Since we know that #x = 3# is a root of the polynomial, we know that the factor #(x-3)# will divide into the polynomial, which provides us with the avenue we need to solve the question. We divide #p(x)# by #(x-3)#, and then we can use the Quadratic Formula to find the remaining two roots (which the problem statement tells us will be complex).
How you divide #p(x)# by #(x-3)# is up to the reader (and you've already done this, but for the sake of others I'll provide this step as well). For this solution I will defer to Synthetic Division since it is generally easier to evaluate for problems like this. If you need more help on synthetic division, there are some great resources here on Socratic that can assist with that.
#3__| color(white)("aaaaaa")4color(white)("aaaaa")-16color(white)("aaaaa")21color(white)("aaaaa")-27#
#color(white)("aaaaaaaa")underline(color(white)("aaaaaaaa|")12color(white)("aa")-12color(white)("aaaaaaaa")27#
#color(white)("aaaaaaaa")4color(white)("aaaaaa")-4color(white)("aaaaaa")9color(white)("aaaaaaaa")0#
The remainder of 0 at the end of the bottom line verifies that #x = 3# is indeed a root, as mentioned. The remaining numbers on the bottom line tell us the result of dividing #p(x)# by #(x-3)# is the following quadratic:
#4x^2-4x+9#
Using the Quadratic Formula leads to our final two roots:
#x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-4)+-sqrt((-4)^2-4(4)(9)))/(2(4)) #
# = (4+-sqrt(16-144))/8 = (4+-sqrt(-128))/8 = (4+-isqrt(128))/8#
# = (4+-isqrt(64*2))/8 = (4+-8isqrt(2))/8 = 1/2+-isqrt(2)#
Thus the remaining two roots are indeed complex, and are #x = 1/2+isqrt(2)# and #x = 1/2-isqrt(2)#
