Solve x³ - 6x² + 11x - 6 = 0 with Steps

Write the equation in standard cubic form $ax^3 + bx^2 + cx + d = 0$.

Apply the Rational Root Theorem (a way to test for simple fraction/whole-number roots before trying harder methods). Testing $x = 1$, we find $f(1) = 0$, which proves that $(x - 1)$ is a factor by the Factor Theorem.

Perform polynomial division (long division / synthetic division) of $x^3 - 6x^2 + 11x - 6$ by the factor $(x - 1)$ to obtain the reduced quotient $x^2 - 5x + 6$. Both Long Division (NCERT standard) and Synthetic Division (shortcut) are shown below:

Express the cubic equation in factored form and isolate the quadratic factor.

Calculate the discriminant $\Delta = b^2 - 4ac$ (tells us how many real roots exist) for $x^2 - 5x + 6 = 0$. Since $\Delta > 0$, there are two distinct real roots.

Apply the quadratic formula $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

Compute the two roots from the quadratic factor:

Final summary: All three real roots of the cubic equation are:

Combine into completely factored form and apply the Zero Factor Principle (if $a \cdot b = 0$, then $a = 0$ or $b = 0$).

Structural Insight (Vieta's Relations for Cubics): Sum of roots = -b/a, Product of roots = -d/a.