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factor m^4-1
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@KimB do you know the expansion of \(a^2-b^2\)??
Try to look at this problem as a difference of squares, with the square root of m^4 being m^2. Thus, you can factor it as: (m^2 +1)(m^2 -1). But, (m^2-1) is also a perfect square, so you can factor that as (m+1)(m-1). Write all three terms to get your final factorization: (m^2+1)(m+1)(m-1).
I am going to provide a little example.\[16x^4 - 81\implies(2x)^4 -(3)^4\implies((2x)^2)^2 - (3^2)^2 \]Now, what @ash2326 provided comes in.
\[((2x)^2 - 3^2)((2x)^2 + 3^2) \implies (2x + 3)(2x - 3)(4x^2 + 9) \]
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