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\begin{tabular}{|l|}\begin{tabular}{|l} 7. Please calculate step by step \\ the remainder that results from the \\ quotient expressed below using \\ the Polynomial Remainder \\ Theorem. \\ \( \left(-5 x^{3}+x^{2}-4 x\right) \div(x-3) \) \\ Please be explicit in writing each \\ step and number each step. Also, \\ please write a brief explanation \\ with each step to explain what you \\ are doing. \\ Please conclude with a statement \\ that explains whether \( (x-3) \) is a \\ factor of the stated polynomial. \end{tabular} \\ \hline\end{tabular}

Ask by Welch Smith. in the United States
Jan 15,2025

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Tutor-Verified Answer

Answer

El residuo es -138, por lo que \( (x - 3) \) no es un factor del polinomio.

Solution

Claro, a continuación se muestra el cálculo paso a paso del residuo al dividir el polinomio \( -5x^{3} + x^{2} - 4x \) entre \( (x - 3) \) utilizando el Teorema del Residuo. Cada paso está numerado y explicado detalladamente. ### Paso 1: Identificar el polinomio y el divisor **Explicación:** Comenzamos identificando el polinomio que vamos a dividir y el divisor. - **Polinomio (f(x))**: \( -5x^{3} + x^{2} - 4x \) - **Divisor**: \( (x - 3) \) ### Paso 2: Aplicar el Teorema del Residuo **Explicación:** El Teorema del Residuo nos indica que el residuo de la división de \( f(x) \) entre \( (x - a) \) es igual a \( f(a) \). En este caso, \( a = 3 \). ### Paso 3: Sustituir \( x = 3 \) en el polinomio **Explicación:** Calculamos el valor de \( f(3) \) sustituyendo \( x \) por 3 en el polinomio dado. \[ f(3) = -5(3)^{3} + (3)^{2} - 4(3) \] ### Paso 4: Calcular las potencias **Explicación:** Resolvemos las potencias de 3. \[ 3^{3} = 27 \quad \text{y} \quad 3^{2} = 9 \] ### Paso 5: Sustituir las potencias calculadas **Explicación:** Reemplazamos los valores obtenidos en la expresión de \( f(3) \). \[ f(3) = -5(27) + 9 - 12 \] ### Paso 6: Realizar las multiplicaciones **Explicación:** Multiplicamos los coeficientes por las potencias de 3. \[ -5(27) = -135 \quad \text{y} \quad -4(3) = -12 \] ### Paso 7: Sumar los términos **Explicación:** Sumamos todos los términos resultantes. \[ f(3) = -135 + 9 - 12 \] \[ f(3) = -138 \] ### Paso 8: Determinar el residuo **Explicación:** El valor obtenido \( f(3) = -138 \) es el residuo de la división. ### Conclusión **Explicación:** Dado que el residuo es \(-138\) y no igual a cero, \( (x - 3) \) **no** es un factor del polinomio \( -5x^{3} + x^{2} - 4x \).

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To find the remainder of the polynomial \( -5x^{3} + x^{2} - 4x \) when divided by \( (x-3) \) using the Polynomial Remainder Theorem, follow these steps: **Step 1:** Identify the polynomial and divisor. We start with the polynomial \( P(x) = -5x^{3} + x^{2} - 4x \) and the divisor \( x - 3 \). We need to evaluate \( P(3) \) to find the remainder. **Step 2:** Substitute \( x = 3 \) into the polynomial. Compute \( P(3) \): \[ P(3) = -5(3)^{3} + (3)^{2} - 4(3) \] Calculate \( (3)^{3} = 27 \), \( (3)^{2} = 9 \), and \( -4(3) = -12 \). **Step 3:** Substitute and calculate step by step. Now plug in these values: \[ P(3) = -5(27) + 9 - 12 \] Calculate \( -5(27) = -135 \): \[ P(3) = -135 + 9 - 12 \] **Step 4:** Perform the final addition. Now, compute \( -135 + 9 = -126 \) and then \( -126 - 12 = -138 \): \[ P(3) = -138 \] **Conclusion:** The remainder when \( -5x^{3} + x^{2} - 4x \) is divided by \( x - 3 \) is \( -138 \). Since the remainder is not zero, \( (x - 3) \) is not a factor of the polynomial \( -5x^{3} + x^{2} - 4x \). --- In history, the Polynomial Remainder Theorem dates back to ancient Greece, with mathematicians like Euclid exploring the properties of polynomial equations. This theorem has evolved over time, shaping modern algebra and calculus. Understanding its root helps anchor our knowledge in mathematical history, showcasing how ancient concepts continue to play a vital role in today's problem-solving techniques! In real-world applications, the Polynomial Remainder Theorem simplifies complex problems, like finding roots in engineering or computer science. When dealing with polynomial models, engineers can predict system behaviors by determining specific outcomes (like the time to failure of a system) without performing long division, making their analyses much more efficient and effective!

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