Question

For 10 and 11 i need to find a polynomial function with real coefficients that has the given zeros. (there's multiple correct answers)

10) 0,2,3i

11) 1,1,2+√3i

Answer 1

See below in bold.

Explanation:

10) Imaginary roots occur in conjugate pairs so the polynomial has another root -3i.

One factor will be x ( because one zero is 0):

In factor form it is:

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

f(x) = x(x - 2)(x^2 - 3ix + 3ix - 9i^2)

= x(x - 2)(x^2 + 9)

= x( x^3 + 9x - 2x^2 - 18)

= x^4 - 2x^3 + 9x^2 - 18x

This function can be multiplied by any non zero constant:

So any function of the form

f(x) = n( x^4 - 2x^3 + 9x^2 - 18x) is the answer.

11).

In factor form this is:

f(x) = (x - 1)^2 (x - (2+ √3i))(x - (2 - √3i))

= (x^2 -2x + 1)(x - 2 - √3i))(x - 2 + √3i))

= (x^2 - 2x + 1)(x^2 - 2x + √3ix - 2x + 4 - 2√3i -√3ix + 2√3i -3i^2)

= (x^2 - 2x + 1)(x^2 - 4x + 7)

= x^4 - 4x^3 + 7x^2 - 2x^3 + 8x^2 - 14x + x^2 - 4x + 7

= x^2 - 6x^3 + 16x^2 - 18x + 7.

f(x) = n(x^2 - 6x^3 + 16x^2 - 18x + 7).

Answer 2

Answer below, is the second part suppose to have two positive 1s, or should one be negative?

Explanation:

For a polynomial to have some set zeroes, the best was I think is to write it in factored form. that is basically (x-a)(x-b) and so on where each x-a represents a zero. specifically a is a zero. So (x-1)(x+1)(x-sqrt(2))(x-i) has four zeroes. 1, -1, sqrt(2) and i. You could expand these parenthesis to get a polynomial in standard form.

Keep in mind (x-a) means positive a is a 0 and (x+a) means negative a is a 0

For the firs one let's write it in factored form.

(x-0)(x-2)(x-3i) = x(x-2)(x-3i) = (x^2-2x)(x-3i) = x^3 - 3ix^2 - 2x^2 + 6ix = x^2 + (-2 - 3i)x^2 + 6ix

Some people don't like having any imaginary units, so they will add a 0. basically if they have (x-ai) they also have a (x+ai) so the two will multiply as follows.

(x-ai)(x+ai) = x^2 +axi - axi - (ai)^2 = x^2 -(-1)a^2 since i^2 = -1

Or if it is a complex number like (x-(a + bi)) where a+bi is the zero you would multiply by (x-(a-bi)) and get:

(x-(a + bi))(x-(a-bi)) = (x - a - bi)(x - a + bi) = x^2 -ax + bix - ax + a^2 - abi - bix + abi - (bi)^2 = x^2 - ax - ax + a^2 - (-1)b^2 = x^2 -2ax + a^2 + b^2

So instead of (x-0)(x-2)(x-3i) you could add one more 0 and have (x-0)(x-2)(x-3i)(x+3i) and this will ilimiate all is in the finction and still get you all necessary 0s.

The next one wants 0s at 1, 1 and 2+sqrt(3)i? I am going to assume one of the 1s is actually -1. if this is wrong let me know.

(x-1)(x+1)(x-(2+sqrt(3)i) and if you want to get rid of all is you could make it one more term longer and have (x-1)(x+1)(x-(2+sqrt(3)i)(x-(2-sqrt(3)i). Can you expand this?