Question
The polynomial of degree \( 5, P(x) \) has leading coefficient 1 , has roots of multiplicity 2 at \( x=4 \) and \( x=0 \) , and a root of multiplicity 1 at \( x=-1 \) Find a possible formula for \( P(x) \). \( P(x)= \)
Ask by Carlson Ward. in the United States
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( P(x) = (x-4)^2 \cdot x^2 \cdot (x+1) \)
Solution
First, we note that the polynomial has the following roots and multiplicities:
- A root at \( x = 4 \) with multiplicity 2, contributing the factor \( (x-4)^2 \).
- A root at \( x = 0 \) with multiplicity 2, contributing the factor \( x^2 \).
- A root at \( x = -1 \) with multiplicity 1, contributing the factor \( (x+1) \).
Since the polynomial is required to be of degree 5 and have a leading coefficient of 1, the polynomial can be written as the product of these factors:
\[
P(x) = (x-4)^2 \cdot x^2 \cdot (x+1)
\]
This expression satisfies all the conditions provided.
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Simplify this solution The Deep Dive
To construct the polynomial \( P(x) \), we start with its roots and their respective multiplicities. Since \( x = 4 \) has a multiplicity of 2, we can write \((x - 4)^2\). The root \( x = 0 \) also has a multiplicity of 2, which gives us \((x - 0)^2\) or \(x^2\). Lastly, for the root \( x = -1 \) with a multiplicity of 1, we include \((x + 1)\). Putting it all together, we have: \[ P(x) = (x - 4)^2 \cdot (x^2) \cdot (x + 1) \] Expanding that gives: \[ P(x) = (x^2 - 8x + 16) \cdot (x^2) \cdot (x + 1) = (x^4 - 8x^3 + 16x^2)(x + 1) \] Now, we can distribute: \[ P(x) = x^5 - 7x^4 + 8x^3 + 16x^2 \] Thus, a possible formula for \( P(x) \) is: \[ P(x) = x^5 - 7x^4 + 8x^3 + 16x^2 \]
