Remainder Theorem says:
If a polynomial p(x)p(x)p(x) is divided by a linear term (x−a)(x-a)(x−a), then the remainder is just the value p(a)p(a)p(a).

Quick Scoop: What is Remainder Theorem?

Think of it as a shortcut for polynomial division:
Instead of doing full long division, you just plug a number into the polynomial.

  • You have a polynomial p(x)p(x)p(x).
  • You divide it by (x−a)(x-a)(x−a).
  • The remainder of this division is p(a)p(a)p(a).

In general form:

p(x)=(x−a)q(x)+rp(x)=(x-a)q(x)+rp(x)=(x−a)q(x)+r

where q(x)q(x)q(x) is the quotient and rrr is the remainder (a constant when divisor is linear).

If you put x=ax=ax=a,

p(a)=(a−a)q(a)+r=rp(a)=(a-a)q(a)+r=rp(a)=(a−a)q(a)+r=r

so remainder =p(a)=p(a)=p(a).

Tiny Example

Find the remainder when
p(x)=x2+4x+4p(x)=x^2+4x+4p(x)=x2+4x+4 is divided by (x−1)(x-1)(x−1).

  • Here, a=1a=1a=1.
  • Compute p(1)=12+4⋅1+4=1+4+4=9p(1)=1^2+4\cdot 1+4=1+4+4=9p(1)=12+4⋅1+4=1+4+4=9.

So the remainder is 999.

Why it’s useful

  • Avoids long division for polynomials.
  • Helps in quickly checking if (x−a)(x-a)(x−a) is a factor:
    • If p(a)=0p(a)=0p(a)=0, then remainder is 0 → (x−a)(x-a)(x−a) is a factor (this idea leads to the factor theorem).
  • Widely used in school algebra and competitive exams for faster calculations.

TL;DR:
To find the remainder when p(x)p(x)p(x) is divided by (x−a)(x-a)(x−a), just calculate p(a)p(a)p(a) and that number is the remainder.

Information gathered from public forums or data available on the internet and portrayed here.