Question

Find the value of a and b so that $x^4+6x^3+9x^2+ax+b$ is exactly divisible by $x^2+3$

• A. $a=-9$ and $b=0$
• B. $a=8$ and $b=0$
• C. $a=0$ and $b=9$
• D. $a=9$ and $b=0$

Answer 1

The correct option is D $a=9$ and $b=0$

Given that $x^2+3$ divides exactly the polynomial $x^3+6x^3+9x^2+ax+b$

$\Rightarrow x^2+3$ is a factor of $x^3+6x^3+9x^2+ax+b$

$\Rightarrow$ by factor theorem, the Remainder$=0$

$\Rightarrow (a-9)x+b=0$

$\Rightarrow a-9=0$ and $b=0$

$\therefore a=9,b=0$