Mathematics, 18.09.2019 17:30 10040816

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)

Answers: 2

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Answer from: littletiger4867

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

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

The roots of this polynomial are

x=2i ------> with multiplicity

x=3i ------> with multiplicity

leading coefficient of

therefore

the case A is a solution

case B

f(x)=(x+2i)(x+3i)

The roots of this polynomial are

x=-2i ------> with multiplicity

x=-3i ------> with multiplicity

leading coefficient of

therefore

the case B is not a solution

case C

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

The roots of this polynomial are

x=2 ------> with multiplicity

x=3 ------> with multiplicity

x=2i ------> with multiplicity

x=3i ------> with multiplicity

leading coefficient of

therefore

the case C is a solution

case D

f(x)=(x+2i)(x+3i)(x-2i)(x-3i)

The roots of this polynomial are

x=-2i ------> with multiplicity

x=-3i ------> with multiplicity

x=2i ------> with multiplicity

x=3i ------> with multiplicity

leading coefficient of

therefore

the case D is a solution

Answer from: 2891643

The correct option is (D) f(x)=(x+2i)(x+3i)(x-2i)(x-3i).

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

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

Thus, (D) is the correct option.

Answer from: Quest

i think it is c, not positive though

step-by-step explanation:

Answer from: Quest

The discount is 8.70 . you dont tell us the percent of tax added so there is no way to find out, unless i'm not seeing anything. without tax its 37.70

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Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1?

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