Question

If $x+1$ is a factor of $a{x}^4+b{x}^3+c{x}^2+dx+e=0$, then

• A. $a+c+e=b+d$
• B. $a+b=c+d$
• C. $a+b+c+d+e=0$
• D. $a+c+b=d+e$

Answer 1

Answer: (A)

$a+c+e=b+d$

Since $x+1$ is a solution of the given equation, $f(-1)$ has to be zero.
$\therefore$ $a\times (-1{)}^4+b\times (-1{)}^3+c\times (-1{)}^2+d\times (-1{)}^1+e=0$
$\therefore a-b+c-d+e=0.$
$\therefore a+c+e=b+d$