Question

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

• A. $-25$
• B. $0$
• C. $10$
• D. $24$

Answer 1

Answer: (C)

$10$

If $1,2,3,4$ are the roots of given equation then

$⟹(x-1)(x-2)(x-3)(x-4)=x^4+a{x}^3+b{x}^2+cx+d$

$⟹(x^2-3x+2)(x^2-7x+12)=x^4+a{x}^3+b{x}^2+cx+d$

$⟹x^4-10x^3+35x^2-50x+24=x^4+a{x}^3+b{x}^2+cx+d$

Comparing the coefficients, we get

$a=-10,b=35,c=-50,d=24$

Then, $a+2b+c=-10+2\times 35-50=10$