Long Mccoy
03/03/2023 · Junior High School
6. \( 6 x^{6}+25 x^{5}+18 x^{4}+18 x^{2}-25 x+6=0 \)
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Para resolver la ecuación dada \(6x^{6}+25x^{5}+18x^{4}+18x^{2}-25x+6=0\), primero observamos que la ecuación es una ecuación polinómica de grado 6.
Para resolver esta ecuación, podemos intentar factorizarla. Si la ecuación se factoriza en factores lineales, podemos resolverla fácilmente.
Vamos a intentar factorizar la ecuación dada.
Factor the expression by following steps:
- step0: Factor:
\(6x^{6}+25x^{5}+18x^{4}+18x^{2}-25x+6\)
- step1: Evaluate:
\(6x^{6}+18x^{4}+25x^{5}+18x^{2}-25x+6\)
- step2: Rewrite the expression:
\(6x^{6}-8x^{4}+13x^{5}+16x^{3}-14x^{2}+3x+12x^{5}-16x^{3}+26x^{4}+32x^{2}-28x+6\)
- step3: Factor the expression:
\(x\left(6x^{5}-8x^{3}+13x^{4}+16x^{2}-14x+3\right)+2\left(6x^{5}-8x^{3}+13x^{4}+16x^{2}-14x+3\right)\)
- step4: Factor the expression:
\(\left(x+2\right)\left(6x^{5}-8x^{3}+13x^{4}+16x^{2}-14x+3\right)\)
- step5: Factor the expression:
\(\left(x+2\right)\left(x+3\right)\left(6x^{4}+7x^{2}-5x^{3}-5x+1\right)\)
- step6: Factor the expression:
\(\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x^{3}+3x-x^{2}-1\right)\)
- step7: Factor the expression:
\(\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)\left(x^{2}+1\right)\)
La ecuación dada se factoriza en:
\((x+2)(x+3)(2x-1)(3x-1)(x^{2}+1) = 0\)
Ahora, podemos resolver la ecuación encontrando los valores de \(x\) que hacen que cada factor sea igual a cero.
Los factores son:
1. \(x+2 = 0\) => \(x = -2\)
2. \(x+3 = 0\) => \(x = -3\)
3. \(2x-1 = 0\) => \(x = \frac{1}{2}\)
4. \(3x-1 = 0\) => \(x = \frac{1}{3}\)
5. \(x^{2}+1 = 0\) => No tiene soluciones reales
Por lo tanto, las soluciones de la ecuación dada son \(x = -2, -3, \frac{1}{2}, \frac{1}{3}\).
Quick Answer
Las soluciones de la ecuación son \(x = -2, -3, \frac{1}{2}, \frac{1}{3}\).
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