\( q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} \), so \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)

February 22, 2026

$q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} \), so \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892$

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\( -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 \), huge positive
\( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)
But \( p \) is large negative denominator: \( p = -\frac{407}{3} \Rightarrow p^3 = -\frac{67432643}{27} \), so \( -4p^3 = -4 \cdot (-\frac{67432643}{27}) = \frac{269730372}{27} \approx 9996788 \)
Thus \( \Delta = 9996788 - 74892 = 9920896 > 0 \), so **three distinct real roots**!
So all three roots \( w_1, w_2, w_3 \) are real. But earlier evaluations showed only one sign change. Contradiction?
Check \( f(0) = -18 \), \( f(1) = 1 - 8 + 9 - 18 = -16 \), \( f(2) = 8 - 32 + 18 - 18 = -24 \), \( f(3) = 27 - 72 + 27 - 18 = -36 \), \( f(4) = 64 - 128 + 36 - 18 = -46 \), \( f(5) = 125 - 200 + 45 - 18 = -48 \), \( f(6) = 216 - 288 + 54 - 18 = -36 \), \( f(6.

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