Hi, Mike!

Since this is a 4th degree polynomial, we know that there must be 4 zeros. But, notice, you have only been given 3. Recall that whenever zeros have imaginary numbers or radicals in them, they occur in both a positive and negative form (this can be seen when using quadratic formula; there is always a plus or minus sign in front of the square root, which is what gives us answers with radicals and imaginary numbers). So, the zeros, -4, 5, and 3+4i must also include the negative version of that imaginary number, 3-4i. Note, I only change the sign right in front of the i.

First, set each zero equal to x: x=-4, x=5, x=3+4i, x=3-4i.

Second, move your information to the left, so that the value zero is isolated on the right:

x+4=0, x-5=0, x-3-4i=0, x-3+4i=0

Third, write each of these as a factor (a value in parentheses) and multiply all factors:

(x+4)(x-5)(x-3-4i)(x-3+4i)

Finally, FOIL each of these factors. Start by foiling the first two factors on their own. Then factor the last two factors on their own (thankfully when you do this all the "i"s will disappear!). And last, multiply these two newly created parentheses together, yielding:

(x^2-x-20)(x^2-6x+25)=x^4-7x^3+11x^2+95x-500

Hope that helps!

Katie