Question

The polynomial $p\left(x\right)=2x^4-x^3-7x^2+ax+b$ is divisible by $x^2-2x-3$ for certain values of $a$ and $b$. The value of $(a+b)$ is:

• A. $-34$
• B. $-30$
• C. $-26$
• D. $-18$

Answer 1

The correct option is A $-34$

Consider $x^2-2x-3=0$
$\Rightarrow x^2-3x+x-3=0$
$\Rightarrow x(x-3)+1(x-3)=0$
$\Rightarrow (x+1)(x-3)=0$
$\Rightarrow x=-1,3$
Since $x^2-2x-3$ divides $p\left(x\right)=2x^4-x^3-7x^2+ax+b$
$\therefore p(-1)=0,p\left(3\right)=0$ $\text{By Factor theorem}$
$p(-1)=0\Rightarrow 2(-1{)}^4-(-1{)}^3-7(-1{)}^2+a(-1)+b=0$
$\Rightarrow 2+1-7-a+b=0$
$\Rightarrow a-b=-4$ $...\left(i\right)$
$Again,p\left(3\right)=0\Rightarrow 2(3{)}^4-(3{)}^3-7(3{)}^2+a(3)+b=0$
$\Rightarrow 162-27-63+3a+b=0$
$\Rightarrow 3a+b=-72$ $...\left(ii\right)$
$\text{On adding (i) and (ii), we get}$
$4a=-76\Rightarrow a=-19$
$\text{From (i)}$
$b=a+4\Rightarrow b=-15$
$\therefore a+b=-19-15=-34$
Option A is correct.