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Question

The function p is a fourth-degree polynomial with x-intercepts 1, 3, and 10 and y-intercept -2. If p(x) is positive only on the interval (3, 10), find p(x).

p(x)=

Answer 1

p(x) = -1/15 * (x - 1)²(x - 3)(x - 10)

Why?
We know that p(x) = a(x - 1)(x - 3)(x - 10)(x - z), where z is one of the 3 zeros 1, 3, or 10, since we know that it is 4th degree (4 zeros), and we know that any polynomial can be factored out as the product of a constant and all its zero terms, including multiplicities. We also know that z must be 1, since we are given that it fails to cross the x-axis at that zero and hence must have an even multiplicity there.

This gives us the function
p(x) = a(x - 1)²(x - 3)(x - 10).

Using the last piece of information:
p(0) = -2 = a(0 - 1)²(0 - 3)(0 - 10)
-2 = a(-1)²(-3)(-10)
-2 = 30a
a = -1/15

So, p(x) = -1/15 * (x - 1)²(x - 3)(x - 10)

Testing:
p(1) = -1/15 * (1 - 1)²(1 - 3)(1 - 10)
= -1/15 * (0)²(-2)(-9) = 0
p(3) = -1/15 * (3 - 1)²(3 - 3)(3 - 10)
= -1/15 * (2)²(0)(-7) = 0
p(10) = -1/15 * (10 - 1)²(10 - 3)(10 - 10)
= -1/15 * (9)²(6)(0) = 0

p(0) = -1/15 * (0 - 1)²(0 - 3)(0 - 10)
= -1/15 * (-1)²(-3)(-10)
= -1/15 * 30 = -2

-1/15 is always negative and (x - 1)² is always nonnegative, so p(x) is negative or zero unless (x - 3) and (x - 10) have opposite signs, which they do only on the interval 3 < x < 10.