I need the steps to the complete factorization of the following problem: 16y^5-8y^3+y.?

Question

I know the answer is: y(2y+1)^2 (2y-1)^2 but I can't figure out the steps.

Answer 1

The first step is to factor out a y, since each part of the problem has a y in it. This gives you y(16y^4-8y^2+1). Now you can factor that into y(4y^2-1)(4y^2-1). The (4y^2-1) can also be factored, giving you y(2y+1)(2y-1)(2y+1)(2y-1) or y(2y+1)^2(2y-1)^2.

Answer 2

first factor out the y. you are left with y(16y^4-8y^2+1). then, you factor the inside. you get y((4y^2-1)^2). Then you factor the inside one more time, and distribute getting the y(2y+1)^2 (2y-1)^2. There, not so hard

Answer 3

16y^5-8y^3+y = y(16y^4 - 8 y^2 +1).

Now 16y^4 - 8 y^2 +1) = (4y^2)^2 - 2* (4y^2)*1 +1^2

= (4y^2 -1) ^2.

and 4y^2 - 1 = (2y)^2 - 1^2 = (2y - 1) (2y + 1).

Joining all these, we get:

16y^5-8y^3+y = y(16y^4 - 8 y^2 +1) = y (4y^2 -1) ^2

= y((2y - 1)^2 (2y + 1)^2 Answer.

Answer 4

(16y^4 - 8 y^2 +1) y

16y^4 - 8 y^2 +1 = (4y^2 - 1)^2

4y^2 - 1 = (2y-1)(2y+1)

Ana