Pregunta
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Write a polynomial \( f(x) \) that satisfies the given conditions. Polynomial of lowest degree with zeros of \( \frac{4}{3} \) (multiplicity 2 ) and \( -\frac{1}{2} \) (multiplicity 1 ) and with \( f(0)=-16 \).

Ask by Osborne Goodwin. in the United States
Mar 15,2025

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The polynomial is \( f(x) = -18 \left(x - \frac{4}{3}\right)^2 \left(x + \frac{1}{2}\right) \).

Solución

We begin by writing the polynomial in factored form based on its zeros. Since the polynomial has a zero at \( x=\frac{4}{3} \) of multiplicity 2 and a zero at \( x=-\frac{1}{2} \) of multiplicity 1, we can write \[ f(x) = a \left(x - \frac{4}{3}\right)^2 \left(x + \frac{1}{2}\right), \] where \( a \) is a constant to be determined. To find \( a \), we use the condition \( f(0) = -16 \). Substitute \( x = 0 \) into the polynomial: \[ f(0) = a \left(0 - \frac{4}{3}\right)^2 \left(0 + \frac{1}{2}\right). \] Simplify each factor: \[ 0 - \frac{4}{3} = -\frac{4}{3}, \quad \left(-\frac{4}{3}\right)^2 = \frac{16}{9}, \quad 0 + \frac{1}{2} = \frac{1}{2}. \] Thus, we have: \[ f(0) = a \cdot \frac{16}{9} \cdot \frac{1}{2} = a \cdot \frac{16}{18} = a \cdot \frac{8}{9}. \] Setting \( f(0) = -16 \): \[ a \cdot \frac{8}{9} = -16. \] Solve for \( a \) by multiplying both sides by \( \frac{9}{8} \): \[ a = -16 \cdot \frac{9}{8} = -18. \] Thus, the polynomial is \[ f(x) = -18 \left(x - \frac{4}{3}\right)^2 \left(x + \frac{1}{2}\right). \]

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Bonus Knowledge

To construct the polynomial \( f(x) \), we start with its zeros and their multiplicities. The zeros given are \( \frac{4}{3} \) with multiplicity 2, and \( -\frac{1}{2} \) with multiplicity 1. Therefore, we can express the polynomial in factored form as: \[ f(x) = k \left( x - \frac{4}{3} \right)^2 \left( x + \frac{1}{2} \right) \] where \( k \) is a constant that we need to determine. Now, let's simplify the factors. Rewrite \( -\frac{4}{3} \) and \( \frac{1}{2} \) to eliminate the fractions. The polynomial becomes: \[ f(x) = k \left( 3x - 4 \right)^2 \left( 2x + 1 \right) \] Next, we need to find the value of \( k \) such that \( f(0) = -16 \). Substituting \( x = 0 \): \[ f(0) = k \left( 3(0) - 4 \right)^2 \left( 2(0) + 1 \right) = k (4)^2 (1) = 16k \] Setting this equal to \(-16\): \[ 16k = -16 \] Dividing both sides by 16 gives: \[ k = -1 \] Now we substitute \( k \) back into our polynomial: \[ f(x) = -1 \left( 3x - 4 \right)^2 \left( 2x + 1 \right) \] Now let's expand this polynomial: 1. Expand \( (3x - 4)^2 \): \[ (3x - 4)(3x - 4) = 9x^2 - 24x + 16 \] 2. Multiply by \( (2x + 1) \): \[ (9x^2 - 24x + 16)(2x + 1) = 18x^3 - 24x^2 + 9x^2 - 48x + 32 + 16 \] This yields: \[ 18x^3 - 24x^2 - 48x + 32 \] Now multiplying the expanded result by -1 gives: \[ f(x) = -18x^3 + 24x^2 + 48x - 32 \] Thus, the polynomial of lowest degree that satisfies the given conditions is: \[ \boxed{-18x^3 + 24x^2 + 48x - 32} \]

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