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A polynomial function has a root of –5 with multiplicity 3, a root of 1 with multiplicity 2, and a root of 3 with multiplicity 7. If the function has a negative leading coefficient and is of even degree, which statement about the graph is true? The graph of the function is positive on (-infinite, –5). The graph of the function is negative on (–5, 3). The graph of the function is positive on (-infinite, 1). The graph of the function is negative on (3, infinite).

Respuesta :

The root of -5 with multiplicity 3 implies that the polynomial is a multiple of

[tex] (x+5)^3 [/tex]

Similarly, the two other roots imply that the polynomial is a multiply of

[tex] (x-1)^2(x-3)^7 [/tex]

So, the minimal polynomial which satisfies your requests on the roots is

[tex] (x+5)^3(x-1)^2(x-3)^7[/tex]

which would be a polynomial of degree 12. This polynomial would be:

  • positive in [tex] (-\infty, -5) [/tex]
  • negative in [tex] (-5, -3) [/tex]
  • positive in [tex] (3, \infty) [/tex]

Since we want a negative leading term, the signs will be opposite: your polynomial is

  • negative in [tex] (-\infty, -5) [/tex]
  • positive in [tex] (-5, -3) [/tex]
  • negative in [tex] (3, \infty) [/tex]

So, the only true statement is the last one.

Answer:

The graph of the function is negative on (3, infinity).

Step-by-step explanation:

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