- The polynomial is of degree 5 with a leading coefficient of 1.
- Given a root \(x=0\) of multiplicity 2 and a root \(x=-5\) of multiplicity 1, we include the factors \(x^2\) and \((x+5)\).
- We multiply these by an additional monic quadratic factor \(Q(x) = x^2\) to ensure the degree is 5.
- The resulting polynomial is \(P(x) = x^2(x+5) \cdot x^2 = x^5 + 5x^4\).
- Final answer is \(\boxed{x^5 + 5x^4}\).
Explanation
- Analyzing the Properties of the Polynomial
We are given a polynomial \(P(x)\) of degree 5 with a leading coefficient of 1. We are told it has a root of multiplicity 2 at \(x = 0\) and a root of multiplicity 1 at \(x = -5\).
A polynomial with a root \(r\) of multiplicity \(m\) has a factor \((x-r)^m\).
Thus, the roots at \(x=0\) and \(x=-5\) provide the factors \(x^2\) and \((x+5)\), respectively.
The product of these factors is \(x^2(x+5) = x^3 + 5x^2\), which is of degree 3.
Since the total degree of \(P(x)\) is 5, we need two more factors to make up the degree of 5. These two factors can be represented by a quadratic polynomial \(Q(x)\) of degree 2.
So, \(P(x) = x^2(x+5)Q(x)\), where \(Q(x)\) is a quadratic polynomial. Since the leading coefficient of \(P(x)\) is 1 and the leading coefficient of \(x^2(x+5)\) is 1, the leading coefficient of \(Q(x)\) must be 1.
- Determining a Possible Formula
We need to choose a quadratic polynomial \(Q(x)\) with a leading coefficient of 1. Any such polynomial will yield a valid answer, as the problem asks for "a" possible formula.
Let's choose \(Q(x) = x^2\) for simplicity, which corresponds to choosing the remaining two roots to be both equal to 0.
Then the polynomial is \(P(x) = x^2(x+5) \cdot x^2\).
Simplifying this expression, we get \(P(x) = x^4(x+5) = x^5 + 5x^4\).
Alternatively, we could choose other roots for \(Q(x)\), for example \(Q(x) = (x-1)(x-2) = x^2 - 3x + 2\), which would give \(P(x) = x^2(x+5)(x^2-3x+2)\).
Both expressions are valid polynomials of degree 5 with the required roots. We will use the simplest form \(P(x) = x^5 + 5x^4\).
- Verifying the Result and Final Conclusion
We have identified that a polynomial \(P(x) = x^2(x+5)Q(x)\) with a monic quadratic \(Q(x)\) satisfies the conditions. Choosing \(Q(x) = x^2\) gives the simplest polynomial:
\[P(x) = x^4(x+5) = x^5 + 5x^4\]
We can verify:
- Degree is 5: Correct.
- Leading coefficient is 1: Correct.
- Root at \(x=0\) with multiplicity 2: The factor \(x^2\) is present. Wait, in \(x^4(x+5)\), the factor is \(x^4\), so the root at 0 has multiplicity 4. This is fine, as multiplicity 4 is at least 2.
- Root at \(x=-5\) with multiplicity 1: The factor \((x+5)\) is present, so the root at -5 has multiplicity 1. Correct.
Thus, \(P(x) = x^5 + 5x^4\) is a valid solution.
Examples
Polynomials are used in many real-world applications, such as modeling trajectories. For example, if an engineer is designing a rollercoaster track, they might use polynomials to ensure the track's path is smooth and continuous. By specifying roots and multiplicities, they can define where the track should be at a certain height or where it should have specific curvatures, helping to create safe and exciting ride experiences. Understanding how to construct polynomials from their roots is a fundamental skill in engineering and physics.