Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

Answer 1

The zeros are: 3, -13, and 5 + 4i
The choices were:
A) f(x) = x^4 - 8x^3 - 12x^2 + 400x - 1599
B) f(x) = x^4 - 200x^2 + 800x - 1599
C) f(x) = x^4 - 98x^2 + 800x - 1599
D) f(x) = x^4 - 8x^3 + 12x^2 - 400x + 1599

Answer 2

I've asked this one before, but I still don't understand it.

Answer 3

Each real zero "a", will give you a factor of (x - a).
Each complex zero a + bi will also have to have its complex conjugate as a zero, adding two more factors:
(x + (a + bi)) and (x - (a - bi))
Then multiply all the factors together to find the polynomial.

Answer 4

In your case, you have two real zeros: 3 and -13.
Based on what I wrote above, which factors do you get from those two zeros?

Answer 5

So I'll have (x-3), (x+13), (x+(5+4i)), and (x-(5+4i)). Right?

Answer 6

Almost.
(x - 3) and (x + 13) are correct.
You have to be careful with the complex zeros.
Since you have complex zero 5 + 4i given, its complex conjugate 5 - 4i is also a zero.
Now write the factors for the complex conjugate zeros.

Answer 7

Oh, so the second one would be (x-(5-4i)) instead?

Answer 8

I'm confused on the next step. How do I multiply all three of them?

Answer 9

Not quite.

The 2 complex roots are:

\(x - (5 + 4i)) and (5 - (5 - 4i))

Answer 10

How did you get the (5-(5-4i))?

Answer 11

The given complex root is 5 + 4i, right?

Answer 12

@mathstudent55 please solve it, I don't get the answer,

Answer 13

Yesh.

Answer 14

I'm getting there. One step at a time.

Answer 15

That automatically gives us a second complex root of 5 - 4i.
The reason for this is that we are told the coefficients of the polynomial are real.
A polynomial with real coefficients can have real roots and/or complex roots, but the complex roots must come in complex conjugate pairs.

Answer 16

Okay. I understand the 5+4i and 5-4i, but not the other half.

Answer 17

Now we know we have complex roots of 5 + 4i and 5 - 4i.
All our roots are: 3, -13, 5 + 4i, 5 - 4i.
Ok?

Answer 18

Got it.

Answer 19

Every root "a" generates a factor of the form (x - a).
We have 4 factors that need to be multiplied together:

\( (x - 3)(x + 13)(x - (5 + 4i)((x - (5 - 4i)) \)

Answer 20

Those are the ones I said the second time. :p -_-

Answer 21

Now we have to multiply together all those factors. We can start by multiplying the first two, but a shortcut is to make sure you multiply the last two together.

Answer 22

I see it now. I wrote 5 instead of x above. Sorry. It was a typo.

Answer 23

I thought so. You're good. I understand the roots now. How do I multiply them though?

Answer 24

The last two factors that have the complex roots in them are of the form of a product of a sum and a difference. They follow this form:

\( (a + b)(a - b) = a^2 - b^2\)

Answer 25

We can take advantage of this as a shortcut.

Answer 26

You're talking about the 5+4i and 5-4i, right?

Answer 27

Yes.

Answer 28

We manipulate the factors with the complex roots into a product of a sum and a difference.

\( (x - 3)(x + 13)\color{red}{(x - (5 + 4i)((x - (5 - 4i))} \)

\(= (x - 3)(x + 13) \color{red}{(x - 5 - 4i)(x - 5 + 4i)} \)

\(= (x - 3)(x + 13) \color{red}{((x - 5) - (4i))((x - 5) + (4i))} \)

Answer 29

Okay... I think I got it so far..

Answer 30

Now we multiply the last two factors together using the product of a sum and a difference.

Answer 31

\(= (x - 3)(x + 13) \color{red}{((x - 5)^2 - (4i)^2)} \)

\(= (x - 3)(x + 13) \color{red}{(x^2 - 10x + 25 - (-16) )} \)

Answer 32

Which parts do we plug into the a^2-b^2? Oh. Nevermind.

Answer 33

\(= (x - 3)(x + 13) (x^2 - 10x + 25 + 16)\)

\(= (x - 3)(x + 13) (x^2 - 10x + 41)\)

Answer 34

So then we need to multiply x^2+10x-39 and x^2-10+41?

Answer 35

Correct.

Answer 36

How do you multiply those.. Is there a thing like FOIL but for three's?

Answer 37

\(= (x^2 +10x - 39) (x^2 - 10x + 41)\)

Multiply every term of the first polynomial by every term of the second polynomial.

Start by taking the first term of the first polynomial and multiplying by all terms of the second one.
Then take the second term of the first polynomial and multiply by every term of the second one.
Then do the same with the third term.
Finally, combine like terms.

Answer 38

\(= (x^2 +10x - 39) (x^2 - 10x + 41)\)

\(= x^4 - 10x^3 + 41x^2 + 10x^3 - 100x^2 + 410x - 39x^2 + 390 x - 1599\)

Answer 39

\(= x^4 - 98x^2 + 800x - 1599\)

Answer 40

That's what I got, too! :D So it's c. Yay!

Answer 41

I have another one. You up for it? :p

Answer 42

the hardest part is finding a quadratic with zero of \( 5 + 4i \) but it is easy if you work backwards
\[x=5+4i\\
x-5=4i\\
(x-5)^2=(4i)^2=16\\
x^2-10x+25=-16\\
x^2-10x+41=0\]

Answer 43

what is even easier is to memorize that if \(a+bi\) is a the root of a quadratic, then it is
\[x^2-2ax+(a^2+b^2)\]

Answer 44

@satellite73 Thanks for the insight.

Answer 45

Okay. :) The other question I have isn't like this one.

Answer 46

yw

Answer 47

Pls start a new post for the new problem.

Answer 48

Fine..