Question

Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and roots of 1-√6 , 1+ √6 , and 7-i

Answer 1

Explanation:

Use this algebra 2 theorem:

If r and q are roots of a polynomial function then

the polynomial function can be expressed as

`(x - r)(x - q)`

Here the roots are

1- root of 6, 1+ root 6, and. 7-i so our. function can be expressed as

`(x - (1 - √(6) ))(x - (1 + √(6) ))(x - (7 - i))`

The first two binomials are difference off squares so the we have

`{x}^(2) + x( - 1 + √(6) ) + x( - 1 - √(6) ) + ( - 5)`

`{x}^(2) - 2x - 5`

The other root is (7-i).

Also note since (7-i) is a root, then (7+I) is also a root.

So

`(x - (7 - i)(x - (7 + i)( {x}^(2) - 2x - 5)`

`( {x}^(2) - 14x + 50)( {x}^(2) - 2x - 5)`

Simplify and it gives us

`{x}^(4) - 16 {x}^(3) + 73 {x}^(2) - 30x - 250`