Find a polynomial function f(x) of degree 3 with real coefficients that satisfies the following conditions.
Zero of 0 and zero of 4 having multiplicity 2; f(5)=20

Respuesta :

Answer:

The polynomial will be [tex]f(x) = \frac{4}{9}x(x - 2)^{2}[/tex].

Step-by-step explanation:

The three degree polynomial function f(x) has zeros at x = 0 and at x = 4 having multiplicity 2.

Therefore, (x - 0) and (x - 4)² will be factors of the polynomial f(x).

Hence, the polynomial will be f(x) = Ax(x - 4)² .............. (1), where A is any constant that we have to evaluate.

Now, given that f(5) = 20

So, from the equation (1) we have

20 = A(5)(5 - 2)²

[tex]A = \frac{4}{9}[/tex]

Therefore, the polynomial will be [tex]f(x) = \frac{4}{9}x(x - 2)^{2}[/tex] (Answer)

Answer:

f(x)=4x(x-4)^2

Step-by-step explanation:

first we have a zero of 0, or (x-0) which can be simplified to just x

then we have a zero of 4 with a multiplicity of 2. thats (x-4) and since there's two we do (x-4)^2

so far we have f(x)=ax(x-4)^2

In order to find f(5)=30, we need to set X to 5. That changes it to a(5)(5-4)^2.

20=a(5)(5-4)^2.

a=4

so we plug the a value in and set X back. That gives us f(x)=4x(x-4)^2