Polynomials Reference
Polynomial Division and Remainder Theorem
Division reveals the quotient, the remainder, and whether a linear factor belongs to a polynomial.
Solved Examples
Jump directly to a worked example by division method or theorem.
Fact Table
| Tool | Use It When | Watch For |
|---|---|---|
| Long division | The divisor has degree two or higher, or synthetic division is awkward. | Insert a zero coefficient for every missing power. |
| Synthetic division | The divisor is $x-c$. | Use $c$, the number that makes $x-c=0$. |
| Remainder Theorem | You need the remainder after division by $x-c$. | The remainder is $f(c)$, not $f(-c)$. |
| Factor Theorem | You need to test whether $x-c$ is a factor. | $x-c$ is a factor exactly when $f(c)=0$. |
Content Formulas
Division Algorithm
$$f(x)=d(x)q(x)+r(x)$$
Remainder Theorem
$$\text{remainder on division by }x-c=f(c)$$
Factor Theorem
$$x-c\text{ is a factor}\iff f(c)=0$$
Write terms in descending powers before you begin. For example, $x^3-5x+6$ must be treated as $x^3+0x^2-5x+6$.
Classic Examples
Use Long Division
Divide $x^3+3x^2-4x-12$ by $x+3$.
Solution Steps
- Divide the leading terms to start the quotient.
- Multiply the divisor by $x^2$ and subtract.
- Repeat with the new leading term.
- State the quotient and remainder.
$$x^3\div x=x^2$$$$\left(x^3+3x^2-4x-12\right)-\left(x^3+3x^2\right)=-4x-12$$$$-4x\div x=-4$$$$\frac{x^3+3x^2-4x-12}{x+3}=x^2-4$$
Use Synthetic Division
Divide $2x^3-3x^2-11x+6$ by $x-3$.
Solution Steps
- Use $3$ because the divisor is $x-3$.
- Bring down the leading coefficient and multiply by $3$.
- Add each column, then multiply by $3$ again.
- Read the quotient coefficients and remainder.
$$\begin{array}{r|rrrr}3&2&-3&-11&6\\&&6&9&-6\\\hline&2&3&-2&0\end{array}$$$$2\downarrow,\qquad 2(3)=6$$$$-3+6=3,\qquad 3(3)=9$$$$2x^2+3x-2\text{ with remainder }0$$
Find a Remainder
Find the remainder when $f(x)=x^4-2x+5$ is divided by $x+1$.
Solution Steps
- Match $x+1$ to $x-c$ to find $c=-1$.
- Apply the Remainder Theorem by evaluating $f(-1)$.
- Evaluate the power and products.
- State the remainder.
$$x+1=x-(-1)$$$$\text{remainder}=f(-1)$$$$f(-1)=(-1)^4-2(-1)+5=1+2+5$$$$\text{remainder}=8$$
Test a Possible Factor
Is $x-2$ a factor of $g(x)=x^3-5x^2+2x+8$?
Solution Steps
- Use $c=2$ from the factor $x-2$.
- Evaluate $g(2)$.
- Simplify to find the remainder.
- Use the Factor Theorem to state the conclusion.
$$x-2\Rightarrow c=2$$$$g(2)=2^3-5(2^2)+2(2)+8$$$$g(2)=8-20+4+8=0$$$$g(2)=0\Rightarrow x-2\text{ is a factor}$$