Question

If a polynomial function f(x) has roots 0, 4, and 3+ sqrt11 what must also be a root of f(x)?

Answer 1

the answer is C.

Explanation:

Answer 2

`3-√(11)`

Explanation:

If you want rational coefficients then you would want the conjugate of any irrational zero given.

The question is equivalent to what is the conjugate of
`3+√(11)`.

The conjugate of
`3+√(11)` is
`3-√(11)`.

In general, the conjugate of a+b is a-b

or the conjugate of a-b is a+b.

Or maybe you like this explanation more:

Let
`x=3+√(11)`

Subtract 3 on both sides:

`x-3=√(11)`

Square both sides:

`(x-3)^2=11`

Subtract 11 on both sides:

`(x-3)^2-11=0`

Use difference of squares to factor. I apply
`u^2-v^2=(u-v)(u+v)`.

`([x-3]-√(11))([x-3]+√(11))=0`

So you have either

`[x-3]-√(11)=0` or
`[x-3}+√(11)=0`

Solve both for x-3 and then x.

Add sqrt(11) on both sides for first equation and subtract sqrt(11) on both sides for second equation:

`x-3=√(11)` or
`x-3=-√(11)`

Add 3 on both sides:

`x=3+√(11)` or
`x=3-√(11)`