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.
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