Algebra 2 ( Write a polynomial function of least degree that has the given roots. )

Question

Write a polynomial function of least degree that has the given roots. 3, 2+sqrt6 I know how to do this, but how do I do this with a sqrt?

Answer 1

x₁ = 3

x₂ = 2 + √6

x₃ = 2 - √6

 

 

Turn these roots into a product of linear factors.

 

f(x) = (x - 3)(x - [2 + √6])(x - [2 - √6])

 

f(x) = (x - 3)[x² - x(2 - √6) - x(2 + √6) + (4 - 6)]

 

f(x) = (x - 3)(x² - 4x - 2)

 

f(x) = x(x² - 4x - 2) - 3(x² - 4x - 2)

 

 

Now just distribute and combine like terms.

Answer 2

If one of the roots for a polynomial function has the form a + sqrt(b), then it will also have a matching root of the form a - sqrt(b). This tells us that, for this particular problem, you're looking for a 3rd degree polynomial with three roots:

1. x₁ = 3
2. x₂ = 2 + sqrt(6)
3. x₃ = 2 - sqrt(6)

Go about the problem in the usual way, setting up your polynomial by multiplying your root factors:

• y = (x - x₁)(x - x₂)(x - x₃)

 

After several lines of algebra, I ended up with the coefficients 1, -7, 10, and 6 for my powers of x. What did you end up with?

Answer 3

a root of 3 means that (x-3) is a factor

 

Since 2+ sqrt(6) is a root, then 2-sqrt(6) is also a factor

 

Therefore (x-3)(x-(2+sqrt(6))(x-(2-sqrt(6))

 

Now take your time multiplying the last 2 factors

 

(x-(2+sqrt(6))(x-(2-sqrt(6)) =

 

x^2 -x(2-sqrt(6)) -x(2+sqrt(6)) + 4 -6

 

x^2 -4x - 2 

 

Now multiply this quadratic by (x-3)

 

(x-3)(x^2 -4x - 2)    

 

I will leave this last part up to you