Question

Write a polynomial function that has a zero at 2 and 1+i

Attempt at solution

Answer 1

(x+1+i)(x-1-i)
x^2 - 2i
(x^2-2i)(x-2) <--This will lead my no where!

Answer 2

The answer is x^3 -4x^2+6x-4 if that could possible help

Answer 3

forever alone -_-

Answer 4

U guys are racist!

Answer 5

*cries*

Answer 6

*sobs*

Answer 7

*stabs self with knife*

Answer 8

will prevent bleeding if helped otherwise death -_-

Answer 9

*bled to death*

Answer 10

*comes back from dead*

Answer 11

*I WILL HAUNT U ALL IF YOU DO NOT ANSWER MY QUESTION BOHOOOHOHOHOHOHOHOH FEAR ME!*

Answer 12

*STABS GHOSTLY SELF IN CHEST ONCE MORE DUE TO A LACK OF ANSWERS*

Answer 13

Hello my name is Steve Jobs
I have stolen this guys soul and I have come back from the dead.

Answer 14

To give this brown guy some pity for his suffering, i would appreciate it if someone answered his question.

Answer 15

OK IF NO ONE ANSWERS HIS SIMPLE QUESTION I WILL PRESS THE SELF DESTRUCT SWITCH ON ALL THE APPLE DEVICES THAT I HAVE MADE

Answer 16

&^#@&^$#&42*^(@#&(@*&)#(*)@# - THIS IS PROFANITY

Answer 17

o_o

Answer 18

what's this all about?

Answer 19

^^^
Write a polynomial function that has a zero at 2 and 1+i

Answer 20

It seems that you realized that if 1+i is a root, then its complex conjugate must also be a root. The complex conjugate of 1+i is 1-i, so you should start first by multiplying:\[(x-(1+i))(x-(1-i))=(x-1-i)(x-1+i)\]

Answer 21

Everything else you did was correct, you just picked the wrong things to multiply in the very first step.

Answer 22

Ohhh thank you, I really appreciate it :)

Answer 23

*Gives joemath314159 <--- Hey THATS PI ! YUM
5 KARMA*
*Takes 14 KARMA away from everyone else*

Answer 24

If \(r_1,\dots r_n\) are the roots of a polynomial \(P\), we have that:
\[P(x)=(x-r_1)\cdots(x-r_n)\]
If \(r_1 \in \mathbb{C}\), then \(\bar{r_1}\in \mathbb{C}\) is also a root. So, while the problem just says 2 and 1+i, 1-i is also included as a root.

So, you have:
\[\begin{align}
P(x)&=(x-2)(x-(1+i))(x-(1-i))\\&=(x-2)(x-1-i)(x+1+i)\\
&=. . .
\end{align}\]

Answer 25

Thanks for helping aswell Ive got my answer and I think getting there will stick due to this episode.
I cant believe people would help out a brown guys.

*Hugs limitless & Joemath*