Without the use of the solve function of a calculator how do I solve the equation: #x^4-5x^3-x^2+11x-30=0#?

#f(x) = x^4-5x^3-x^2+11x-30#

We are told that #(x-5)# is a factor, so separate it out:

#x^4-5x^3-x^2+11x-30 = (x-5)(x^3-x+6)#

We are told that #(x+2)# is also a factor, so separate that out:

#x^3-x+6 = (x+2)(x^2-2x+3)#

The discriminant of the remaining quadratic factor is negative, but we can still use the quadratic formula to find the Complex roots:

#x^2-2x+3# is in the form #ax^2+bx+c# with #a=1#, #b=-2# and #c=3#.

The roots are given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (2+-sqrt((-2)^2-(4*1*3)))/(2*1)#

#= (2+-sqrt(4-12))/2#

#= (2+-sqrt(-8))/2#

#= (2+-sqrt(8)i)/2#

#= (2+-2sqrt(2)i)/2#

#=1+-sqrt(2)i#

Let us try without knowing that #(x-5)# and #(x+2)# are factors.
The constant term equals the roots product,so
#30 = r_1*r_2*r_3*r_4#.

This coefficient is an integer value whose factors are #pm 1, pm 2,pm 5, pm3# Trying those values we can see that
#p(-2) = p(5) = 0# obtaining two roots.
We can represent the polynomial as
#x^4 - 5 x^3 - x^2 + 11 x - 30=(x-5)(x+2)(x² + a x + b)#

Calculating the right side and comparing both sides we obtain
#-5=a-3#
#-1=b-3a-10#
#11=-10a-3b#
#-30=-10b#

Solving for #(a,b)# we get #a=-2,b=3#
Evaluating the roots of #x^2-2x+3=0# we get #1 - i sqrt[2],1 + i sqrt[2]#