if ax3+bx3+cx+d is exactly divisible by (x+1) and (x+2)
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A/c to question, If ax3 + bx2+ cx + d is exactly divisible by (x + 1) and (x + 2)
so, x = -1 and -2 are the roots of given expression ax³ + bx² + cx + d .
put x = -1,
a(-1)³ + b(-1)² + c(-1) + d = 0
-a + b - c + d = 0...........(1)
put x = -2,
a(-2)³ + b(-2)² + c(-2) + d = 0
-8a + 4b - 2c + d = 0.........(2)
multiply , 2 with equation (1) and then subtracting from equation (2),
-2a + 2b - 2c + 2d - (-8a + 4b - 2c + d) = 0
-2a + 2b - 2c + 2d + 8a - 4b + 2c - d = 0
6a - 2b + d = 0
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