(a) \( x^{3}+4 x^{2}+x-6 \)

1. We start with the cubic polynomial: \[ x^{3} + 4x^{2} + x - 6. \] 2. We look for possible rational roots using the Rational Root Theorem. The possible roots are factors of the constant term (\(-6\)) divided by factors of the leading coefficient (\(1\)). Hence, the candidates are: \[ \pm 1,\ \pm 2,\ \pm 3,\ \pm 6. \] 3. Test \( x = 1 \): \[ 1^{3} + 4(1)^{2} + 1 - 6 = 1 + 4 + 1 - 6 = 0. \] Since the result is zero, \( x = 1 \) is a root. 4. Since \( x = 1 \) is a root, we can factor the polynomial by dividing by \( (x - 1) \). 5. Use synthetic division to divide \( x^{3}+4x^{2}+x-6 \) by \( (x-1) \): - Coefficients: \( 1,\ 4,\ 1,\ -6 \). - Bring down the \( 1 \). - Multiply: \( 1 \times 1 = 1 \), add to \( 4 \) to get \( 5 \). - Multiply: \( 5 \times 1 = 5 \), add to \( 1 \) to get \( 6 \). - Multiply: \( 6 \times 1 = 6 \), add to \(-6\) to get \( 0 \). The quotient is: \[ x^{2} + 5x + 6. \] 6. Now, factor the quadratic: \[ x^{2} + 5x + 6. \] Identify two numbers that multiply to \( 6 \) and add to \( 5 \) (these numbers are \( 2 \) and \( 3 \)). Thus: \[ x^{2} + 5x + 6 = (x + 2)(x + 3). \] 7. Combine the factors, the complete factorization is: \[ x^{3} + 4x^{2} + x - 6 = (x - 1)(x + 2)(x + 3). \]