Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. -1, 2, and 1 - i

Answer 1

so, just double checking here, but the factors (aka zeros) of this polynomial are -1, 2 and (1-i), the last factor being a complex number?

Answer 2

Yes.

Answer 3

Hint: complex zeros always come in conjugate pairs (a+bi, a-bi)

Answer 4

at least that's true if you want to have only real coefficients....

Answer 5

Yeah i got that, i got to the point that i know it would be something like this, (x−1)(x+2)(x−(1+i))(x−(1−i). But i have no idea on how to actually write it in standard form.

Answer 6

Multiply that sucker out! :-)

Answer 7

Hint: do the two complex binomials first

Answer 8

But before you do, remember that the zeros are the values of x that make one of the (x-zn) terms = 0. So if x = -1 is a zero, that means that the binomial is actually (x+1) so that (-1 + 1) = 0

Answer 9

Well im not going to lie, i havent been paying any attention in math, and thats why im here haha. So i would appreciate a bit more guidance for this because i honestly have no idea how to multiply the two complex binomials and where to even start to do it.

Answer 10

In other words, to construct your list of product terms, it's always (x - <zero>), not (x + <zero>)

Answer 11

Okay, we can walk through it. But first, let's construct the right expression to multiply so that we don't have to do it over :-)

Answer 12

So, just the real roots, x = -1, and x = 2, what would the polynomial look like (before multiplication)?

Answer 13

Haha i have no idea. Im pretty much 100% lost in math right now, but im going to take a guess and say, (x−1)(x+2)?

Answer 14

Nope, but good try :-)

Remember, the reason they are called zeros is that the polynomial has a value of 0 at that point. Any polynomial can be written as

\[(x - z_0)(x-z_1)(x-z_2)... = 0\]
where \(z_0, z_1, z_2, ...\) are the zeros.

Now, for the polynomial to equal 0, obviously one of those \(x-z_n)\) suckers has to equal 0, right?

Answer 15

Oops, make that \((x-z_n)\)

Answer 16

Yeah makes sense

Answer 17

So if we a function that = 0 when x = -1, and when x = 2, that means there's an \((x+1)\) and a \((x-2)\) in there to cause that.

Answer 18

Okay.

Answer 19

So our polynomial having roots x=-1, x = 2, x = 1-i and x = 1+i is going to look like this before multiplication:
\[(x+1)(x-2)(x-(1-i))(x-(1+i))\]

Answer 20

Alright

Answer 21

You know how to multiply \[(x+1)(x-2) = x^2 - 2x + x -2 = x^2 -x -2\] so that isn't a big deal. But how to multiply \[(x-(1-i))(x-(1+i))\]?

Answer 22

Yeah i have no idea how to do that second part

Answer 23

Hint: Don't get rid of the brackets inside.

Answer 24

paranthesis*

Answer 25

Well, same way, just somewhat more painful!

\[(x-(1-i))(x-(1+i)) = x^2 - (1+i)x -(1-i)x -(-(1-i)(1+i))\]
Now let's take a little side trip and evaluate
\[(1+i)(1-i) = 1^2 -i + i - i^2 = 1-i^2\]but remember \(i^2 = -1\) so that becomes
\[1 -(-1) = 2\]
\[x^2 -(1+i)x - (1-i)x + 2\]
\[x^2 - x -ix -x + ix + 2\]
\[x^2 -2x + 2\]

Answer 26

Now we multiply that with our two real zeros:
\[(x^2 - 2x + 2)(x^2 - x -2)\]

Answer 27

Give it a shot and let's see what you get!

Answer 28

(1+i)(1−i)=12−i+i−i2=1−i2 I understand how you did that,

and this, but remember i2=−1 so that becomes
1−(−1)=2,

but i have no idea how you got this, x2−(1+i)x−(1−i)x+2. and everything afterwards

Answer 29

Here, let's try it a slightly different way. Let's set c = 1+i and d = 1-i, then our expression is
\[(x-c)(x-d) = x^2 -cx -dx + cd = x^2 -(c+d)x + cd\]Right?

Answer 30

Oh so this is like a equation already setup? And we just plug the numbers in?

Answer 31

But if c = 1+i and d = 1-i, then \[c+d = 1+i + 1 - i = 2\]and\[cd = (1+i)(1-i) = 1-i+i-i^2 = 2\]

Answer 32

so that makes the product of \[(x-c)(x-d) = x^2 - (2)x + 2\]
Much easier that way!

Answer 33

What remains is to multiply\[(x^2-2x+2)(x^2-x-2)\]

Answer 34

\[x^4-x^3-2x^2-2x^3+2x^2+4x+2x^2-2x-4 = x^4-3x^3+2x^2+2x-4\]

Answer 35

Yeah that last step is easy

Answer 36

Now the fun begins :-) We ought to check our work!

x=2
\[2^4-3(2^3)+2(2)^2+2(2)-4 = 16 - 24 + 8 + 4 - 4 = 0 \]That one works!
x=-1
\[(-1)^4-3(-1)^3+2(-1)^2 + 2(-1)-4 = 1 -(-3) + 2(1) - 2 -4 = 0 \]That one works!

Answer 37

The other two are a bit more tedious:

\[(1-i)^2=(1 -2i + i^2) = -2i\]\[(1-i)^3 = -2i(1-i) = -2i + 2i^2 = -2i -2\]\[(1-i)^4 = ((1-i)^2)^2 = (-2i)^2 = 4i^2 = -4\]
\[(1-i)^4 - 3(1-i)^3 + 2(1-i)^2 + 2(1-i) - 4 = \]\[-4 -3(-2i-2) + 2(-2i) + 2(1-i) - 4 = -4 +6i + 6 -4i + 2 - 2i - 4 = \]\[-4+6+2-4+6i-4i+2i =0\]

I'll let you do the other one as a learning exercise :-)

Answer 38

.This is soo much for one problem...

Answer 39

@#$#@% I typed the wrong sign in front of that last 2i in the final line

Answer 40

It's good for you, it grows hair on your chest as my old math teacher used to say :-)

Answer 41

Haha, well im going to take a break. I feel like just getting this answer without knowing anything of nothing, is good enough for now. I appreciate your help a lot! Thanks!

Answer 42

Sure, I'm ready for a break too! Shoot message any time if you need some help. I'm not always online, but if you're interested in learning, I'm interested in helping.

Answer 43

"shoot me a message"

Answer 44

Will do! :)