The polynomial of degree 3, P(x), has a root of multiplicity 2 at x=4 and a root of multiplicity 1 at x= -1. The y intercept is y= -6.4

Answer 1

P(x) = a(x - 4)²(x + 1)

-6.4 = a(x - 4)²(0 + 1)

-6.4 = 16a

-0.4 = a

P(x) = -0.4(x - 4)²(x + 1)

Answer 2

The polynomial P(x) has roots at x=4 and x=-1, which means that these are solutions to the polynomial when P(x) = 0. It will therefore benefit us to begin building the polynomial from its factored form, in the inverse fashion of how we find the roots when we know the polynomial:

P(x) = 0 = (x-4)(x+1)

Note also that the root x=4 has multiplicity 2, meaning it appears twice.

P(x) = 0 = (x-4)(x-4)(x+1)

Rearranging as a polynomial, we have:

P(x) = 0 = x³ - 8x² + 16x + x² - 8x + 16

Collecting terms, we get:

P(x) = 0 = x³ - 7x² + 8x + 16

We are also told that the y-intercept (when x=0) is P(0) = -6.4. The y-intercept is represented by the constant term in the polynomial, which is 16 in the one we have constructed. Luckily, we can easily transform this constant to any number we want by dividing the function, P(x), by the same number on both sides.

When P(x) = 0, the value of the function remains unchanged, so the roots we have "inserted" are still valid. How much should we divide by? The answer is given by the following expression:

16 / a = -6.4 , where a is some number.

In this case, a = -5/2.

So, we have:

P(x) = (x³ - 7x² + 8x + 16) / (-5/2)

P(x) = -0.4x³ + 2.8x² - 3.2x - 6.4