OneFactoring Question: Solving x^3 + x + 2=0

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To solve the equation x^3 + x + 2 = 0, the correct factorization is (x - 2)(x + 1)^2. When factoring, it's beneficial to use the rational root test to identify potential roots by substituting values into the polynomial. Grouping terms can simplify the factoring process, as demonstrated by rewriting the polynomial as x^3 - x + 2x + 2. After identifying one root, such as x = -1, polynomial division or equating coefficients can help find the remaining factors. Understanding these techniques can make factoring cubic equations more efficient.
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Homework Statement


factor
<br /> x^3 + x +2 = 0 <br />



Homework Equations



I know the answer is
<br /> (x-2)(x+1)^2



The Attempt at a Solution



My question is when factoring problems like this - what should I look for or what should I group or separate to make the factoring easier?

Are some problems like this simply trial and error or with some tips and tricks (from experience) do they become faster?

Thanks
Sparky_
 
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Usually you go for the http://www.purplemath.com/modules/rtnlroot.htm" first. Sometimes you can play around with it and try to factor by grouping. I got the answer by first writing x3 + x + 2 as x3 - x + 2x + 2 and then factor them by grouping.
 
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Since you know that it can be factored into (x-2)(x+1)^2 then you know that if you plug x=2 or x=-1 into the polynomial, you will get zero. The idea of the rational root test allows you to find all possible rational roots by plugging in certain values and seeing if that value will return 0.

So if you plugged in x=2, you will get 23+2+2=12 so I can tell you know that it can't be factored into what you gave me. Plugging in x=-1 gives (-1)3-1+2=0 so we can take out a factor (x+1).

Now to find out what is left, you can use polynomial division or equate coefficients of each side by expanding the left side:

(x+1)(x^2+ax+2)=x^3+x+2

Do you see how this works and how to solve for a?
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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