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?
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.
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 |
Oct 21 at 1:44
the function was originally the velocity function and I wish to integrate into displacement hence the absolute value.
Oct 21 at 1:46
@dxiv lol so if just actually attempted the rational zero theorem the first guess wouldve cleared
Oct 21 at 2:30