Given:

#f(x) = x^5+8x^3+2x-15#

If we can find the zeros of #f(x)# then we can deduce its factors.

#color(white)()#

**Magnitude analysis**

Note that:

#color(blue)(3)^5 = 243 > 237 = 216+6+15 = 8(color(blue)(3)^3)+2(color(blue)(3))+15#

Hence we can deduce that any zeros of #f(x)# lie inside the circle of radius #3# in the Complex plane. That is: #abs(x) < 3#

#color(white)()#

**Descartes' Rule of Signs**

The pattern of signs of the coefficients of #f(x)# is #+ + + -#. With one change, that means that #f(x)# has exactly one positive Real zero.

The pattern of signs of the coefficients of #f(-x)# is #- - - -#. With no changes, that means that #f(x)# has no negative Real zeros.

#color(white)()#

**Rational roots theorem**

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-15# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-3, +-5, +-15#

We have already eliminated the possibility of #abs(x) >= 3# and of #x < 0#, so the only rational possibility is #x=1#:

#f(1) = 1+8+2-15 = -4#

#color(white)()#

So #f(x)# has no rational zeros. It has an irrational zero in #(1, 3)# and two complex pairs of zeros, all satisfying #abs(x) < 3#.

#color(white)()#

#f(x)# is a typical quintic, with zeros that are not expressible using elementary functions such as #n#th roots, trigonometric, logarithmic or exponential functions. It is possible to find numerical approximations to the zeros, but there is no sensible algebraic solution.