What is a polynomial function of least degree with a leading coefficient of 1 that has the zeros 4 and 3 + √2?

Question

need asap

Answer 1

When you have roots that are conjugates, we can use the formula to find a quadratic factor.

 

x² - (sum of roots)x + product of roots

 

sum of roots = (3 + √2) + (3 - √2) = 6

product of roots = (3 + √2)(3 - √2) = 9 - 2 = 7

 

Your quadratic factor is (x² - 6x + 7).  Multiply this by (x - 4).

 

f(x) = (x - 4)(x² - 6x + 7)

f(x) = x³ - 6x² + 7x - 4x² + 24x - 28

f(x) = x³ - 10x² + 31x - 28

Answer 2

If you need to have rational coefficients, then use 3 - √2 as another root.

 

Then, a possible polynomial is (x-4)[x-(3+√2)]{x - (3-√2)]

 

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

 

                                          = (x-4)[(x-3)² - 2]

 

                                          = (x-4)(x²-6x+7) = x³-6x²+7x-4x²+24x-28

 

                                                                     = x³ - 10x² + 31x - 28

 

Answer 3

Hi Logan, first find the values of factors based on the information given. Remember the zero product rule!

 

x = 4

 

x - 4 = 0

 

One factor is (x - 4)

 

x = 3 + √2

 

x - (3 + √2) = 0

 

Second factor is (x - (3 + √2))

 

To get the polynomial, multiply the two factors:

 

(x - 4) * (x - (3 + √2))

 

Questions?

Answer 4

So the polynomial of least degree with shown zeros is a polynomial of degree 2!