Find a cubic polynomial such that P(-3) =0, P(2)=-4 and P(x)>0 only when x>5

Question

PLEASE HELP DUE TODAY This question has been stumping me for the longest time

Answer 1

Because the curve touches the x-axis when x=-3 we know (x+3) is a root but since it can't cross the axis, it is a double root. (x+3)^2. Since P(x) > 0 only for x>5, it only crosses the x-axis at x=5 s0 (x-5) must be a root. Then you solve for the coefficient a

P(x)=a(x+3)^2 (x-5) P(2)= -4= a(2+3)^2 (2-5) -4= a(5)^2 (-3) -4= 25a * -3 -4= -75a a= 4/75

So the cubic polynomial that would answer this question is P(x)= (4/75(x+3)^2)(x-5)

Hope that helps and makes some sense

Answer 2

There are roots at 5 and -3. Since the function is only greater than 0 in one interval, x > 5, the root at -3 is a double root.

The function is of the form

P(x) = a(x+3)²(x-5)

As P(2) = -4,

a(2+3)²(2-5) = -4

-75a = -4

a = 4/75

P(x) = 4/75(x+3)²(x-5) = 4/75x³ + 4/75x² - 28/25 - 12/5