Respuesta :

I think it is the last one b.c. the first part of the equation//(2x+5)(2x-5) = 16x^4-625 b.c of the fomrmula x^2-y^2...
the last part i is most likely imaginary number which has a value of -1...but you want x^2 - y^2 remember (16x^2-625) so then its the last one
Hope that helps

Answer:      (2x+5)(2x-5)(2x+i5)(2x-i5)

Step-by-step explanation:   [tex]16x^4-625[/tex]

                         =  [tex]2^4x^4-5^4[/tex]

                         =  [tex](2x)^4-5^4[/tex]

                         =  [tex]((2x)^2-5^2)((2x)^2+5^2)[/tex]   (∵ a²-b²=(a-b)(a+b))

                         =   (2x+5)(2x-5)(2x+i5)(2x-i5)

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