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Find a polynomial function of degree 3 such that f (2) = 24 and x– 1, x and x+ 2

are factors of the polynomial.

Respuesta :

f(x) = 3x³ + 3x² - 6x

given the factors of the polynomial , then

f(x) =ax(x - 1)(x + 2) ( a is a multiplier )

to find a use f(2) = 24

f(2) = 2a(1)(4) = 24 ⇒8a = 24 ⇒ a = 3, thus

f(x) = 3x(x² + x - 2) = 3x³ + 3x² - 6x






The polynomial function is [tex]f(x) = 3x^3 + 3x^2 - 6x[/tex].

Given that,

  • f(2) = 24.
  • And, x– 1, x and x+ 2 represents the factors of the polynomial.

Now based on the above information, the calculation is as follows:

f(x) =ax(x - 1)(x + 2)

Here a is a  multiplier.

Now

f(2) = 24

f(2) = 2a(1)(4) = 24

8a = 24

a = 3

It can be write as

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

[tex]f(x) = 3x^3 + 3x^2 - 6x[/tex].

Therefore we can conclude that the polynomial function is [tex]f(x) = 3x^3 + 3x^2 - 6x[/tex].

Learn more: brainly.com/question/12976257