- Start with the factored form of the polynomial using given roots: \(P(x) = a \cdot x \cdot (x + 3) \cdot (x - 2)^2\).
- Use the point \((5, 36)\) to solve for the leading coefficient: \(36 = a \cdot 5 \cdot 8 \cdot 3^2\), which gives \(a = \frac{1}{10}\).
- Assemble the final polynomial: \(P(x) = \frac{1}{10}x(x + 3)(x - 2)^2\).
- The final answer is \(\boxed{\frac{1}{10}x(x + 3)(x - 2)^2}\).
Explanation
- Formulating the General Expression of P(x)
The polynomial \(P(x)\) has a degree of 4. We are given its roots and their multiplicities:
- A root of multiplicity 2 at \(x = 2\) means that \((x - 2)^2\) is a factor of the polynomial.
- A root of multiplicity 1 at \(x = 0\) means that \(x\) is a factor.
- A root of multiplicity 1 at \(x = -3\) means that \((x + 3)\) is a factor.
Given these, we can write the general form of the polynomial as:
\[P(x) = a \cdot x \cdot (x + 3) \cdot (x - 2)^2\]
where \(a\) is a non-zero leading coefficient. The degree of this expression is \(1 + 1 + 2 = 4\), which is consistent with the given information.
- Finding the Leading Coefficient a
We are given that the graph of \(P(x)\) passes through the point \((5, 36)\), which means that when \(x = 5\), \(P(5) = 36\). We substitute these values into our general equation:
\[36 = a \cdot 5 \cdot (5 + 3) \cdot (5 - 2)^2\]
Now we simplify the expression:
\[36 = a \cdot 5 \cdot 8 \cdot (3)^2\]
\[36 = a \cdot 40 \cdot 9\]
\[36 = 360a\]
To find the value of \(a\), we divide both sides by 360:
\[a = \frac{36}{360} = \frac{1}{10}\]
- Final Formula and Verification
Now that we have the value of the leading coefficient \(a = \frac{1}{10}\), we can write the final formula for the polynomial:
\[P(x) = \frac{1}{10}x(x + 3)(x - 2)^2\]
We can verify this result by checking the condition \(P(5) = 36\):
\[P(5) = \frac{1}{10}(5)(5 + 3)(5 - 2)^2 = \frac{1}{10}(5)(8)(3^2) = \frac{1}{10}(40)(9) = \frac{360}{10} = 36\]
The condition is satisfied, so the formula is correct. The final answer is \(\boxed{\frac{1}{10}x(x + 3)(x - 2)^2}\).
Examples
Polynomials are essential in modeling real-world physical systems. For example, engineers use them to describe the trajectory of a projectile or to model the stress distribution in a beam under load. By identifying specific points the curve passes through and knowing its roots, they can construct the exact equation that governs the system's behavior, allowing them to predict outcomes at any given input value. This algebraic method of fitting a curve to data points is a fundamental skill in data analysis and physics.