Give a polynomial of degree 3 that has zeros of 18, 10i, and −10i, and has a value of −3434 when x=1. Write the polynomial in standard form axn+bxn−1+….

The polynomial in standard form is:

The zeros of the polynomial are 18, 10i and -10i.

So, the polynomial is:

Open bracket

Apply difference of two squares

Expand

In complex numbers, i^2 = -1.

So, we have:

Expand

From the question, P(1) = -34.

So, we have:

Divide both sides by -17

Make (a) the subject

Substitute in

Expand

Open bracket

Hence, the polynomial in standard form is:

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Given:

Polynomial of degree 3 that has zeros of 18, 10i, and −10i, and has a value of −34 when x=1.

To find:

The polynomial in standard form.

Solution:

If c is a zero of a polynomial then (x-a) is a factor of that polynomial.

18, 10i, and −10i are zeroes, so (x-18), (x-10i) and (x+10i) respectively are the factors of given polynomial.

So, the three degree polynomial is

where, a is a constant.

      ...(i)

The value of P(x) is -34 at x=1.

Divide both sides by 1717.

Put this value in (i).

Therefore, the required polynomial is .