Polynomial Division: The Reverse Process
If (x-2) is a factor, multiplying by the quotient must give back the original polynomial.
Equation
y = (x - 2)*(x^2 + 3*x + 1) - (x^3 + x^2 - 2*x - 8)
Graph
Table
| x | y |
|---|---|
| -2 | 12 |
| -1 | 9 |
| 0 | 6 |
| 1 | 3 |
| 2 | 0 |
| 3 | -3 |
| 4 | -6 |
| 5 | -9 |
What this lesson covers
What you do
You shape the function y = (x - 2)*(x^2 + 3*x + k) - (x^3 + x^2 - 2*x - 8) and watch the graph answer.
Challenges to clear
- Division in reverse: slide k until (x−2)(x²+3x+k) matches the original polynomial — 0 at x = 1.
- Still 0 at x = 3 — the quotient ends in k = 4, remainder zero.
- Divide x³ + 3x² − 10x − 24 by (x − 3). The quotient starts x² + mx + … — slide m to 6.
- Now the last term of the quotient. Slide k until the difference is 0 at x = 1 — remainder zero means the division was exact.
Check yourself
At x=2, the function value is 0. What does this confirm about the polynomial x^3 + x^2 - 2x - 8?
- It confirms that x=2 is a root of the polynomial. — correct
- It confirms that the polynomial has no real roots.
- It confirms that the leading coefficient is 2.
- It confirms that the remainder of the division is 2.
Think about it
- x³ + x² − 2x − 8 ÷ (x − 2): the quotient is x² + 3x + k. Multiplying back must give the original. What k?