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

Step-by-step 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?

See below in bold.

Step-by-step 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).