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1. \( x^{3}-3 x^{2}+2 x=0 \) 2. \( 3 x^{3}-2 x^{2}-8 x=0 \) 3. \( 6 x\left(x^{2}-36\right)=0 \)

Ask by Jimenez Clark. in South Africa
Feb 20,2025

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The solutions are: 1. \( x=0, 1, 2 \) 2. \( x=-\frac{4}{3}, 0, 2 \) 3. \( x=-6, 0, 6 \)

Solución

Solve the equation \( x^{3}-3x^{2}+2x=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}-3x^{2}+2x=0\) - step1: Factor the expression: \(x\left(x-2\right)\left(x-1\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x=0\\&x-2=0\\&x-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=0\\&x=2\\&x=1\end{align}\) - step4: Rewrite: \(x_{1}=0,x_{2}=1,x_{3}=2\) Solve the equation \( 3x^{3}-2x^{2}-8x=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(3x^{3}-2x^{2}-8x=0\) - step1: Factor the expression: \(x\left(x-2\right)\left(3x+4\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x=0\\&x-2=0\\&3x+4=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=0\\&x=2\\&x=-\frac{4}{3}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{4}{3},x_{2}=0,x_{3}=2\) Solve the equation \( 6x(x^{2}-36)=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(6x\left(x^{2}-36\right)=0\) - step1: Elimination the left coefficient: \(x\left(x^{2}-36\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x=0\\&x^{2}-36=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=0\\&x=6\\&x=-6\end{align}\) - step4: Rewrite: \(x_{1}=-6,x_{2}=0,x_{3}=6\) The solutions to the given equations are: 1. \( x^{3}-3x^{2}+2x=0 \) has solutions \( x_{1}=0, x_{2}=1, x_{3}=2 \). 2. \( 3x^{3}-2x^{2}-8x=0 \) has solutions \( x_{1}=-\frac{4}{3}, x_{2}=0, x_{3}=2 \). 3. \( 6x(x^{2}-36)=0 \) has solutions \( x_{1}=-6, x_{2}=0, x_{3}=6 \).

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Bonus Knowledge

Let's consider the first equation, \( x^{3}-3 x^{2}+2 x=0 \). To solve it, we can factor out an \( x \), giving us \( x(x^{2}-3x+2)=0 \). This further factors to \( x(x-1)(x-2)=0 \), leading to the solutions \( x=0, 1, 2 \). Now for the second equation, \( 3 x^{3}-2 x^{2}-8 x=0 \), we can factor out \( x \) as well, leaving us with \( x(3x^{2}-2x-8)=0 \). We now solve \( 3x^{2}-2x-8=0 \) using the quadratic formula \( x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \). Plugging in \( a=3, b=-2, c=-8 \), we find two additional solutions to the equation along with \( x=0 \). For the third equation, \( 6 x\left(x^{2}-36\right)=0 \), we can set each factor to zero: either \( 6x=0 \) (yielding \( x=0 \)) or \( x^{2}-36=0 \), which leads us to \( x=6 \) and \( x=-6 \). In total, you would find the roots for each of these equations to be \( x=0, 1, 2, 6, -6 \). Insights like these can help illuminate patterns in polynomial behavior and solutions!

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