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Mathematics 13 Online
OpenStudy (anonymous):

Factor completely. x2 - x - 35 A. prime B. (x - 35)(x + 1) C. (x - 5)(x + 7) D. (x + 5)(x - 7)

OpenStudy (anonymous):

A. Prime

OpenStudy (anonymous):

y?

OpenStudy (anonymous):

Because B = x^2 -34x -35 C = x^2 +2x -35 D = x^2 - 2x -35 There is no way to factor that equation

OpenStudy (anonymous):

okk(:

OpenStudy (anonymous):

\(x^2-x-35\) has roots at \(x=\frac{1}{2}\left(1+\sqrt{141}\right)\) and \(x=\frac{1}{2}\left(1-\sqrt{141}\right)\). So this can be factored as: \[\left(x-\frac{1}{2}\left(1+\sqrt{141}\right) \right)\left(x-\frac{1}{2}\left(1-\sqrt{141}\right) \right)\] However, factorization is considered "prime" if the roots are not integers. Our roots are not integers. Thus, \(x^2-x-35\) is prime.

OpenStudy (anonymous):

ans is prime..

OpenStudy (anonymous):

A) prime

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