MATH SOLVE

4 months ago

Q:
# If a polynomial function f(x) has roots 1+square root of 2 and-3, what must be a factor of f(x)?

Accepted Solution

A:

To find the factor a a polynomial from its roots, we are going to seat each one of the roots equal to[tex]x[/tex], and then we are going to factor backwards.

We know for our problem that one of the roots of our polynomial is -3, so lets set -3 equal to[tex]x[/tex] and factor backwards:

[tex]x=-3[/tex]

[tex]x+3=0[/tex]

[tex](x+3)[/tex] is a factor of our polynomial.

We also know that another root of our polynomial is [tex]1+ \sqrt{2} [/tex], so lets set [tex]1+ \sqrt{2} [/tex] equal to [tex]x[/tex] and factor backwards:

[tex]x=1+ \sqrt{2} [/tex]

[tex]x-1= \sqrt{2} [/tex]

[tex]x-1- \sqrt{2}=0 [/tex]

[tex](x-(1+ \sqrt{2})=0 [/tex]

([tex](x-(1+ \sqrt{2} ))[/tex] is a factor of our polynomial.

We can conclude that there is no correct answer in your given choices.

We know for our problem that one of the roots of our polynomial is -3, so lets set -3 equal to[tex]x[/tex] and factor backwards:

[tex]x=-3[/tex]

[tex]x+3=0[/tex]

[tex](x+3)[/tex] is a factor of our polynomial.

We also know that another root of our polynomial is [tex]1+ \sqrt{2} [/tex], so lets set [tex]1+ \sqrt{2} [/tex] equal to [tex]x[/tex] and factor backwards:

[tex]x=1+ \sqrt{2} [/tex]

[tex]x-1= \sqrt{2} [/tex]

[tex]x-1- \sqrt{2}=0 [/tex]

[tex](x-(1+ \sqrt{2})=0 [/tex]

([tex](x-(1+ \sqrt{2} ))[/tex] is a factor of our polynomial.

We can conclude that there is no correct answer in your given choices.