Can #y=x^5 - x^3 + x^2 - 2 x - 5# be factored? If so what are the factors

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Question

Can #y=x^5 - x^3 + x^2 - 2 x - 5# be factored?  If so what are the factors

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Abigail Allen · Accepted Answer

In theory yes, in practice not really.

#f(x) = x^5-x^3+x^2-2x-5# Using the rational root theorem, the only possible rational roots of #f(x) = 0# are: #+-1#, #+-5#. We find: #f(1) = 1-1+1-2-5 = -6# #f(-1) = -1+1+1+2-5 = -2# #f(5) = 3125-125+25-10-5=3010# #f(-5) = -3125+125+25+10-5 = -2970# So there are no rational roots. In fact, this particular quintic seems to have no roots expressible in terms of ordinary radicals so about the best you can do is find the roots it does have using numerical methods. Basically we know it has at least one Real (algebraic) root, but we can't find it algebraically (practically speaking). So we're left in the frustrating position of knowing that in theory this quintic can be factored into a linear factor and two quadratic factors all with Real coefficients, but not being able to specify them algebraically.

Savannah Anderson · Answer

undefined Yes, the polynomial $ y = x^5 - x^3 + x^2 - 2x - 5 $ can be factored. One possible factorization is $ (x - 1)(x + 1)(x^3 + x^2 - 5) $.