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

Step-by-step 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

$(x - ( - 2))(x - \sqrt{5} )(x - \frac{10}{3} )$

$(x + 2)(x - \sqrt{5} )(3x - 10)$

$(3 {x}^{2} - 4x - 20)(x - \sqrt{5} )$

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

$(3 {x}^{2} - 4x - 20)(x - \sqrt{5} )(x + \sqrt{5} )$

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

Which will gives us a rational coeffeicent of degree 4.

Why we didn't do

$(x - \sqrt{5} )$

?

Because

$(x - \sqrt{5} ) {}^{2} = {x}^{2} - 2 \sqrt{5} + 5$

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