<|im_start|>user
How do I solve the polynomial equation \( x^4 - 5x^3 + 6x^2 = 0 \)?<|im_end|>
<|im_start|>assistant
To solve the polynomial equation \( x^4 - 5x^3 + 6x^2 = 0 \), we first factor out the common term \( x^2 \), which gives us \( x^2(x^2 - 5x + 6) = 0 \). This allows us to find one root immediately: \( x^2 = 0 \) gives us \( x = 0 \) with multiplicity 2. Next, we can solve the quadratic \( x^2 - 5x + 6 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -5, c = 6 \). This gives us \( x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} \). Thus, the roots are \( x = 3 \) and \( x = 2 \). Therefore, the complete set of roots for the equation is \( x = 0, 0, 2, 3 \).<|im_end|>

<|im_start|>user
How do I solve the polynomial equation \( x^4 - 5x^3 + 6x^2 = 0 \)?<|im_end|>
<|im_start|>assistant
To solve \( x^4 - 5x^3 + 6x^2 = 0 \), I will use the quadratic formula directly on the polynomial. The coefficients are \( a = 1, b = -5, c = 6 \). Plugging these into the formula, I get \( x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2} \). This leads to some complicated calculations, and I end up with roots that don't seem to fit the original polynomial. Thus, I cannot find the correct roots.<|im_end|>

How do I solve the polynomial equation \( x^4 - 5x^3 + 6x^2 = 0 \)?