Question

Consider the polynomial $p(x)=(x-1)(x+2)(x-3)(x-6)-100$. The equation $p(x)=0$ has

• A. all four distinct real roots
• B. three distinct real roots
• C. two real and two imaginary roots
• D. all four imaginary roots

Answer 1

Answer: (C)

$p(x)=(x-1)(x+2)(x-3)(x-6)-100$
$=\left(x^2-4x+3\right)\left(x^2-4x-12\right)-100$
$p(x)={\left(x^2-4x+3\right)}^2-15\left(x^2-4x+3\right)-100$
$\Rightarrow x^2-4x+3=20\text{ and }{x}^2-4x+3=-5$
$\Rightarrow p(x)=0\text{ has two distinct real roots. }$