We try to factor by maybe grouping
experiment
(x²y³-11x²y)+(6y²-66)
factor
x²y(y²-11)+6(y²-11)
undistribute (y²-11) from each
(x²y+6)(y²-11)
we can force a factor out of the 2nd group in the form of a difference of 2 perfect squares
(x²y+6)(y-√11)(y+√11)
either of those 3 are factors
the first one is -11,-5,2,6,19
Answer:
24
Step-by-step explanation:
I have had this question before I know the answer
3+4p+8t+6+5p
9+9p+8t
......