Question

find a third degree polynomial having only real coefficients, 2 and "-2i" are zeros and f(3)=-15

Answer 1

Answer:

$f(x)=-\frac{15}{13}x^3+\frac{30}{13} x^2-\frac{60}{13} x+\frac{120}{13}$

Step-by-step explanation:

$f(x)=a(x-2)(x+2i)(x-2i)$

$f(x)=a(x-2)(x^2+4)$

$-15=a(3-2)(3^2+4)$

$-15=13a$

$-\frac{15}{13}=a$

$f(x)=-\frac{15}{13}(x-2)(x^2+4)$

$f(x)=-\frac{15}{13}(x^3-2x^2+4x-8)$

$f(x)=-\frac{15}{13}x^3+\frac{30}{13} x^2-\frac{60}{13} x+\frac{120}{13}$