Question:

Step 1: From $\lim_{x\to\infty}\frac{x^2 P(x)}{x^5+5x+6} = 2$, the degree of $x^2 P(x)$ must equal 5, so $\deg P(x) = 3$. Also, the leading coefficient: if $P(x) = ax^3 + \ldots$, then $\frac{x^2 \cdot ax^3}{x^5} = a = 2$. So $P(x) = 2x^3 + bx^2 + cx + d$. Step 2: Use the given values. $P(1) = 2+b+c+d = 2 \Rightarrow b+c+d = 0$. $P(2) = 16+4b+2c+d = 16 \Rightarrow 4b+2c+d = 0$. $P(3) = 54+9b+3c+d = 54 \Rightarrow 9b+3c+d = 0$. Step 3: Solve the system: $b+c+d=0$, $4b+2c+d=0$, $9b+3c+d=0$. Subtract first from second: $3b+c=0$. Subtract second from third: $5b+c=0$. Subtract: $2b=0 \Rightarrow b=0$, then $c=0$, then $d=0$. Step 4: So $P(x) = 2x^3$. Then $P(4) = 2(64) = 128$. Option (b) is correct. Step 5: Compute the area bounded by $y = P(x) = 2x^3$, $x=0$, $x=2$ and the $x$-axis. $\int_0^2 2x^3\,dx = 2\cdot\frac{x^4}{4}\Big|_0^2 = \frac{x^4}{2}\Big|_0^2 = \frac{16}{2} = 8$ sq. units. Option (d) is correct.