Question 219501
You are very close. Most of your +'s should be -'s. The general form would be:
{{{P(x) = (x - r[1])(x - r[2])(x - r[3])}}}... where {{{r[n]}}} is a root. We want the minus signs so that when x is one of the roots, the minus makes the factor with that root after the minus sign zero which, in turn, makes the entire product zero.<br>
So your function should be:
{{{P(x) = (x - (-1))(x - (1 - 2i))(x - (1 + 2i))(x - (3i))(x - (-3i))}}}
Simplifying within the parentheses we get:
{{{P(x) = (x + 1)(x - 1 + 2i)(x - 1 - 2i)(x - 3i)(x + 3i)}}}<br>

To multiply this out take advantage of the pattern: {{{(a + b)(a - b) = a^2 - b^2}}}. The second and third factors fit this pattern with "a" = "x-1" and "b" = 2i. The fourth and fifth factors also fit this pattern more obviously. So we'll start by multiplying the 2nd and 3rd factors and multiplying the 4th and 5th factors. (Remember multiplication is commutative and associative so we can multiply in any order we choose.)
{{{P(x) = (x + 1)((x - 1)^2 - (2i)^2)(x^2 - (3i)^2)}}}
Simplifying we get:
{{{P(x) = (x + 1)(x^2 -2x + 1 - 4i^2)(x^2 - 9i^2)}}}
{{{P(x) = (x + 1)(x^2 -2x + 1 - 4(-1))(x^2 - 9(-1))}}}
{{{P(x) = (x + 1)(x^2 -2x + 1 + 4)(x^2 + 9)}}}
{{{P(x) = (x + 1)(x^2 -2x + 5)(x^2 + 9)}}}
At this point the i's have disappeared. We still have some multiplications to go. I think the easiest way is to multiply the 1st and third factors first:
{{{P(x) = (x^3 + x^2 + 9x + 9)(x^2 -2x +5)}}}
And finally we multiply each term of the first factor by each term of the second factor:
{{{P(x) = x^5 + x^4 + 9x^3 + 9x^2 - 2x^4 - 2x^3 - 18x^2 - 18x + 5x^3 + 5x^2 + 45x + 45}}}
Adding like terms:
{{{P(x) = x^5 - x^4 + 12x^3 - 4x^2 + 27x + 45}}}