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: (B)

$P(x)=(x-1)(x+2)(x-3)(x-6)-100$ $=\left(x^2-4x+3\right)\left(\underline{x^2-4x}-12\right)-100$ $={\left(x^2-4x\right)}^2-9\left(x^2-4x\right)-136$ $P(x)=\left(x^2-4x-17\right)\left(x^2-4x+8\right)$ $P(x)=0$ $\therefore \text{ equation has two distinct real and two imaginary roots. }$