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Write a polynomial f(x)that satisfies the given conditions,
Polynomial of lowest degree with zeros of - 4 (multiplety 1), 3 (multiplicly 3), and with f(0)= 216,

Respuesta :

Alfpfeu

Answer:

[tex]\boxed{\sf \ \ \ -2(x+4)(x-3)^3 \ \ \ }[/tex]

Step-by-step explanation:

Hello,

let's note k a real, we can write the polynomial as

[tex]k(x-(-4))^1(x-3)^3=k(x+4)(x-3)^3[/tex]

and we know that f(0)=216 so

[tex]216=k(0+4)(0-3)^3=k*4*(-1)^3*3^3=-27*4*k=-108k\\\\<=> k=-\dfrac{216}{108}=-2[/tex]

So the solution is

[tex]-2(x+4)(x-3)^3[/tex]

hope this helps

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