Write a polynomial [tex]f(x)[/tex] that satisfies the given conditions.
Polynomial of lowest degree with zeros of [tex]-\frac{5}{6}[/tex] (multiplicity 2) and [tex]\frac{1}{4}[/tex] (multiplicity 1) and with [tex]f(0)=75[/tex].
[tex]f(x)=[/tex] $\square$
Answer (1)
The polynomial has zeros at x = − 6 5 (multiplicity 2) and x = 4 1 (multiplicity 1), so f ( x ) = a ( x + 6 5 ) 2 ( x − 4 1 ) .
Use the condition f ( 0 ) = 75 to find a : 75 = a ( 6 5 ) 2 ( − 4 1 ) , which gives a = − 432 .
Substitute a back into the polynomial: f ( x ) = − 432 ( x + 6 5 ) 2 ( x − 4 1 ) .
Expand and simplify the polynomial: f ( x ) = − 432 x 3 − 612 x 2 − 120 x + 75 .
f ( x ) = − 432 x 3 − 612 x 2 − 120 x + 75
Explanation
Understanding the Problem We are given that the polynomial f ( x ) has zeros at x = − 6 5 with multiplicity 2 and x = 4 1 with multiplicity 1. Also, we know that f ( 0 ) = 75 . Our goal is to find the polynomial f ( x ) of the lowest degree that satisfies these conditions.
General Form of the Polynomial Since the polynomial has zeros at x = − 6 5 with multiplicity 2 and x = 4 1 with multiplicity 1, the general form of the polynomial is given by f ( x ) = a ( x + 6 5 ) 2 ( x − 4 1 ) , where a is a constant.
Using the Condition f(0) = 75 We are given that f ( 0 ) = 75 . Substituting x = 0 into the expression for f ( x ) , we get f ( 0 ) = a ( 0 + 6 5 ) 2 ( 0 − 4 1 ) = a ( 6 5 ) 2 ( − 4 1 ) = 75.
Solving for the Constant a Now we solve for a :
a ( 36 25 ) ( − 4 1 ) = 75 a = 75 ⋅ − 25 36 ⋅ 4 = 75 ⋅ − 25 144 = 3 ⋅ 1 144 ⋅ ( − 1 ) = − 432.
Substituting a into the Polynomial Substituting the value of a back into the general form of the polynomial, we get f ( x ) = − 432 ( x + 6 5 ) 2 ( x − 4 1 ) .
Expanding the Polynomial Expanding the polynomial, we have \begin{align*} f(x) &= -432\left(x^2 + \frac{10}{6}x + \frac{25}{36}\right)\left(x - \frac{1}{4}\right) &= -432\left(x^2 + \frac{5}{3}x + \frac{25}{36}\right)\left(x - \frac{1}{4}\right) &= -432\left(x^3 - \frac{1}{4}x^2 + \frac{5}{3}x^2 - \frac{5}{12}x + \frac{25}{36}x - \frac{25}{144}\right) &= -432\left(x^3 + \left(-\frac{1}{4} + \frac{5}{3}\right)x^2 + \left(-\frac{5}{12} + \frac{25}{36}\right)x - \frac{25}{144}\right) &= -432\left(x^3 + \left(-\frac{3}{12} + \frac{20}{12}\right)x^2 + \left(-\frac{15}{36} + \frac{25}{36}\right)x - \frac{25}{144}\right) &= -432\left(x^3 + \frac{17}{12}x^2 + \frac{10}{36}x - \frac{25}{144}\right) &= -432\left(x^3 + \frac{17}{12}x^2 + \frac{5}{18}x - \frac{25}{144}\right) &= -432x^3 - 432\left(\frac{17}{12}\right)x^2 - 432\left(\frac{5}{18}\right)x + 432\left(\frac{25}{144}\right) &= -432x^3 - 36 \cdot 17 x^2 - 24 \cdot 5 x + 3 \cdot 25 &= -432x^3 - 612x^2 - 120x + 75. \end{align*}
Final Polynomial Therefore, the polynomial is f ( x ) = − 432 x 3 − 612 x 2 − 120 x + 75.
Examples
Polynomials are used to model curves and shapes in various fields, such as engineering, computer graphics, and economics. For example, engineers might use a polynomial to model the trajectory of a projectile, while economists might use a polynomial to model the growth of a company's revenue. Understanding how to construct polynomials with specific properties, like given zeros and function values, is crucial for creating accurate and useful models in these fields. This skill allows professionals to make predictions and informed decisions based on mathematical representations of real-world phenomena.
