How do I find a polynomial of degree 4 with the following zeros: 1+i, -2-i?

Question

Find a Polynomial of degree 4 with the following zeros:1+i, -2-i
I need help with this problem on my math hw.
Thanks

Answer 1

If the polynomial is to have real coefficients, which I assume it does, then it must also have, among its roots, the complex conjugates of the given roots: 1-i, and -2+i.

Then multiply (x-1+i)(x-1-i)(x+2+i)(x+2-i) (x - each root)

Shortcut: to multiply (x+a+bi)(x+a-bi) = x^2 + 2ax + a^2 + b^2.

Answer 2

This has real coeffivients 1 +i is a root so is 1-i complex conjugates
(x-1) ^2 = -1 or x^2-2x+2 = 0

-2 -i is a root

so (x+2) ^2 = -1 or x^2+4x+ 5 =0

multiply(x^2-2x+2)(x^2+4x+5) = 0
you may or may not simplify as per requirements

Answer 3

the zeros are
1 + i, 1 - i, -2 - i, and -2 + i

(x - (1 + i))(x - (1 - i))(x - (-2 - i))(x - (-2 + i))

(x^2 - (1 - i)x - (1 + i)x + ((1 - i)(1 + i)))(x^2 - (-2 + i)x - (-2 - i)x + ((-2 - i)(-2 + i)))

(x^2 - x + ix - x - ix + (1 + i - i - i^2))(x^2 + 2x - ix + 2x + ix + (4 - 2ix + 2ix - i^2)

(x^2 - 2x + (1 - (sqrt(-1))^2))(x^2 + 4x + (4 - (sqrt(-1))^2))

(x^2 - 2x + (1- (-1))) * (x^2 + 4x + (4 - (-1)))

(x^2 - 2x + 2)(x^2 + 4x + 5)

x^4 + 4x^3 + 5x^2 - 2x^3 - 8x^2 - 10x + 2x^2 + 8x + 10

x^4 + (4 - 2)x^3 + (5 - 8 + 2)x^2 + (-10 + 8)x + 10

x^4 + 2x^3 - x^2 - 2x + 10

ANS : x^4 + 2x^3 - x^2 - 2x + 10

for proof, go to www.quickmath.com, look on the left side for the word Solve under Equations, then type in

x^4 + 2x^3 - x^2 - 2x + 10 = 0

then click solve, and you will get your zeros.

Answer 4

properly, you have 2 of the words, from the given zeros: [x - i][x -a million -2i] Now, on account which you desire genuine coefficients, multiply them by making use of their complicated conjugates [x - i][x + i][x -a million -2i][x-a million+i] = definitely multiplying this out is left as an workout for the asker....

Answer 5

(x - 1 - i)(x + 2 + i)^3 = 0

or
(x - 1 - i)^2 (x + 2 + i)^2 = 0

or
(x - 1 - i)^3 (x + 2 + i) = 0

^_^