Question

If ${ax}^3+{bx}^2+cx+d$ 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

The correct option is D $6a-2b+d=0$

Since $a{x}^3+b{x}^2+cx+d$ is exactly divisible by $(x+1)$ and $(x+2)$

$\therefore P(-1)=P(-2)=0$

$\therefore a(-1{)}^3+b(-1{)}^2+c(-1)+d=0$

$\therefore -a+b-c+d=0$

$b+d=a+c$........... $\left(1\right)$

$\therefore -8a+4b-2c+d=0$

$\therefore 4b+d=8a+2c$........... $\left(2\right)$

Eliminating $\left(c\right)$ from $\left(1\right)$ & $\left(2\right)$

$\therefore 2b+2d=2a+2c$

$4b+d=8a+2c$

$-2b+d=-6a$

$\therefore 6a-2b+d=0$