Solve tor \( x \) in each of the following expressions. \( 3 x^{3}-7 x^{2}+4=0 \) \( x-1 \) \( x^{3}-x^{2}-22 x+40=0 \)

Solve the equation \( 3x^{3}-7x^{2}+4=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(3x^{3}-7x^{2}+4=0\) - step1: Factor the expression: \(\left(x-2\right)\left(x-1\right)\left(3x+2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-2=0\\&x-1=0\\&3x+2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=2\\&x=1\\&x=-\frac{2}{3}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{2}{3},x_{2}=1,x_{3}=2\) Solve the equation \( x-1=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x-1=0\) - step1: Move the constant to the right side: \(x=0+1\) - step2: Remove 0: \(x=1\) Solve the equation \( x^{3}-x^{2}-22x+40=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}-x^{2}-22x+40=0\) - step1: Factor the expression: \(\left(x-4\right)\left(x-2\right)\left(x+5\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-4=0\\&x-2=0\\&x+5=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=4\\&x=2\\&x=-5\end{align}\) - step4: Rewrite: \(x_{1}=-5,x_{2}=2,x_{3}=4\) The solutions to the given equations are: 1. \( 3x^{3}-7x^{2}+4=0 \) has solutions \( x=-\frac{2}{3}, x=1, x=2 \). 2. \( x-1=0 \) has solution \( x=1 \). 3. \( x^{3}-x^{2}-22x+40=0 \) has solutions \( x=-5, x=2, x=4 \).