Problem

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 $x=-4$. It goes through the point $(5,40.5)$. Find a formula for $P(x)$. \[ P(x)= \]

Solution

Step 1 :We are given a polynomial of degree 4, P(x), with a root of multiplicity 2 at x=2 and roots of multiplicity 1 at x=0 and x=-4. It also passes through the point (5,40.5).

Step 2 :We can write the polynomial in the form P(x) = a*(x-2)^2*x*(x+4).

Step 3 :We can find the value of 'a' by substitifying the point (5,40.5) into the equation. This gives us 40.5 = a*(5-2)^2*5*(5+4).

Step 4 :Solving this equation gives us a value of a = 0.1.

Step 5 :Substituting a = 0.1 back into the equation gives us the polynomial P(x) = 0.1x(x - 2)^2(x + 4).

Step 6 :\(\boxed{P(x) = 0.1x(x - 2)^2(x + 4)}\) is the final answer.

From Solvely APP
Source: https://solvelyapp.com/problems/7309/

Get free Solvely APP to solve your own problems!

solvely Solvely
Download