Remainder = p(2). Quick check without long division.

Remainder Theorem — Find Remainder When p(x) Divided by (x-2)

Question

Given $p(x) = x^3 - 3x^2 + 4x - 2$ , find the remainder when $p(x)$ is divided by $(x - 2)$ .

Solution — Step by Step

If a polynomial $p(x)$ is divided by $(x - a)$ , the remainder is $p(a)$ . Here the divisor is $(x - 2)$ , so $a = 2$ . We don’t need long division at all — just evaluate the polynomial at $x = 2$ .

p(2) = (2)^3 - 3(2)^2 + 4(2) - 2

Work each term separately to avoid sign errors:

$(2)^3 = 8$
$3(2)^2 = 3 \times 4 = 12$
$4(2) = 8$

p(2) = 8 - 12 + 8 - 2

= (8 + 8) - (12 + 2)

= 16 - 14 = 2

The remainder when $p(x)$ is divided by $(x - 2)$ is 2.

Quick sanity check: if the remainder were 0, then $x = 2$ would be a root. Since $p(2) = 2 \neq 0$ , $(x-2)$ is not a factor — which makes sense.

Why This Works

The Remainder Theorem comes from polynomial long division. When we divide $p(x)$ by $(x - a)$ , we always get:

p(x) = (x - a) \cdot q(x) + r

where $q(x)$ is the quotient and $r$ is a constant remainder (degree less than 1).

Now substitute $x = a$ on both sides: the $(x - a)$ term vanishes, giving $p(a) = r$ . That’s it — the remainder equals $p(a)$ , no long division required.

This is why this theorem is a massive time-saver in board exams. A problem that would take 2-3 minutes with long division takes 20 seconds with the Remainder Theorem. In CBSE Class 9 and 10, this appears almost every year — either directly or as part of a Factor Theorem question.

Alternative Method — Polynomial Long Division

If you want to verify, you can actually divide $x^3 - 3x^2 + 4x - 2$ by $(x - 2)$ :

x³ - 3x² + 4x - 2  ÷  (x - 2)

Step-by-step:

$x^3 \div x = x^2$ . Multiply: $x^2(x-2) = x^3 - 2x^2$ . Subtract → $-x^2 + 4x - 2$
$-x^2 \div x = -x$ . Multiply: $-x(x-2) = -x^2 + 2x$ . Subtract → $2x - 2$
$2x \div x = 2$ . Multiply: $2(x-2) = 2x - 4$ . Subtract → remainder = 2 ✓

Both methods give $r = 2$ . Use Remainder Theorem in exams — long division is only for verification or when you need the quotient too.

Common Mistake

Students substitute $x = -2$ instead of $x = 2$ when the divisor is $(x - 2)$ .

The divisor is $(x - a)$ , so set $x - a = 0$ → $x = a$ . For $(x - 2)$ : $x - 2 = 0$ gives $x = 2$ . The sign flips only when the divisor is $(x + a)$ — for example, $(x + 3)$ gives $x = -3$ . Don’t mix these up.

This is a scoring topic — NCERT Exercise 2.3 has several direct questions of this type. Once you’re comfortable with it, Factor Theorem (which checks if the remainder is 0) is just one more line. Both together take about 10 minutes to master and regularly appear in board papers for 2-3 marks.

Question

Solution — Step by Step

Why This Works

Alternative Method — Polynomial Long Division

Common Mistake

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