The expression does not factorise and the roots seem to be t=-0.16669 and 4.48653 approx. Check that you haven't missed out a term (e.g., t^2).
If the equation was meant to factorise completely, then the product of the factors=-50. The factors must therefore consist of 1, 2, 5, 5. The factors of the 3t^4 must consist of 3t, t, t, t. The minus sign in front of the constant 50 implies that there must be an odd number of pluses and minuses amongst the factors: one minus and 3 pluses, or three minuses and one plus.
Let the coefficient of the apparently missing t^2 be k. The equation then reads:
3t^4+2t^3+kt^2-300t-50=0.
If (t-1) is a factor then 3+2+k-300-50=0 and k=345.
If (t+1) is a factor then 3-2+k+300-50=0 and k=52-303=-251.
If (t-2) is a factor then 48+16+4k-600-50 and k=(1/4)(650-64)=293/2.
If (t+2) is a factor then 48-16+4k+600-50 and k=(1/4)(66-648)=-291/2.
It seems improbable that (t-2) or (t+2) could be a factor because k would be a mixed number (or improper fraction), so k=345 or -251.
If (t-5) is a factor then 1875+250+25k-1500-50=0 and k=(1550-2125)/25=-23.
If (t+5) is a factor then 1875-250+25k+1500-50=0 and k=(300-3375)/25=-123.
We know that (3t+f) must be included as a factor, but if f=-5 or +5 then, since we need two 5's within the set of factors, so if one contains 3t, the other can only contain t. That means that (t-5) or (t+5) must be a factor.
There's no point trying other factors like (t-25) or (t+10), because they would require (t-2) or (t+2), or (t-5) or (t+5), and (t-1) and (t+1) as other factors, which have already been covered.
The exercise shows that no one value of k allows us to find more than one root. Had any two values of k above been the same we would have had at least two roots and a quadratic, which could have been solved, if it had had real roots. The conclusion is that finding one root only leaves us with a cubic equation that has no rational, real roots. This means that the question has been wrongly stated, and the inclusion of kt^2 doesn't help to solve it satisfactorily.