Dividing polynomials?

(t⁴-11t³+8t²+36t+36) ÷ (t-3) = (t³-8t²-16t-12)

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Using synthetic division : -


3 l 1______- 11 ______8______36______36
-- I_________ 3______-24____-48_____- 36
__ I 1________-8_____-16____- 12______ 0

x³ - 8 x² - 16 x - 12

First, check that P(3)=0. I've checked, it works.
Then write P(X)=(X^3+aX^2+bX+c)(X-3)
It can be developed to X^4-11X^3+8X^2+36X+36 = X^4+(a-3)X^3+(b-3a)X^2+(c-3b)X-3c
Then you identify term-by-term, which gives a system of four equations:
-11=a-3
8=b-3a
36=c-3b
36=3c
One of them should be redundant, so we should choose three of them to solve.
Of course the first two I choose are the two trivial ones, a=-8 and c=12. Then as the third let's choose b=8+3a=-16
So X^3-8X^2-16X+12 should be the quotient you're looking for.