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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).

User Bart C
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2 Answers

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Answer:

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

Explanation:

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


(x+5)^3

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


(x-1)^2(x-3)^7

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


(x+5)^3(x-1)^2(x-3)^7

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

  • positive in
    (-\infty, -5)
  • negative in
    (-5, -3)
  • positive in
    (3, \infty)

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

  • negative in
    (-\infty, -5)
  • positive in
    (-5, -3)
  • negative in
    (3, \infty)

So, the only true statement is the last one.

User Khangharoth
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