This chapter introduces two connected ideas for polynomial division: Remainder Theorem and Factor Theorem. They help you test factors quickly, find roots, and break a cubic (or higher) polynomial into simpler factors.
A polynomial in \(x\) is an expression like \(P(x)=2x^3-3x^2+5x-7\). If you divide \(P(x)\) by \((x-a)\), you get:
\[ P(x) = (x-a)Q(x) + R, \]
where \(Q(x)\) is the quotient and \(R\) is a constant remainder.
Remainder Theorem
\[ \text{When } P(x)\text{ is divided by }(x-a),\ \text{remainder}=P(a). \]
Example 1. Find the remainder when \(P(x)=x^3-4x+1\) is divided by \((x-2)\).
By the theorem, remainder \(=P(2)=2^3-4(2)+1=8-8+1=1\).
Example 2. Find the remainder when \(2x^3+x^2-5\) is divided by \((x+1)\).
\((x+1)=(x-(-1))\), so use \(a=-1\): \(P(-1)=2(-1)^3+(-1)^2-5=-2+1-5=-6\).
Factor Theorem
\[ (x-a)\text{ is a factor of }P(x)\iff P(a)=0. \]
This is a direct consequence of the Remainder Theorem: factor means remainder \(0\).
Example 3. Is \((x-3)\) a factor of \(P(x)=x^3-4x^2-3x+18\)?
\(P(3)=27-36-9+18=0\), so yes, \((x-3)\) is a factor.
Example 4. Solve \(x^3-6x^2+11x-6=0\).
Try integer candidates \(\pm1,\pm2,\pm3,\pm6\). We get \(P(1)=0\), so \((x-1)\) is a factor. Dividing gives \(x^2-5x+6\), which factorises to \((x-2)(x-3)\). Hence:
\[ x^3-6x^2+11x-6=(x-1)(x-2)(x-3), \]
so roots are \(x=1,2,3\).
Synthetic division is a compact way to divide by \((x-a)\). It is fast for checking many candidates in exam settings.
Example 5. Divide \(x^3-6x^2+11x-6\) by \((x-1)\):
\[ \begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6\\ & & 1 & -5 & 6\\ \hline & 1 & -5 & 6 & 0 \end{array} \]
Quotient is \(x^2-5x+6\), remainder \(0\).