Question

If $1,2,3$ and $4$ are the roots of the equation $x^4+a{x}^3+b{x}^2+cx+d=0$, then $a+2b+c$ is equal to

• A. $-25$
• B. 0
• C. 10
• D. 24

Answer 1

Answer: (C)

If $1,2,3,4$ are the roots of the equation
$x^4+a{x}^3+b{x}^2+cx+d=0,$ then
$(x-1)(x-2)(x-3)(x-4)$
$=x^4+a{x}^3+b{x}^2+cx+d$
$\Rightarrow \left(x^2-3x+2\right)\left(x^2-7x+12\right)$
$=x^4+a{x}^3+b{x}^2+cx+d$
$\Rightarrow x^4-10x^3+35x^2-50x+24$
$=x^4+a{x}^3+b{x}^2+cx+d$
$\Rightarrow a=-10,b=35,c=-50,d=24$
Now, $a+2b+c=-10+2\times 35-50$
$=10$