Question

If a polynomial function, f(x), with rational coefficients has roots 0, 4, and 3 startroot 11 endroot, what must also be a root of f(x)?

Answers

1. thuthao
3 – √11  also be a root of f(x).

### What is Polynomial function ?

A polynomial function is a function that involves only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc.

### According to the given Information:

Get the conjugate of any irrational zero if you want rational coefficients.
The conjugate of 3 + √11  is  3 – √11
x = 3 +√ 11
Subtract 3 on both sides
x minus 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
u² minus v² = (u minus v) (u plus 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 minus 3 = √11 or x minus 3 = – √11
By adding 3 on both sides
x = 3 + √11 or x = 3 – √11
As a result,
the root of 3 – √11 must likewise be f(x).
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2. Acacia
Answer:
-3√11
Step-by-step explanation:
If the coefficients of a polynomial are rational, any irrational root will have a conjugate that is also a root.

### Irrational roots

The root 3√11 is irrational, so its conjugate, -3√11, will also be a root.
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Additional comments
The conjugate of a root of the form a+√b or a+bi will be the same form with the sign changed: a-√b or a-bi.
The conjugate of 3√11 = 0 +√99 will be 0 -√99 = -3√11.
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A quadratic with rational coefficients can only have irrational roots of the form a±√b, where ‘a’ and ‘b’ may be any rational number.

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