Question

What is the value of the leading coefficient a, if the polynomial function P(x) = a(x−b)²(x−c) has multiplicity of 2 at the point (2,0) and also passes through the points (5,0) and (0,60)?

Answer 1

Answer:

a=-3

Step-by-step explanation:

The graph of P passes through the point (2,0), thus P(2)=0. This means that 2 is a root of this polynomial, and it has multiplicity 2. Therefore (x-2)² is a factor of P and it's the maximum power of (x-2) that divides P, then b=2.

The graph of P also passes through (5,0), so P(5)=0. Then 5 is another root of p, that is, (x-5) is a factor of P. From this, c=5.

At this point, we have that P(x)=a(x-2)²(x-5). Again, because the graph of the function P(x) passes through (0,60) we obtain that P(0)=60. Substituting in our previous equation, 60=P(0)=a(-2)²(-5)=-20a. Dividing both sides by -20, we conclude that a=-3.