How do you solve #x^3+3x^2-10x-150# using the fundamental theorem of algebra?

1 Answer
Apr 18, 2016

Three roots of #x^3+3x^2-10x-150# are #{5,-4-isqrt14,-4+isqrt14)#

Explanation:

According to fundamental theorem of algebra, any polynomial of degree #n# with complex coefficients has exactly #n# roots, counted with multiplicity.

As #f(x)=x^3+3x^2-10x-150# is a polynomial of degree #3#, it has #n# roots.

It is apparent that #5# is one of the root as

#f(5)=5^3+3xx5^2-10xx5-150=125+75-50-150=0# and hence #(x-5)# divides #x^3+3x^2-10x-150#.

Dividing by #(x-5)#, we get #x^2+8x+30# or

#x^3+3x^2-10x-150=(x-5)(x^2+8x+30)#

using quadratic formula complex roots of #x^2+8x+30# are #(-8+-sqrt(8^2-4xx1xx30))/(2xx1)# or #(-8+-sqrt(64-120))/2=-4+-isqrt56/2#

or #-4+-isqrt14#

Hence, three roots of #f(x)=x^3+3x^2-10x-150# are #{5,-4-isqrt14,-4+isqrt14)#