MATH SOLVE

2 months ago

Q:
# Write a polynomial function of least degree with integral coefficients that has the given zeros. -5, 3i

Accepted Solution

A:

Solving for the polynomial function of least degree with
integral coefficients whose zeros are -5, 3i
Β
We have:

x = -5 Then x + 5 = 0 Therefore one of the factors of the polynomial function is (x + 5)

Also, we have:

x = 3i

Which can be rewritten as:

x = Sqrt(-9)

Square both sides of the equation:

x^2 = -9

x^2 + 9 = 0 Therefore one of the factors of the polynomial function is (x^2 + 9)

The polynomial function has factors: (x + 5)(x^2 + 9)

= x(x^2 + 9) + 5(x^2 + 9) = x^3 + 9x + 5x^2 = 45 Therefore, x^3 + 5x^2 + 9x β 45 = 0 f(x) = x^3 + 5x^2 + 9x β 45 Β The polynomial function of least degree with integral coefficients that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x β 45

x = -5 Then x + 5 = 0 Therefore one of the factors of the polynomial function is (x + 5)

Also, we have:

x = 3i

Which can be rewritten as:

x = Sqrt(-9)

Square both sides of the equation:

x^2 = -9

x^2 + 9 = 0 Therefore one of the factors of the polynomial function is (x^2 + 9)

The polynomial function has factors: (x + 5)(x^2 + 9)

= x(x^2 + 9) + 5(x^2 + 9) = x^3 + 9x + 5x^2 = 45 Therefore, x^3 + 5x^2 + 9x β 45 = 0 f(x) = x^3 + 5x^2 + 9x β 45 Β The polynomial function of least degree with integral coefficients that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x β 45