Question

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

4, -2, and -1 + 2i

How would I solve this?

Answer 1

so the polynomial has 4 roots..

the real roots are x = 4 and x = -2

the complex roots are

\[x = -1 \pm i\]

Answer 2

so to find the complex roots

start with

\[x = -1 \pm i\]

add 1 to both sides

\[x + 1 = \pm i\]

square both sides of the equation

\[(x + 1)^2 = i^2\]

remember i^2 = -1

then

\[x^2 + 2x + 1 = -1\]

add 1 to both sides of the equation and you will have the quadratic factor of the polynomial...

so the polynomial is

\[P(x) = (x^2 + 2x +1)(x -4)(x+1)\]

just distribute for the equation

Answer 3

Okay.

Answer 4

2i, 5, -5?

Answer 5

is that another question...?

Answer 6

No, is that the correct answers @campbell_st

Answer 7

well the polynomial is

\[P(x) = (x^2 + 2x + 2)(x -4)(x +2)\]

which becomes

\[P(x) = (x^2+ 2x +2)(x^2 -2x - 8)\]

you just need to multiply the 2 quadratic factors together

Answer 8

The answer I come up with is not one of the options.

X^4-10x^2-20x-16

Answer 9

f(x)=(x-4)(x+2)[x-(-1+2i)][x-(-1-2i)]

Answer 10

Multiply the four factors and that will be your answer.

Answer 11

Okay I got x^4+6.5x^2-26x

Answer 12

Is that correct?

Answer 13

f(x)=(x^2-2x-8)(x^2+5)=x^4-2x^3-8x^2+5x^2-10x-40=x^4-2x^3-3x^2-10x-40

Answer 14

That's what I get from multiplying those 4 factors. Is that one of your choices?

Answer 15

Yes, I must have done it wrong. Thank you.

Answer 16

Wait, no.

Answer 17

what the hell

Answer 18

If 4is a root, then x-4 is a factor.
If -2 is a root then x-(-2)=x+2 is a factor
If -1+2i is a root then -1 - 2i is also a root
If -1+2i is a root then x-(-1+2i) is a factor
If -1-2i is a root then x - (-1-2i) is a factor

Answer 19

So now we have 4 factors of the polynomial:
(x-4)(x+2)[x-(-1+2i)][x-(-1-2i)]
(x-4)(x+2)=x^2-2x-8
[x-(-1+2i)][x-(-1-2i)]=x^2-x(-1-2i)-x(-1+2i)+(-1+2i)(-1-2i)=
x^2+x+2xi+x-2xi+5=x^2+2x+5

Answer 20

So now we must multiply (x^2-2x+8)(x^2+2x+5)=
x^4+2x^3+5x^2-2x^3-4x^2-10x+8x^2+16x+40=
x^4+9x^2+6x+40

Answer 21

Is that one of your choices?

Answer 22

No.

Answer 23

@Mertsj

Answer 24

What are the answer choices?

Answer 25

a. f(x) = x4 - 7x2 - 26x - 40
b. f(x) = x4 - 3x3 - 8x2 - 13x - 40
c. f(x) = x4 + 6.5x2 - 26x - 40
d. f(x) = x4 - 3x3 + 8x2 + 13x + 40

Answer 26

I made a typo when I multiplied:
(x^2-2x-8)(x^2+2x+5)=x^4+2x^3+5x^2-2x^3-4x^2-10x-8x^2-16x-40=
x^4-7x^2-26x-40

Answer 27

Before I used x^2-2x+8 instead of x^2-2x-8

Answer 28

Ah I see.

Answer 29

Thanks.

Answer 30

yw