Mark M. answered • 12/07/15
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Retired math prof. Very extensive Precalculus tutoring experience.
Since 2i is a root of f(x) and f(x) has real coefficients, , -2i is also a root.
Since f(x) has degree 3, there can't be more than 3 roots.
The roots are 2i, -2i, and -4
f(x) = a(x-2i)[x-(-2i)](x-(-4)]
= a(x-2i)(x+2i)(x+4)
= a(x2-4i2)(x+4)
= a(x2+4)(x+4)
= a(x3+4x2+4x+16)
Since f(-1) = -45, 15a = -45
So a = -3
f(x) = -3(x3+4x2+4x+16) = -3x3-12x2-12x-48
Navi M.
Also, how would I go about solving this problem if I have 4 roots?
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04/18/20
Mark M.
tutor
If a number, c, is a root, then x - c is a factor. You can't have 4 roots for a polynomial of degree < 4.
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04/19/20
Where N.
How did the -4i^2 became +4? Did I missing something?
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07/20/20
Mark M.
tutor
i^2 = -1. So, -4(i^2) = -4(-1) = +4
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07/20/20
Where N.
Thank you very much! Appreciate it
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07/22/20
Emily B.
Where did the 15 from the equation, 15a = -45, come from?
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12/03/21
Navi M.
Why does it become f(x) = a(x-2i)[x-(-2i)](x-(-4)]?04/18/20