Question

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