1. ## Complex zeros

hey I'm having trouble with a concept of these. I know know how to form a polynomial with real and complex zeros when I'm given them, but I dont know what to do unless it's a zero in the form of A+Bi.

Form a polynomial ${f{{\left({x}\right)}}}$ with real coefficients having the given degree and zeros. Degree: 4 Zeros: $-{2},{2},-{9}{i}$
So 4th degree (x-2)(x+2)(-9i)(9i)? I tried this and got:
$\displaystyle{{x}}^{{4}}+{77}{{x}}^{{2}}-{162}{x}+{324}$...... wrong

Given the following complex zero of the polynomial ${f{{\left({x}\right)}}}={{x}}^{{3}}+{83}{{x}}^{{2}}+{620}{x}+{1500}$ find all remaining real zeros and express ${f{{\left({x}\right)}}}$ in factored form with real coefficients.
Complex zero = $-{4}-{2}{i}$
I have no idea where to start on this.. normally we can just factor out the zero and go from there, but this is different...

Form a polynomial ${f{{\left({x}\right)}}}$ with real coefficients having the given degree and zeros.
Degree: 4
Zeros: ${5}{i},-{2}+{3}{i}$
So... again (5i)^2 (2-3i)(2+3i) or would it be (5i)(-5i)?

Thanks

2. Originally Posted by xsavethesporksx
hey I'm having trouble with a concept of these. I know know how to form a polynomial with real and complex zeros when I'm given them, but I dont know what to do unless it's a zero in the form of A+Bi.

Form a polynomial ${f{{\left({x}\right)}}}$ with real coefficients having the given degree and zeros. Degree: 4 Zeros: $-{2},{2},-{9}{i}$
So 4th degree (x-2)(x+2)(x-9i)(x+9i)? I tried this and got:
$\displaystyle{{x}}^{{4}}+{77}{{x}}^{{2}}-{162}{x}+{324}$...... wrong
The factor theorem says that if x=a is a zero then x–a is a factor. That applies whether a is real or complex. So if –9i is a zero then x+9i is a factor. If you want the coefficients to be real then the complex conjugate expression x–9i must also be a factor. Therefore $(x+9i)(x-9i) = x^2 + 81$ is a factor. The other two zeros are x=2 and x=–2, so x–2 and x+2 are also factors and hence so is $(x-2)(x+2) = x^2-4$. Multiplying both pairs of factors together, you get $(x^2 + 81)(x^2-4) = x^4 + 77x^2 -324$ (so your answer was close but not quite exact).

Originally Posted by xsavethesporksx
Given the following complex zero of the polynomial ${f{{\left({x}\right)}}}={{x}}^{{3}}+{83}{{x}}^{{2}}+{620}{x}+{1500}$ find all remaining real zeros and express ${f{{\left({x}\right)}}}$ in factored form with real coefficients.
Complex zero = $-{4}-{2}{i}$
I have no idea where to start on this.. normally we can just factor out the zero and go from there, but this is different...
Try this one for yourself, along the same lines. If x = –4–2i is a zero, then x–(–4–2i) = x+4+2i is a factor ... .

Originally Posted by xsavethesporksx
Form a polynomial ${f{{\left({x}\right)}}}$ with real coefficients having the given degree and zeros.
Degree: 4
Zeros: ${5}{i},-{2}+{3}{i}$
So... again (x–5i)(x+5i)(x–(2-3i))(x–(2+3i))