Quick Algebra problem

[RA1fu]

Will someone please show me the steps for this problem:

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

Please help me out.

 

[kill dank]

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?

 

[RA1fu]

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 :\

 

[Morally Skilled]

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.

 

[wackywamba]

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

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

Answer

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

Alternatively,

Answer

(Y^2 - 4)^2

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

--------------------------------------------------

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

Set

x = y^2

So you land up with

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

 

[RA1fu]

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