How do you find all the real and complex roots of #x^4 + 7x^3 + 31x^2 + 175x + 150 = 0#?

1 Answer
Feb 23, 2016

Answer:

Use the rational root theorem to find the two Real roots #-1#, #-6# and factor the quartic to find the remaining quadratic factor and Complex roots #+-5i#

Explanation:

#f(x) = x^4+7x^3+31x^2+175x+150#

By the rational root theorem, any rational roots of #f(x) = 0# must be expressible in the form #p/q# where #p# and #q# are integers, #p# a divisor of the constant term #150# and #q# a divisor of the coefficient #1# of the leading term.

In addition, since all of the coefficients of #f(x)# are positive, it has no positive roots.

So the only possible rational roots are:

#-1, -2, -3, -5, -6, -10, -15, -25, -30, -50, -75, -150#

Trying each in turn, we find:

#f(-1) = 1-7+31-175+150 = 0#
...
#f(-6) = 1296-1512+1116-1050+150 = 0#

So #x=-1# and #x=-6# are roots and #(x+1)# and #(x+6)# are factors of #f(x)#

#x^4+7x^3+31x^2+175x+150#

#=(x+1)(x^3+6x^2+25x+150)#

#=(x+1)(x+6)(x^2+25)#

The remaining quadratic factor has zeros #+-5i#

So the roots of #f(x) = 0# are #-1#, #-6#, #+-5i#