2. Given that \( (\mathrm{x}+1) \) and \( (\mathrm{x}-2) \) are factors of \( x^{4}-3 \mathrm{x}^{3}+a \mathrm{x}^{2}+\mathrm{bx}+4 \) find the values of \( a \) and \( b \).

To find the values of \( a \) and \( b \) given that \( (\mathrm{x}+1) \) and \( (\mathrm{x}-2) \) are factors of \( x^{4}-3 \mathrm{x}^{3}+a \mathrm{x}^{2}+\mathrm{bx}+4 \), we can use the fact that if \( x + 1 \) and \( x - 2 \) are factors, then the polynomial should equal zero for \( x = -1 \) and \( x = 2 \). First, substitute \( x = -1 \): \[ (-1)^{4} - 3(-1)^{3} + a(-1)^{2} + b(-1) + 4 = 0 \] \[ 1 + 3 + a - b + 4 = 0 \implies a - b + 8 = 0 \implies a - b = -8 \quad \text{(Equation 1)} \] Next, substitute \( x = 2 \): \[ (2)^{4} - 3(2)^{3} + a(2)^{2} + b(2) + 4 = 0 \] \[ 16 - 24 + 4a + 2b + 4 = 0 \implies 4a + 2b - 4 = 0 \implies 4a + 2b = 4 \quad \text{(Equation 2)} \] Now we have a system of equations: 1. \( a - b = -8 \) 2. \( 4a + 2b = 4 \) From Equation 1, we can express \( a \) in terms of \( b \): \[ a = b - 8 \] Now, substitute \( a \) in Equation 2: \[ 4(b - 8) + 2b = 4 \] \[ 4b - 32 + 2b = 4 \implies 6b - 32 = 4 \implies 6b = 36 \implies b = 6 \] Substituting \( b = 6 \) back into the expression for \( a \): \[ a = 6 - 8 = -2 \] Thus, the values are: \[ \boxed{a = -2} \quad \text{and} \quad \boxed{b = 6} \]