Question

pls tell me the answer​

Answer 1

Answer:

${x}^{3} - 4 {x}^{2} + ax + b \\ x - 2 \: and \: x + 1 \: are \: the \: factor \\ x - 2 = 0 \\ x = 2 \\ {2}^{3} - 4( {2}^{2} ) + a \times 2 + b = 0 \\ 8 - 16 + 2a + b = 0 \\ 2a + b = 8.......(1) \\ x + 1 = 0 \\ x = - 1 \\ {( - 1)}^{3} - 4( { - 1})^{2} + a( - 1) + b = 0\\ - 1 - 4 - a + b = 0 \\ - 5 - a + b = 0 \\ b = 5 + a \\ put \: b \: in \: (1) \\ 2a + 5 + a = 8 \\ 3a = 8 - 5 \\ 3a = 3 \\ a = 1 \\ put \: a \: in \: (1) \\ 2(1) + b = 8 \\ b = 8 - 2 \\ b = 6$