When | x | < 1 2 , the coefficient of x 4 in the expansion of 3 x 2 - 5 x + 3 ( x - 1 ) ( 2 x + 1 ) ( x + 3 ) is

Step-by-step solution

Rewrite the given expression F = - x 2 - 5 3 x + 1 ( 1 - x ) - 1 ( 1 + 2 x ) - 1 1 + x 3 - 1 = - x 2 - 5 3 x + 1 1 + x + x 2 + x 3 + x 4 1 - 2 x + 4 x 2 - 8 x 3 + 16 x 4 1 - x 3 + x 2 9 - x 3 27 + x 4 81 On neglecting higher degree terms = - 1 + x + x 2 + x 3 + x 4 - 5 3 x + 5 3 x 2 + 5 3 x 3 + 5 3 x 4 + x 2 + x 3 + x 4 Simplify the above expression. = - 1 - 2 3 x + x 2 3 + x 3 3 + x 4 3 1 - 7 x 3 + 43 9 x 2 - 259 27 x 3 + 1555 81 x 4 = - 2166 81 = - 722 27