Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 2, and the given zeros. -4,0,2,4

A least degree polynomial, having rational coefficients and a leading coefficient of 2, with,-4, 0, 2, and 4 as the zeros of the polynomial is;

f(x) = 2·x⁴ - 4·x³ - 32·x² + 64·x

Step-by-step explanation:

The given parameters of the polynomial are;

The leading coefficient of the polynomial = 2

The zeros of the polynomial = -4, 0, 2, 4

We note that zeros of -4, and 4 gives a factor of the form, (x² - 4²)

For a zero of the polynomial equal to 0, one of the factors of the polynomial is equal to 'x'

To have a leading coefficient of 2, we can add '2' as a factor of the polynomial

Therefore, we can have the factors of the polynomial as follows;

(x² - 4²)·2·x×(x - 2) = 0

From the above equation, using a graphing calculator, we get the following possible polynomial;

(x² - 4²)·2·x×(x - 2) = 2·x⁴ - 4·x³ - 32·x² + 64·x = 0

Therefore, a polynomial, function of least degree that has rational coefficients, a leading coefficient of 2,and the zeros, -4, 0, 2, and 4 is  f(x) = 2·x⁴ - 4·x³ - 32·x² + 64·x.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2