Can someone help me find the zeros to this polynomial function?

Question

Function: f(x)= x³+x²+9x+9

Zero: 3i

Answer 1

Zero of 3i means also zero of -3i

Find the polynomial that gives these 2 roots, and divide the original f(x) by this polynomial to find the third factor/zero.

x=3i gives factor of x-3i

x=-3i gives factor of x-(-3i) = x+3i

(x-3i)(x+3i) = x² - (3i)² = x² + 9

x²+9 divided into x³+x²+9x+9 = x + 1, so the last zero is x=-1

Could also solve this by "grouping": (x³+x²) +(9x+9) = x²(x+1) + 9(x+1) = (x+1)(x²+9)

Answer 2

The first thing to notice is that some terms have factorable/repeating terms:

f(x) = x³ + x² + 9x + 9

= x³ + 9x + x² + 9

= x(x² + 9) + (x² + 9)

Now that we have grouped and factored, we can add the factors on the outside of each (x² + 9):

f(x) = x(x² + 9) + (x² + 9)

= (x+1)(x² + 9)

One of the zeroes for (x² + 9) would then be 3i, but another would also be -3i. The final remaining zero for the (x+1) portion is -1.