Quick Algebra problem

Will someone please show me the steps for this problem:

Y^4-8Y^2+16=0
Factor.

Please help me out.

 

Simplifying
Y4 + -8Y2 + 16

Reorder the terms:
16 + -8Y2 + Y4

Factor a trinomial.
(4 + -1Y2)(4 + -1Y2)

Factor a difference between two squares.
((2 + Y)(2 + -1Y))(4 + -1Y2)

Factor a difference between two squares.
((2 + Y)(2 + -1Y))(2 + Y)(2 + -1Y)

Final result:
(2 + Y)(2 + -1Y)(2 + Y)(2 + -1Y) or (2+y)^2 (2-y)^2



That's what I got from this calculator online... I was never good at those either and it doesn't look right to me, but maybe you could see?

 

Thanks for trying to help but it is supposed to end with two sets of parenthesis

for example (x-4)(x+3)

and once I factor those together it is supposed to equal out to original problem.

Then I set (x-4)=0 and (x+3)=0 and the final answer would be x=-3 and x=4

That is how the answer is supposed to look :\

 

http://tinypic.com/r/1z6tlyr/5

There's how I solved it.

I don't know how to describe it, but how I learned it was mostly a guess&check style.

The two exponents are ^4 and ^2 and the integer is positive so it's either two positive or two negative signs, and since the 8y^2 is negative, it's two negatives.

From then you just find numbers that will combine into the 8y^2 (4) as well as multiply into 16.

Hope this helps.

 

Code:

Y^4 - 8Y^2 + 16 = 0

You treat it like you would a quadratic equation, you land up with:

Answer

Code:

(Y^2 - 4)(Y^2 - 4)

Alternatively,

Answer

Code:

(Y^2 - 4)^2

Try factoring out, you'll see you land up with the original answer.

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How to get there in case you don't know:

Set

Code:

x = y^2

So you land up with

Code:

x^2 - 8x + 16 = 0

Hopefully you can solve this, then you substitute y^2 back in for x.

Obviously your root is, y = 2

 

Thank you all for the feedback, I think I got this stuff down now.