Factoring a polynomial with 4 terms where factoring by grouping does not work? [on hold]

Ok so I am instructed to factor t3−10t2+27t−18t^3 -10t^2 +27t -18
However, factoring by grouping does not work in this case?
I know I can guess and check via the rational zero theorem. BUT THAT SEEMS SO TEDIOUS. Are there any other methods possible?

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1 Answer
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While in this case you’ll hit the jackpot if you try t=1t=1, the rational root theorem works well even if the smallest root would have large prime factors and would not come up a long time after the end of the universe when trying out the candidates one by one. To speed up the process, you should consider a change of variable after each negative result. E.g. substituting t=−1t=-1 yields -56. This means that substituting t=u−1t = u – 1 and rewriting the polynomial in powers of uu will yield a constant term of −56-56. The roots in terms of uu must thus be divisors of 56=7∗856 = 7*8, In terms of tt these possible roots are thus of the form ±2n7m−1\pm 2^n 7^m – 1 with 0≤n≤30\leq n\leq 3 and 0≤m≤10\leq m\leq 1. But any root must also be a divisor of 18, this leaves you with the candidates:

t=−9,−3,−2,1,3,6t = -9,-3,-2,1,3,6

which is a considerably shorter list than the original list of all the divisors of 18.

  

 

+1 for an attempted serious answer to an otherwise borderline trolling question. Except t=−1t=-1 gives −56-56.
– dxiv
Oct 21 at 1:38

  

 

@dxiv Thanks,I’ll correct the answer.
– Count Iblis
Oct 21 at 1:43

  

 

It’s not a troll question, I am finding the roots so i can integrate | t^3-10t^2+27t-18 |
– Prandals
Oct 21 at 1:44

  

 

the function was originally the velocity function and I wish to integrate into displacement hence the absolute value.
– Prandals
Oct 21 at 1:46

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@dxiv lol so if just actually attempted the rational zero theorem the first guess wouldve cleared
– Prandals
Oct 21 at 2:30