Question 709498: y^3-11y^2+28y Factor completely.
Answer by josh_jordan(263)   (Show Source):
You can put this solution on YOUR website! To factor this completely, let's first look at what each of the terms have in common. You'll notice that each term has a y in common. So, we can start by factoring the y. This gives us:
Next, we need to see if we can factor what's in the parenthesis. To do so, we need to figure out if there are two numbers, that when multiplied together, give us 28, and when added together, give us -11. We know that both numbers must be negative, because 2 negatives multiplied together give us a positive, and 2 negative numbers added together, will give us a negative. Let's find these two numbers:
-1 x -28 = 28; -1 + -28 = -29 WILL NOT WORK
-2 x -14 = 28; -2 + -14 = -16 WILL NOT WORK
-4 x -7 = 28; -4 + -7 = -11 WORKS!!!
Now that we know -4 and -7 works, we will put these two numbers into factor form:
(y - 4)(y - 7)
Tack on the y that we factored out originally, to the beginning of these factors, and we will have our final answer:
y(y - 4)(y - 7)
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