Factoring Cubic Polynomials/Function

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To factor the cubic polynomial t^3 - 6t^2 - 36t - 40 = 0, it is suggested to first identify potential roots by testing factors of -40. The Newton-Raphson method can be used for finding roots, but plotting the function often yields quicker results. Once a root, such as -2 or 10, is found, the polynomial can be divided by the corresponding linear factor to simplify it into a quadratic equation. If the polynomial were modified to include +36t, the sum of cubes formula could be applied for further factorization. Ultimately, trial and error with the factors of -40 is the recommended approach for solving this cubic equation.
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Homework Statement


Doing my vector mechanics dynamics homework and I cannot believe I am stuck on this part.

trying to factor

t^3 - 6t^2-36t - 40 = 0

Homework Equations




The Attempt at a Solution


I honestly do not remember where to begin. I remember there is a formula for perfect cubes, but do not remember what is needed for this situation. Any tips is greatly appreciated.
 
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There is a formula but it's nasty and you could always just try Newton-Raphson method for the first factor then long division?
 
You can find all the factors of 40 to try guess how the polynomial factors. Generally, though, I find plotting the function is the quickest way of finding its roots, at least for homework problems.
 
Easy way is to find one root and divide the cubic by it.

i.e. in this situation you can see -2 is a root, so divide the polynomial by t+2, then factor the result to get the other roots.
 
If the problem were changed to +36t, then you would be able to factorize by using the perfect cube.
You would end up having (t-6)^3+176=0 and then you could use the sum of two cubes formula to factorize further (or even just solve the equation for t directly).

But this is not the case :smile:
 
Well the simplest way is to find factors of -40. Replace each in the original expression one by one and when the result is zero: Bingo! You've found a root. In this case 10 is one root. next, divide the original expression by x-r (r = root found by trial and error) to get a quadratic equation. solve it and you've got your roots:smile:
 
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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