Ugh college algebra question!? step by step pleeease!?

Question

ugh my book is stupid it doesnt show me how to get answers step by step! please help
Find a polynomial function P of the lowest possible degree, having real coefficients, with the given zeros.
3+3i, -2, and 1
The polynomial of lowest degree is P(x)=

Answer 1

If the coefficients are real then the polynomial mst also have 3 - 3i (the conjugate of 3 + 3i) as a zero (as (3 - 3i) + (3 + 3i) = 6 and (3 - 3i)*(3 + 3i) = 18. both real numbers)

So the polynomial is P(x) = (x-1)(x+2)(x-(3+3i))(x-(3-3i))
= (x² +x - 2)(x² - 6x + 18)
=x^4 - 5x³ + 10x² + 30x - 36

Answer 2

There is the theorem that if r satisfies f(x), then x-r is a factor of f(x). So form the product

(x-(3+3i)) (x-2)(x+1)

This will give imaginary coefficients. You have to include the conjugate 3-3i as well. So the product is

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

This is the required equation, and the polynomial on the left is the required polynomial P. Next step is to multiply it out. Since 3+3i +3-3i = 6, and (3+3i)(3-3i) = 3^2-(3i)^2 = 9 - (-9) = 18, the product of the first two factors is x^2-6x+18. The rest is straightforward polynomial multiplication.

Answer 3

You cannot have only the given zeros. You must have the conjugate for each imaginary zero or you would get into unpaired imaginaries which voilates several theorems.
Therfore you must also include 3-3i
Give a term (x-z) where z each zero, and multiply the FOUR terms together:
[x-(3+3i)][x-(3-3i)](x+2)(x-1)
Do not remove the brackets and change the signs; instead, multiply the two imagniary terms together to get:
[x^2-x(3-3i)-x(3+3i)+(9-9i+9i+9)](x+2)(x-1)
Then distribute, at the same time you may wish to multiply out the second two terms:
(x^2-3x+3i-3x-3i+18)(x^2+x-2)
Collect like terms:
(x^2-6x+18)(x^2+x-2)
Distribute again and collect like terms:
x^4-5x^3+10x^2+30x-36

Done!

Answer 4

off the bat, p has to equal 4, because there must be an unrealzero and there are already two real zeros.
since it has real coefficients, 3-3i is also a 0
(x-1)(x+2)(x-(3+3i))(x-(3-3i))
(x^2+x-2)
*
(x^2+-3x+-3x+9)

Answer 5

If there's a root of 3+3i, there must also be a root of 3-3i.
Thus, P(x)=(x+2)(x-1)(x-3-3i)(x-3+3i)
=(x^2+x-2)(x^2-6x+18)
Then expand.

Answer 6

for real coeefficients you nals need the complex conjugate of 3+3i as zero

P(x) = ( x - 3 -3i)(x-3+3i) * (x+2)(x-1);

degree is 4

Answer 7

Hey passed out long back but would that be [z-(3+3i)][z+2]{z-1}
or replace z by x if the variable is not complex.Im me if u need more info and think I may have lol

Answer 8

you are able to make up 3 distributive residences by ability of multiplying the 2d polynomial by ability of each and every term interior the 1st polynomial. x^2(x^2 - 6x - 18) + x(x^2 - 6x - 18) -2(x^2 - 6x - 18)= x^4 - 6x^3 - 18x^2 + x^3 - 6x^2 - 18x - 2x^2 + 12x + 36 Then combine like words. x^4 -5x^3 -26x^2 -6x +36