How do I do this math problem?

Question

I am taking Algebra 2

Write the simplest polynomial function with integral coefficients that has the given zeros?

1. 2, 1 - i

the zeros are

2 and 1 - i

how do you get the x-1 and x + i?

How did you get this: (x-2)(x-1+i) ?


I am so confused :[

Where did you get the (x-1)?

Answer 1

I'll assume you mean 1, 2, and 1-i are three roots of the polynomial. Imaginary roots always come in pairs, a +- bi. So, the simplest one is:

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

Multiply it all out to get the polynomial.

Answer 2

The zeros of the polynomial are roots of the given poynomial. So the roots are 1, 2, (1-i)

But complex roots always occur in pairs. Since (1-i) is one the roots , (1+i) is also a root.

so the factors of the given polynomial P(x) are

(x-1), (x-2), (x -(1-i)) and (x - (1+i))

Polynomial is the product of all the factors

P(x) = (x-1)(x-2)[x -(1-i)][x - (1+i)]

P(x) = (x^2 -3x + 2)[(x -1) + i][(x-1)-i)]

P(x) = (x^2 -3x + 2)[(x-1)^2 - i^2]

P(x) = (x^2 -3x + 2)(x^2 + 1 -2x +1)(since i^2 = -1)

P(x) = (x^2 -3x + 2)(x^2 -2x +2)

P(x) = (x^4 - 2x^3 + 2x^2 -3x^3 + 6x^2 -6x + 2x^2 - 4x + 4)

P(x) = x^4 -5x^3 + 10x^2 - 10x + 4

Answer 3

Because you have i in there, I assume that you are working with complex numbers.

A theorem in Complex numbers states that the number of roots of a polynomial is always equal to the highest power (some roots may be repeating).

To create a polynomial that has 'a' as a root, simply multiply any polynomial by (x-a).

So, to create a polynomial that has three roots, we use three factors:
(x-2)(x-1)(x+i)
This will give you a third degree complex polynomial with three roots: +2, +1 and -i

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Well, your line of numbers could be interpreted this way;

1. This is question number 1
2, is a desired root
1-i is the other desired root.

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

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

Answer 4

For a polynomial function with integer coefficients, complex roots *always* come in conjugate pairs. Since we have one complex root, we require another complex root, (1 + i).

To create the polynomial, convert each root to (x - r) for each root r.

f(x) = (x - 1)(x - 2) (x - (1 - i)) (x - (1 + i))

f(x) = (x ^2 - 3x + 2) (x^2 - x(1 + i) - x(1 - i) + (1 - i)(1 + i))

f(x) = (x^2 - 3x + 2) (x^2 - (x + xi) - (x - xi) + (1 - i^2))

f(x) = (x^2 - 3x + 2) (x^2 - 2x + 1 - (-1))

f(x) = (x^2 - 3x + 2) (x^2 - 2x + 2)

f(x) = x^4 - 2x^3 + 2x^2 - 3x^3 + 6x^2 - 6x + 2x^2 - 4x + 4

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

(give or take some algebraic errors)

Answer 5

12

Answer 6

use a calculator or ASK YOUR TEACHER!!!!!!!!!!!!

Answer 7

CHEESE!!!!