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)

Step-by-step explanation:

Let's check each case to determine the solution to the problem

case A

The roots of this polynomial are

------> with multiplicity

------> with multiplicity

leading coefficient of

therefore

the case A is a solution

case B

The roots of this polynomial are

------> with multiplicity

------> with multiplicity

leading coefficient of

therefore

the case B is not a solution

case C

The roots of this polynomial are

------> with multiplicity

------> with multiplicity

------> with multiplicity

------> with multiplicity

leading coefficient of

therefore

the case C is a solution

case D

The roots of this polynomial are

------> with multiplicity

------> with multiplicity

------> with multiplicity

------> with multiplicity

leading coefficient of

therefore

the case D is  a solution

The correct option is (D)

Step-by-step explanation:  We are given to select the correct polynomial function that has a leading co-efficient of 1 and roots 2i and 3i with multiplicity 1.

We know that the complex roots of a polynomial function always occur in pairs.

That is, if (a + bi) is a root of a polynomial function P(x), then its conjugate (a - bi) will also be a root of P(x).

Since, 2i and 3i are the roots of the polynomial function, so their conjugates - 2i and - 3i will also be the roots.

Also, we know that if x = a is aroot of a polynomial, then (x - a) is a factor of the polynomial.

Therefore, (x -2i), (x - 3i), (x + 2i) and (x + 3i) are the factors of the given polynomial.

Also, since the leading co-efficient is 1, so the required polynomial will be

Thus, (D) is the correct option.

3 is greater then both of those numbers

16

step-by-step explanation: