SOLUTION: Two roots of a cubic polynomial equation with real coefficients are �3 and �4i. If the leading coefficient of the polynomial is 1, what is the equation? Not really sure how to d

Question 986063: Two roots of a cubic polynomial equation with real coefficients are �3 and �4i. If the leading coefficient of the polynomial is 1, what is the equation?
Not really sure how to do this?
Answer by srinivas.g(540) (Show Source): You can put this solution on YOUR website!
If a polynomial has real coefficients, then if a complex
number a + bi is a root, then so is its conjugate a - bi is also the root
given roots are -3 and -4i
conjugate -4i = +4i
Hence 3 roots are -3,-4i,4i
If a polynomial f(x) has roots r1, r2, r3, ��, rn
and leading coefficient a, then
f(x) = a(x-r1)(x-r2)��(x-rn)
for this problem a=1
f(x)= (x-(-3))(x-(-4i))(x-4i)
f(x)= (x+3)(x+4i)(x-4i)
f(x)= (x+3)(x(x-4i)+4i(x-4i)

(Sine i^2=-1)

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