Question

if the polynomial p(x)=(x^3 + ax^2 + bx+ 6) has (x-2) as a factor and leaves a remainder 3 when divided by (x-3) then find the values of a and b. ​

Answer 1

$p(x) = {x}^{3} + a {x}^{2} + bx + 6 \\ putting \: 2 \: in \: place \: of \: x \: \\ p(2) = {2}^{3} + a( {2}^{2} ) + b(2) + 6 \\ \: \: \: \: \: \: \: \: \: = > 8 + 4a + 2b + 6 = 0 \: (since \: x - 2 \: is \: a \: factor \: ) \\ \: \: \: \: \: \: \: \: \: = > 4a + 2b = - 14 \\ \: \: \: \: \: \: \: \: \: = > 2a + b = - 7 \: \: \: \: \: \: ....(i) \\ putting \: 3 \: in \: place \: of \: x \\ p(3) = {3}^{3} + a( {3}^{2}) + b(3) + 6 \\ \: \: \: \: \: \: \: \: \: = > 27 + 9a + 3b + 6 = -3 \: (since \: x - 3 \: leaves \: a \: remainder \: of \: 3) \\ \: \: \: \: \: \: \: \: \: = > 9a + 3b = - 18 \\ \: \: \: \: \: \: \: \: \: = > 3a + b = - 6 \: \: \: \: \: \: \: \: .....(ii) \\ solving \: (i) \: and \: (ii) \\ a = 1 \: and \: b = - 9$