If #1+3i# is a zero of #f#, what are all the zeros of #f(x)=x^4–3x^3+6x^2+2x-60#?

1 Answer
Apr 14, 2015

If #1+3i# is a zero of #f(x)=x^4–3x^3+6x^2+2x−60#, then so is #1-3i#. (Complex conjugates theorem) So, you should be able to obtain the quadratic that has #1+3i and 1-3i# as its zeros, then perform long division on the polynomial to find another factor and the remaining zeros!

There are two ways:

  1. Expand: #(x-(1+3i))(x - (1-3i))# (factor theorem)
    #x^2-(1+3i)x-(1-3i)x+10 = x^2 - 2x + 10#

  2. Use the sum and product of roots : #1+3i+1 - 3i= 2# and #(1+3i)(1-3i)= 1 + 9 = 10#. Sum is equal to #-b/a#, so #-b/a=2/1# and #c/a=10/1#. One can solve to find a = 1, b = -2 and c = 10.

Now, for the long division :
screen shot

FINALLY, we can now factor #x^2-x-6=(x+2)(x-3)# and find the remaining, Real zeros, x = -2, x = 3.