SOLUTION: This is factoring the problem again. The problem is 16y^2=25. The answer I got was 16y^2(y-8). Though I am thinking I am way off on that. Please help me Thank You=)

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Question 147540This question is from textbook algebra
: This is factoring the problem again. The problem is 16y^2=25. The answer I got was 16y^2(y-8). Though I am thinking I am way off on that. Please help me Thank You=) This question is from textbook algebra

Found 2 solutions by jim_thompson5910, mangopeeler07:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Are you trying to factor ? If you are, then it is not factorable.


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Are you trying to factor ?

Start with the given expression

Rewrite as

Rewrite as


Now use the difference of squares. Remember, the difference of squares formula is where in this case and

Plug in and

So the expression



factors to



Notice that if you foil the factored expression, you get the original expression. This verifies our answer.
Answer by mangopeeler07(462)   (Show Source): You can put this solution on YOUR website!
. I am assuming that you have to solve for y. First, you factor. First, move 25 over to the other side so that it is all set equal to zero (since y is sqaured, you must do this). . Now you see that is the difference between two perfect squares. so factor it out as the difference of the square roots times the sum of the square roots. . That is how to factor the original equation. Then plug in the values of y that would make each expression (individually) zero, because it is set equal to zero. Ask yourself, what minus five would give me zero? You got five right? Good. Now ask yourself, four times what would give me five? You should have gotten 5/4. Now let's move to the other expression. This time, ask, what plus five would give me zero? You should get -5. Now ask, 4 times what would give me -5? You should get -5/4. These are your two values for y. There are two values because in the original equation, y is squared. So y=5/4;-5/4.