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# If a polynomial function f(x) has roots 0, 4, and 3 + √11 , what must also be a root of f(x)?

**Solution:**

If you want rational coefficients then you should get the conjugate of any irrational zero.

The conjugate of 3 + √11 is 3 - √11

**Explanation:**

x = 3 + √11

**Subtract 3 on both sides**

x - 3 = 3 + √11 - 3

x - 3 = √11

**By squaring on both sides**

(x - 3)^{2} = 11

**Now subtract 11 on both sides**

(x - 3)^{2} - 11 = 0

**To factor use the difference of squares**

u^{2} - v^{2} = (u - v) (u + v)

[(x - 3) - 11][(x - 3) + 11] = 0

We get,

(x - 3) - 1 = 0 or (x - 3) + 11 = 0

Solve for x - 3 and x

**Add √11 on both sides of first equation and subtract √11 on both sides of second equation**

x - 3 = √11 or x - 3 = - √11

By adding 3 on both sides

x = 3 + √11 or x = 3 - √11

**Therefore, 3 - √11 must also be a root of f (x).**

## If a polynomial function f(x) has roots 0, 4, and 3 + √11 , what must also be a root of f(x)?

**Summary:**

If a polynomial function f(x) has roots 0, 4, and 3 + √11 , 3 - √11 is also a root of f(x).

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