\ 5x\ -\ 30]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\left(x\ -\ 6\right)\ +\ 5x\ -\ 30]


This is looking promising because if you take a 5 out of the other two terms you get *[tex \Large x\ -\ 6] again.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\left(x\ -\ 6\right)\ +\ 5\left(x\ -\ 6\right)]


Now you can factor out *[tex \Large (x\ -\ 6)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x^2\ +\ 5\right)\left(x\ -\ 6\right)]


And all is well with the world.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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