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Delicious77 [7]
2 years ago
11

Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1? f(x) = (x – 2i)(x – 3i) f(x) =

(x + 2i)(x + 3i) f(x) = (x – 2)(x – 3)(x – 2i)(x – 3i) f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)
Mathematics
2 answers:
kykrilka [37]2 years ago
8 0

Answer:

P(x)=(x-2i)(x-3i)

Step-by-step explanation:

Build a Polynomial Knowing its Roots

If we know a polynomial has roots x1, x2, ..., xn, then it can be expressed as:

P(x)=a(x-x1)(x-x2)...(x-xn)

Where a is the leading coefficient.

Note the roots appear with their signs changed in the polynomial.

If the polynomial has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1, then:

P(x)=1(x-2i)(x-3i)

\mathbf{P(x)=(x-2i)(x-3i)}

jeyben [28]2 years ago
4 0

Answer:

a

Step-by-step explanation:

edge

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