Question

If a polynomial function f(x) has roots 3 and radical 7 what must also be a root of f(x)?

Answer 1

Answer:

The answer is -√7

Step-by-step explanation:

∵ The polynomial has roots 3 and √7

∴ It must have root -√7 the conjugate of √7

∴ The roots of f(x) are 3 , √7 , -√7

   (one rational and two irrational conjugate to each other)

Answer 2

Answer:   -√7 must also be a root of f(x).

Step-by-step explanation:  Given that a polynomial function f(x) has roots 3 and radical 7.

We are to find the other number that must be a root of f(x).

We know that

irrational roots of a polynomial function always occur in conjugate pairs.

That is, if (a + √b) is a root of a polynomial function, then its conjugate pair (a - √b) is also a root of the polynomial function.

For the given polynomial f(x), the given roots are 3 and √7.

Now, $\sqrt7=0+\sqrt 7.$

Then, the other root will be

$0-\sqrt7=-\sqrt7.$  

Thus, -√7 must also be a root of f(x).