Question

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).

Answer 1

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.