Polynomial Puzzle Worksheet
Polynomial Puzzle
Algebra 2: Solving Higher-Order Functions
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Factoring Techniques
- • Greatest Common Factor (GCF)
- • Factor by Grouping (4 terms)
- • Sum/Difference of Cubes
Finding Roots
- • Rational Root Theorem (\( \pm \frac{p}{q} \))
- • Synthetic Division
- • Quadratic Formula for remainders
Pro Tip
The degree of the polynomial tells you exactly how many solutions (real or complex) you should find!
Level 1: Factoring Foundations
Solve by factoring. Find all zeros.
1
\( f(x) = x^3 - 2x^2 - 9x + 18 \)
Zeros:
2
\( g(x) = x^4 - 13x^2 + 36 \)
Zeros:
Level 2: Synthetic Strategy
Use the Rational Root Theorem and Synthetic Division.
3
\( h(x) = x^3 - 7x - 6 \)
Zeros:
4
\( p(x) = x^3 - 4x^2 + x + 6 \)
Zeros:
Level 3: The Complex Challenge
Watch out for imaginary/complex roots!
5
\( f(x) = x^3 - 3x^2 + x - 3 \)
Zeros:
6
\( g(x) = x^4 - 1 \)
Zeros:
Ultra Bonus
A 3rd-degree polynomial has zeros at \( x = 2 \) and \( x = 1+i \). Write the function in factored form and expanded form.
Polynomial Puzzle Answer Key
Answer Key
Polynomial Puzzle Solutions
Teacher Resource
Level 1: Factoring Foundations
1 \( x^3 - 2x^2 - 9x + 18 \)
Group: \( (x^3 - 2x^2) - (9x - 18) \)
Factor: \( x^2(x - 2) - 9(x - 2) \)
Result: \( (x^2 - 9)(x - 2) \)
Final: \( (x - 3)(x + 3)(x - 2) \)
Zeros: \( 3, -3, 2 \)
2 \( x^4 - 13x^2 + 36 \)
Factor like quadratic: let \( u = x^2 \)
Factor: \( (x^2 - 4)(x^2 - 9) \)
Result: \( (x - 2)(x + 2)(x - 3)(x + 3) \)
Zeros: \( 2, -2, 3, -3 \)
Level 2: Synthetic Strategy
3 \( x^3 - 7x - 6 \)
Test roots: \( f(-1) = (-1)^3 - 7(-1) - 6 = 0 \)
Synt. Div by -1: \( 1, -1, -6 \)
Quotient: \( x^2 - x - 6 = (x - 3)(x + 2) \)
Zeros: \( -1, 3, -2 \)
4 \( x^3 - 4x^2 + x + 6 \)
Test roots: \( f(2) = 8 - 16 + 2 + 6 = 0 \)
Synt. Div by 2: \( 1, -2, -3 \)
Quotient: \( x^2 - 2x - 3 = (x - 3)(x + 1) \)
Zeros: \( 2, 3, -1 \)
Level 3: The Complex Challenge
5 \( x^3 - 3x^2 + x - 3 \)
Group: \( x^2(x - 3) + 1(x - 3) \)
Factor: \( (x^2 + 1)(x - 3) \)
Solve \( x^2 + 1 = 0 \rightarrow x = \pm \sqrt{-1} \)
Zeros: \( 3, i, -i \)
6 \( x^4 - 1 \)
Diff. of Squares: \( (x^2 - 1)(x^2 + 1) \)
Result: \( (x - 1)(x + 1)(x^2 + 1) \)
Zeros: \( 1, -1, i, -i \)
Bonus Solution
Step 1: Identify all roots. If \( 1+i \) is a root, its conjugate \( 1-i \) must also be a root.
Step 2 (Factored Form): \( f(x) = (x - 2)(x - (1 + i))(x - (1 - i)) \)
Step 3 (Expand Complex Factors): \( (x - 1 - i)(x - 1 + i) = ((x - 1) - i)((x - 1) + i) = (x - 1)^2 - i^2 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2 \)
Step 4 (Final Expansion): \( f(x) = (x - 2)(x^2 - 2x + 2) = x^3 - 2x^2 + 2x - 2x^2 + 4x - 4 \)
Expanded Form: \( f(x) = x^3 - 4x^2 + 6x - 4 \)
