How do you factor completely #-4y^3+3y^2+8y-6#?

1 Answer

Answer:

Correct answer is #-(4y-3)(y^2-2)#

Explanation:

So you start off with the first two terms which are #-4y^3# and #3y^2#. Now you must think what numbers could we possibly have that will multiply to them.

Only the factors #4, 1, 3# and #1# work but we cannot ignore the negative. If we were to place the negative in the other set of ( ), the answer would not come out right.

So for now, it should be #-(4y-3)(y^2)#

So what number will give us #-6#? Since #-3# times something gives us #-6#, it has to be #-2# because once we do the FOIL method and multiply it by the negative, we are not going to get #-6# because if it were to be a positive two, it would still give us a negative #6# but since the negative is being multiplied to it, we would receive #-6#.

So our final answer has to be: #-(4y-3)(y^2-2)#