🧭 Problem

The roots of the cubic

are \(p, q, r\). Find the value of

1️⃣ Solution 1 — Complex plug-in \(f(2i)f(-2i)\)

Let \[ f(x) = x^3 + 2x^2 - x + 3 = (x-p)(x-q)(x-r). \]

Notice that \[ p^2 + 4 = p^2 - (-4) = (p-2i)(p+2i). \] Similarly for \(q\) and \(r\).

So the product becomes

Now just evaluate the polynomial at \(x = 2i\) and \(x = -2i\).

Compute \(f(2i)\)

Compute \(f(-2i)\)

Multiply

Hence \[ (p^2+4)(q^2+4)(r^2+4) = 125, \] so the answer is (D) 125.

2️⃣ Solution 2 — Pure Vieta / symmetric-sum expansion

From \[ x^3 + 2x^2 - x + 3 = 0, \] by Vieta we have:

• \(p + q + r = -2\) (negative of coeff. of \(x^2\))
• \(pq + pr + qr = -1\) (coeff. of \(x\))
• \(pqr = -3\) (negative of constant term)

We want \[ (p^2+4)(q^2+4)(r^2+4). \]

Expand step by step:

Now express each symmetric piece:

1. \(p^2 q^2 r^2 = (pqr)^2 = (-3)^2 = 9.\)

2. \(p^2 q^2 + q^2 r^2 + r^2 p^2 = (pq + qr + rp)^2 - 2pqr(p+q+r).\)

3. \(p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq+qr+rp).\)

Substitute back

Again we get 125.

This method is longer but fully algebraic — good for CAT mainsheet or if evaluator wants to see Vieta usage.

🧠 Why the complex trick is so powerful here

Whenever you see \((p^2 + \alpha^2)(q^2 + \alpha^2)(r^2 + \alpha^2)\) and \(p, q, r\) are roots of \(f(x)=0\), try to write \((x^2+\alpha^2) = (x-\alpha i)(x+\alpha i)\). Then the whole product becomes \(f(\alpha i)f(-\alpha i)\). Here \(\alpha = 2\), so answer is \(f(2i)f(-2i)\). 1 line.

📝 Quick Quiz

For the same cubic \(x^3 + 2x^2 - x + 3 = 0\), what is \((p^2 + 1)(q^2 + 1)(r^2 + 1)\)? (Hint: follow the same complex trick.)