how do you factor y^4-1 completely

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how do you factor y^4-1 completely

how do you factor y^4-1 completely?

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Answer 1

ReportAnswer #1

first, it is a difference of two squares:

(y^2-1)(y^2+1)
Next, the first term is again the difference of two squares. If you allow imaginary numbers, then the second term is a difference of two squares (y+i)(y-i)

Answer 2

ReportAnswer #2

you should recognize the difference of squares

y^4-1
= (y^2 + 1)(y^2 - 1) , ahh, once more
= (y^2 + 1)(y + 1)(y-1)

Answer 3

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how do you factor y^4-1 completely

Homework Answers

Add Answer of:
how do you factor y^4-1 completely

A) Factor each polynomial completely, given that the binomial following it is a factor of the polynomial

Factor Completely 2x^3-40x^2-192x I have a common factor of x so I rewrote the problem as x(x^2-40x+192) then trying to use FOIL rewrote as (x ) (x ) And that is where i'm stuck

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factor completely, factor GCF first 8t^4+75t^3+27t^2

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factor this polynomial function completely f(x)=12x^4+10x^3-214x^2+198x+90 I found all the possible rational zeros but just cant seem to factor it all

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use the factor theorem to determine whether x-c is a factor of f(x) f(x)=x^3+8x^2-18x+20; x-10 I have a doubt in my answer, i determined that x-10 is not a factor

how do you factor y^4-1 completely

Homework Answers

Add Answer of: how do you factor y^4-1 completely

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how do you factor y^4-1 completely