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?
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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)
