Question

Suppose a polynomial of degree 4 with rational coefficients has the given numbers as zeros. Find the other zero

-2, square root 5, 10/3

Answer 1

Explanation:

The root is

-sqr root of 5.

First, we put these roots in the forn of

`(x - a)`

where a is the root

So we have

`(x - ( - 2))(x - √(5) )(x - (10)/(3) )`

`(x + 2)(x - √(5) )(3x - 10)`

`(3 {x}^(2) - 4x - 20)(x - √(5) )`

To get rid of that square root, let have another root that js the conjugate posive root of 5.

`(3 {x}^(2) - 4x - 20)(x - √(5) )(x + √(5) )`

`(3 {x}^(2) - 4x - 20)(x {}^(2) + 5)`

Which will gives us a rational coeffeicent of degree 4.

Why we didn't do

`(x - √(5) )`

?

Because

`(x - √(5) ) {}^(2) = {x}^(2) - 2 √(5) + 5`

If we foiled out we will still have a irrational coeffceint.