How do you find all the zeros of #f(x)=x^4+5x^3-2x^2-18x-12#?

1 Answer
Aug 4, 2016

Answer:

#f(x)# has zeros #-1, 2, -3-sqrt(3)# and #-3+sqrt(3)#

Explanation:

#f(x) = x^4+5x^3-2x^2-18x-12#

Note that:

#f(-1) = 1-5-2+18-12 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#x^4+5x^3-2x^2-18x-12#

#=(x+1)(x^3+4x^2-6x-12)#

Next try substituting #x=2# in the remaining cubic to get:

#8+4(4)-6(2)-12 = 8+16-12-12 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#x^3+4x^2-6x-12#

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

#=(x-2)((x+3)^2-9+6)#

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

#=(x-2)((x+3)-sqrt(3))((x+3)+sqrt(3))#

#=(x-2)(x+3-sqrt(3))(x+3+sqrt(3))#

Hence zeros #x = -3+-sqrt(3)#