✍️ Practice Set: Polynomial Challenges
Test your skills using the Horner Method, Remainder Theorem, and Factor Theorem. Try to solve these on paper first, then click the "Solution" toggle to check your work.
🟢 Level 1: The Basics
Problem 1: Use the Horner Method to find the quotient and remainder of $(x^3 - 4x^2 + 2x + 5) \div (x - 3)$.
Click to see Solution
Horner Table: | | 1 | -4 | 2 | 5 | | :--- | :--- | :--- | :--- | :--- | | (3) | ↓ | 3 | -3 | -3 | | | 1 | -1 | -1 | (2) |
- Quotient: $x^2 - x - 1$
- Remainder: $2$
🟡 Level 2: The Missing Term
Problem 2: Find the remainder of $(2x^4 - 5x^2 + 6) \div (x + 2)$. Hint: Don't forget the placeholders for $x^3$ and $x$!
Click to see Solution
Coefficients: $2, 0, -5, 0, 6$
$k$ value: $-2$
Horner Table: | | 2 | 0 | -5 | 0 | 6 | | :--- | :--- | :--- | :--- | :--- | :--- | | (-2)| ↓ | -4 | 8 | -6 | 12 | | | 2 | -4 | 3 | -6 | (18) |
- Remainder: $18$
🟠 Level 3: The Factor Theorem
Problem 3: Is $(x - 1)$ a factor of $P(x) = x^3 + 2x^2 - x - 2$? Prove it using the Factor Theorem.
Click to see Solution
According to the Factor Theorem, if $P(1) = 0$, then $(x-1)$ is a factor.
$$P(1) = (1)^3 + 2(1)^2 - (1) - 2$$ $$P(1) = 1 + 2 - 1 - 2$$ $$P(1) = 0$$
Conclusion: Yes, $(x-1)$ is a factor because the remainder is zero.
🔴 Level 4: Solving for $k$
Problem 4: The polynomial $P(x) = x^3 + kx^2 - 4x + 1$ has a remainder of $5$ when divided by $(x - 2)$. Find the value of $k$.
Click to see Solution
Use the Remainder Theorem: $P(2) = 5$.
$$(2)^3 + k(2)^2 - 4(2) + 1 = 5$$ $$8 + 4k - 8 + 1 = 5$$ $$4k + 1 = 5$$ $$4k = 4$$ $k = 1$
🟣 Level 5: Horner-Kino Challenge
Problem 5: Divide $P(x) = x^4 - 3x^3 + 5x^2 - 2x + 1$ by $(x^2 - x - 2)$.
Click to see Solution
Multipliers: $m_1 = 1, m_2 = 2$ (derived from $x^2 - x - 2$).
Quotient Result: $x^2 - 2x + 5$
Remainder Result: $-x + 11$
(This uses the two-row multiplication method discussed in the Horner-Kino section.)