How do you find all the zeros of #F(x)=x^3-3x^2+9x+13#?

1 Answer
Mar 5, 2016

Answer:

Identify #x=-1# as a zero by examining the coefficients, separate out the corresponding factor #(x+1)#, then solve the remaining quadratic factor by completing the square.

Explanation:

First notice that reversing the signs of the coefficients of the terms with odd degree results in coefficients that sum to zero.

That is #-1-3-9+13 = 0#.

So #F(-1) = 0# and #(x+1)# is a factor:

#x^3-3x^2+9x+13#

#= (x+1)(x^2-4x+13)#

#=(x+1)(x^2-4x+4+9)#

#=(x+1)((x-2)^2-(3i)^2)#

#=(x+1)((x-2)-3i)((x-2)+3i)#

#=(x+1)(x-2-3i)(x-2+3i)#

So the zeros of #F(x)# are #x=-1# and #x=2+-3i#