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

Step-by-step 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

$(x - (1 - \sqrt{6} ))(x - (1 + \sqrt{6} ))(x - (7 - i))$

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

${x}^{2} + x( - 1 + \sqrt{6} ) + x( - 1 - \sqrt{6} ) + ( - 5)$

${x}^{2} - 2x - 5$

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

${x}^{4} - 16 {x}^{3} + 73 {x}^{2} - 30x - 250$