Question

If ax3 + bx2+ cx + d is exactly divisible by (x + 1) and (x + 2), then which of the following is true
(A) 3a – 3b + d = 0
(B) 8a – b + 2d = 0
(C) 5a – 2b + 3d = 0
(D) 6a – 2b + d = 0

Answer 1

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

hence, option (d) is correct.

Answer 2

Answer:

(d) 6a-2b+d=0

Step-by-step explanation:

Factor theorem:

If (x+a) is a factor of f(x), then f(-a)=0

$f(x)=ax^3+bx^2+cx+d$

since (x+1) is a factor of f(x),

f(-1)=0

a(-1)+b(1)+c(-1)+d=0

-a+b-c+d=0.....(1)

since (x+2) is a factor of f(x),

f(-2)=0

a(-8)+b(4)+c(-2)+d=0

           -8a+4b-2c+d=0......(2)

(1)*2=> -2a+2b-2c+2d=0

subtracting we get,

=> -6a+2b-d=0

=> 6a-2b+d=0