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)?

Answer 1

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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Answer 2

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.