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(Valor \( 10 \% \) ) Considérense los polinomios \( P(x)=\frac{1}{2} x^{2}+\frac{3}{4} x-5 \) y \( Q(x)=\frac{2}{3} x^{2}-\frac{1}{3} x \). Encontrar \( P(x) \cdot Q(x) \)

Ask by Carlson Carter. in Colombia
Mar 08,2025

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El producto de los polinomios \( P(x) \) y \( Q(x) \) es: \[ P(x) \cdot Q(x) = \frac{1}{3}x^4 + \frac{1}{3}x^3 - \frac{43}{12}x^2 + \frac{5}{3}x \]

Solution

Multiplicaremos los polinomios \[ P(x)=\frac{1}{2}x^2+\frac{3}{4}x-5 \qquad \text{y} \qquad Q(x)=\frac{2}{3}x^2-\frac{1}{3}x \] Paso 1. Se escribe el producto: \[ P(x)\cdot Q(x)=\left(\frac{1}{2}x^2+\frac{3}{4}x-5\right)\left(\frac{2}{3}x^2-\frac{1}{3}x\right) \] Paso 2. Se distribuye término a término. \[ \begin{array}{rcl} \frac{1}{2}x^2\cdot\frac{2}{3}x^2 &=& \frac{1}{2}\cdot\frac{2}{3}\, x^4=\frac{1}{3}x^4,\\[1mm] \frac{1}{2}x^2\cdot\left(-\frac{1}{3}x\right) &=& -\frac{1}{2}\cdot\frac{1}{3}\, x^3=-\frac{1}{6}x^3,\\[1mm] \frac{3}{4}x\cdot\frac{2}{3}x^2 &=& \frac{3}{4}\cdot\frac{2}{3}\, x^3=\frac{1}{2}x^3,\\[1mm] \frac{3}{4}x\cdot\left(-\frac{1}{3}x\right) &=& -\frac{3}{4}\cdot\frac{1}{3}\, x^2=-\frac{1}{4}x^2,\\[1mm] -5\cdot\frac{2}{3}x^2 &=& -\frac{10}{3}x^2,\\[1mm] -5\cdot\left(-\frac{1}{3}x\right) &=& \frac{5}{3}x. \end{array} \] Paso 3. Se agrupan los términos semejantes según el grado. - Término de \(x^4\): \[ \frac{1}{3}x^4 \] - Términos de \(x^3\): \[ -\frac{1}{6}x^3+\frac{1}{2}x^3=\left(-\frac{1}{6}+\frac{1}{2}\right)x^3=\left(-\frac{1}{6}+\frac{3}{6}\right)x^3=\frac{2}{6}x^3=\frac{1}{3}x^3 \] - Términos de \(x^2\): \[ -\frac{1}{4}x^2-\frac{10}{3}x^2. \] Para combinarlos usamos el mínimo común denominador, que es \(12\): \[ -\frac{1}{4}x^2=-\frac{3}{12}x^2 \quad \text{y} \quad -\frac{10}{3}x^2=-\frac{40}{12}x^2. \] Sumando: \[ -\frac{3}{12}x^2-\frac{40}{12}x^2=-\frac{43}{12}x^2. \] - Término de \(x\): \[ \frac{5}{3}x. \] Paso 4. Se escribe el polinomio resultante: \[ P(x)\cdot Q(x)=\frac{1}{3}x^4+\frac{1}{3}x^3-\frac{43}{12}x^2+\frac{5}{3}x. \]

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Para encontrar el producto \( P(x) \cdot Q(x) \), multiplicamos cada término de \( P(x) \) por cada término de \( Q(x) \). \[ P(x) = \frac{1}{2} x^{2} + \frac{3}{4} x - 5 \] \[ Q(x) = \frac{2}{3} x^{2} - \frac{1}{3} x \] Multiplicamos \( P(x) \) por \( Q(x) \): 1. \( \frac{1}{2} x^{2} \cdot \frac{2}{3} x^{2} = \frac{1}{3} x^{4} \) 2. \( \frac{1}{2} x^{2} \cdot \left(-\frac{1}{3} x\right) = -\frac{1}{6} x^{3} \) 3. \( \frac{3}{4} x \cdot \frac{2}{3} x^{2} = \frac{1}{2} x^{3} \) 4. \( \frac{3}{4} x \cdot \left(-\frac{1}{3} x\right) = -\frac{1}{4} x^{2} \) 5. \( -5 \cdot \frac{2}{3} x^{2} = -\frac{10}{3} x^{2} \) 6. \( -5 \cdot \left(-\frac{1}{3} x\right) = \frac{5}{3} x \) Ahora sumamos todos los términos: \[ P(x) \cdot Q(x) = \frac{1}{3} x^{4} + \left(-\frac{1}{6} x^{3} + \frac{1}{2} x^{3}\right) + \left(-\frac{1}{4} x^{2} - \frac{10}{3} x^{2}\right) + \frac{5}{3} x \] Simplificando los coeficientes: - Para \( x^{3} \): \( -\frac{1}{6} + \frac{3}{6} = \frac{2}{6} = \frac{1}{3} \) - Para \( x^{2} \): \( -\frac{1}{4} - \frac{10}{3} = -\frac{3}{12} - \frac{40}{12} = -\frac{43}{12} \) Por lo tanto, \( P(x) \cdot Q(x) \) se expresa como: \[ P(x) \cdot Q(x) = \frac{1}{3} x^{4} + \frac{1}{3} x^{3} - \frac{43}{12} x^{2} + \frac{5}{3} x \]

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