factors of 2x^3 - 4x^2 +x - 3
You asked:
Factor the expression: \(\displaystyle 2 \cdot {x}^{3} - 4 \cdot {x}^{2} + x - 3\)
MathBot Answer:
Factorization over the real numbers:
\[4 \cdot \left(0.5 x^{3} - x^{2} + 0.25 x - 0.75\right)\]Factorization over the complex numbers:
\[2 \left(x - \left(\frac{2}{3} + \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{5 \sqrt{39}}{36} + \frac{95}{108}} + \frac{5}{18 \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{5 \sqrt{39}}{36} + \frac{95}{108}}}\right)\right) \left(x - \left(\frac{2}{3} + \frac{5}{18 \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{5 \sqrt{39}}{36} + \frac{95}{108}}} + \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{5 \sqrt{39}}{36} + \frac{95}{108}}\right)\right) \left(x - \left(\frac{5}{18 \sqrt[3]{\frac{5 \sqrt{39}}{36} + \frac{95}{108}}} + \frac{2}{3} + \sqrt[3]{\frac{5 \sqrt{39}}{36} + \frac{95}{108}}\right)\right)\]\(i\) is the imaginary unit, defined as \(i^2 = -1\).