Apolynomial 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).

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

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

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

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

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

So, the only true statement is the last one.

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step-by-step explanation:

-2 is the right