Question

Find the value of a using long division method: $\sqrt{x^4+2x^3+5x^2+ax+4}$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

$÷$	$x^2+x+2$ <= Quotient
$x^2$	$x^4+2x^3+5x^2+ax+4$ $x^4$
$2x^2+x$	$2x^3+5x^2$ $2x^3+x^2$
$2x^2+2x+2$	$4x^2+ax+4$ $4x^2+4x+4$
Remainder $\Rightarrow 0$

Since the given polynomial is a perfect square, we must have $a=4$.