Question

The polynomial of degree 4, P(x) has a root of multiplicity 2 at x= 2 and roots of multiplicity 1 at

X = 0 and x= -2. It goes through the point (5, 220.5).
Find a formula for P(x).

Answer 1

The polynomial P(x) with the given roots and point is P(x) = 0.7(x - 2)²(x)(x + 2).

Step-by-step explanation:

To find the formula for a polynomial P(x) of degree 4 with the given roots and a point through which it passes, we can use the fact that a polynomial can be expressed as a product of its roots.

The polynomial has a root of multiplicity 2 at x=2, a single root at x=0, and another single root at x=-2.
Thus, the general form of the polynomial is:

P(x) = a(x - 2)²(x)(x + 2)

Where a is a constant we need to find. Since the polynomial passes through the point (5, 220.5), we substitute these values into the polynomial to solve for a:

220.5 = a(5 - 2)²(5)(5 + 2)

After calculating:

220.5 = 9a * 5 * 7

220.5 = 315a

Dividing both sides by 315 gives:

a = 220.5 / 315

a = 0.7

Thus, the specific polynomial that meets the criteria is:

P(x) = 0.7(x - 2)²(x)(x + 2)