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)

THIS USER ASKED 👇

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)

THIS IS THE BEST ANSWER 👇

The right choice is (D)

Step – by – step explanation: We are given the choice of the correct polymorphic function with initial coefficient 1 and roots 2i and 3i with multiplicity 1.

We know that complex roots of polypial function always occur in pairs.

That is, if (a + bi) is a root of the polymath function P (x), its conjugate (a – bi) will also be a root of P (x).

Since then, 2i and 3i have been the roots of the political function, so their conjunctions – 2i and – 3i will also be rooted.

Also, we know that if x = a is located on a polyimol, (x – a) is a factor of the polyimial.

Thus, (x -2i), (x – 3i), (x + 2i) and (x + 3i) are the given polymial factors.

Also, since 1 is the initial prime coefficient, so the polynomial will be necessary

So (D) is the right choice.