Solve \( x^{3}-2 x^{2}-15 x+36=0 \) by factorising.

To factorize the polynomial \( x^{3}-2 x^{2}-15 x+36=0 \), we can use the Rational Root Theorem to test for possible rational roots. Testing \( x = 3 \): \[ 3^{3} - 2(3^{2}) - 15(3) + 36 = 27 - 18 - 45 + 36 = 0 \] So, \( x - 3 \) is a factor. Now, we can perform polynomial long division or synthetic division to divide \( x^{3}-2 x^{2}-15 x+36 \) by \( x - 3 \). The result is \( x^{2} + x - 12 \). Next, we factor \( x^{2} + x - 12 \): \[ x^{2} + x - 12 = (x - 3)(x + 4) \] Now we can write the original polynomial as: \[ (x - 3)^{2}(x + 4) = 0 \] Setting each factor to zero gives us the roots: 1. \( x - 3 = 0 \) → \( x = 3 \) (with multiplicity 2) 2. \( x + 4 = 0 \) → \( x = -4 \) Thus, the solutions to the equation \( x^{3}-2 x^{2}-15 x+36=0 \) are \( x = 3 \) (with multiplicity 2) and \( x = -4 \).