Theory:

Theorem
If a polynomial \(p(x)\) is divided by \((x-a)\), then \(p(a)\) is the remainder.
Explanation:
 
The theorem states that, any polynomial \(p(x)\) when divided by an expression of the form \((x-a)\) leaves a remainder \(p(a)\).
Example:
1. Find the remainder when \(x^{2} + 3x + 2\) is divided by \(x + 2\).
 
Given:
 
The polynomial \(p(x) = x^{2} + 3x + 2\).
 
To find:
 
The remainder when \(p(x) = x^{2} + 3x + 2\) is divided by \(x + 2\).
 
Theorem used:
 
If a polynomial \(p(x)\) is divided by \((x-a)\), then \(p(a)\) is the remainder.
 
Solution:
 
Step 1: Find the zero of the polynomial \(x = a\).
 
Equate \(x + 2\) to zero and solve for \(x\).
 
\(x\) \(+\) \(2\) \(=\) \(0\)
 
\(x\) \(=\) \(-2\)
 
Step 2: Find the remainder \(p(-2)\).
 
Substitute \(x = -2\) in (p(x)\).
 
\(p(-2) = (-2)^{2} + 3(-2) + 2\)
 
\(= 4 - 6 + 2\)
 
\(= 0\)
 
 
2. Find the remainder when \(x^{2} + 3x - 2\) is divided by \(x + 1\).
 
Given:
 
The polynomial \(p(x) = x^{2} + 3x - 2\).
 
To find:
 
The remainder when \(p(x) = x^{2} + 3x - 2\) is divided by \(x + 1\).
 
Theorem used:
 
If a polynomial \(p(x)\) is divided by \((x-a)\), then \(p(a)\) is the remainder.
 
Solution:
 
Step 1: Find the zero of the polynomial \(x = a\).
 
Equate \(x + 1\) to zero and solve for \(x\).
 
\(x\) \(+\) \(1\) \(=\) \(0\)
 
\(x\) \(=\) \(-1\)
 
Step 2: Find the remainder \(p(-1)\).
 
Substitute \(x = -1\) in \(p(x)\).
 
\(p(-1) = (-1)^{2} + 3(-1) - 2\)
 
\(= 1 - 3 - 2\)
 
\(= -4\)
Important!
In Example (1), \(x + 2\) is the factor of the polynomial \(x^{2} + 3x + 2\) as it satisfies the equation  \(p(x) = 0\).
 
In Example (2), \(x + 1\) is not the factor of the polynomial \(x^{2} + 3x - 2\) as it does not satisfies the equation  \(p(x) = 0\).