haydenn621 haydenn621
  • 16-12-2021
  • Mathematics
contestada

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

Respuesta :

goddessboi goddessboi
  • 16-12-2021

Answer:

[tex]f(x)=-\frac{15}{13}x^3+\frac{30}{13} x^2-\frac{60}{13} x+\frac{120}{13}[/tex]

Step-by-step explanation:

[tex]f(x)=a(x-2)(x+2i)(x-2i)[/tex]

[tex]f(x)=a(x-2)(x^2+4)[/tex]

[tex]-15=a(3-2)(3^2+4)[/tex]

[tex]-15=13a[/tex]

[tex]-\frac{15}{13}=a[/tex]

[tex]f(x)=-\frac{15}{13}(x-2)(x^2+4)[/tex]

[tex]f(x)=-\frac{15}{13}(x^3-2x^2+4x-8)[/tex]

[tex]f(x)=-\frac{15}{13}x^3+\frac{30}{13} x^2-\frac{60}{13} x+\frac{120}{13}[/tex]

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