# homework help: Pre-Calculus/Algebra Help Please! - Help.com

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!!! I needed some help on homework within which are problems my teacher had never explained how to do if anyone can help I would greatly appreciate it here they are:

Use the given zero to find the remaining zero of the given function:
f(x)=x^4-9x^3+21x^2+21x-130; zero: 3-2i
I was never told how to do long division including an i term

Write a polynomial that meets the given conditions: fourth degree polynomial with real coefficients; zeros -2, 0, 3+5i

Form a polynomial with the given information:
Degree 5; zeros:5, 6+5i, -5i

Again thank you for a response in advance

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For the first question, since it is a fourth order polynomial, we can expect that there will be four solutions. We can get these by factorising the equation - no long division necessary :) I’m not an expert so hope the following isn’t too confusing!

I could be wrong but maybe you should have been given two zeros: 3-2i and 3+2i.
[If not, remember that we need to end up with real coefficients: if we are multiplying (x+m+ni)(x+r+si) where m,n,r and s are real constants, then we get x^2+(r+m+si+ni)x+br+sni^2 = f(x) where f(x) is real. This means that si+ni=0, so in this case, s=-2 then n=2. One way to think of this is that each solution that includes an imaginary number must have a counterpart solution that helps it to get rid of the imaginary part.]

Back to the problem.
We know that when x=3-2i, f(x)=0 and when x=3+2i, f(x)=0.
So we can rewrite the equation as:
f(x) = (x-3+2i)(x-3-2i)(ax^2+bx+c) where a, b, and c are unknown constants.
Expanding the first two brackets gives us:
f(x) = (x^2+9-4i^2)(ax^2+bx+c)
Substituting for f(x):
x^4-9x^3+21x^2+21x-130=(x^2+13)(ax^2+bx+c)
Now, try expanding the two brackets on the right hand side and compare both sides…this should tell you the values of a, b, and c.

For the following two questions, we can do something similar. There are two steps:
(1) If you are given a ‘zero’ with an imaginary part, remember that you need to get rid of it in order to have real coefficients in the polynomial (I tried to explain this in the second paragraph) - so each of the zeros that have an imaginary part in them will tell you another of the zeros. For the second question, you are given two zeros but one has an imaginary part which tells you a third. Since it’s a fourth order polynomial, this leaves you with just one zero left to find before you have the function.
(2) If you know all but one of the zeros, then, for a polynomial with the form ..x^5 + ..x^4 + ..x^3 + ..x^2 + ..x + .. (where .. represents various constants), you can write the function in a factorised form (with some unknowns) and multiply out the brackets to get an equation with some unknowns.

Hope this is helpful, and good luck!

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