Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 8, -14, and 3 + 9i

Options: f(x) = x4 - 11x3 + 72x2 - 606x + 10,080
f(x) = x4 - 303x2 + 1212x - 10,080
f(x) = x4 - 11x3 - 72x2 + 606x - 10,080
f(x) = x4 - 58x2 + 1212x - 10,080

28

step-by-step explanation:

8+8+6+6

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.