What is Polynomial Theorems and Roots?

Fundamental Theorem of Algebra and finding roots

How can I study Polynomial Theorems and Roots effectively?

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

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What course covers Polynomial Theorems and Roots?

Polynomial Theorems and Roots is part of the Algebra 2 course on Study Mondo, specifically in the Polynomial Functions section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Polynomial Theorems and Roots?

Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Polynomial Theorems and Roots

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Polynomial Theorems and Roots | Study Mondo

📚 Practice Problems

1Problem 1easy

❓ Question:

Use the Remainder Theorem to find the remainder when P(x) = 2x³ - 5x² + 3x - 7 is divided by (x - 2).

💡 Show Solution

Step 1: Recall the Remainder Theorem: When P(x) is divided by (x - a), the remainder is P(a)

Step 2: Identify a = 2 (from x - 2): We need to find P(2)

Step 3: Evaluate P(2): P(2) = 2(2)³ - 5(2)² + 3(2) - 7 P(2) = 2(8) - 5(4) + 6 - 7 P(2) = 16 - 20 + 6 - 7 P(2) = -5

Answer: The remainder is -5

2Problem 2easy

❓ Question:

List all possible rational roots: $x^3 - 5x^2 + 2x - 6 = 0$

💡 Show Solution

Use the Rational Root Theorem:

Factors of constant term ( $-6$ ): $\pm 1, \pm 2, \pm 3, \pm 6$ Factors of leading coefficient ( $1$ ):

3Problem 3easy

❓ Question:

Determine if (x + 3) is a factor of P(x) = x³ + 4x² - 3x - 18.

💡 Show Solution

Step 1: Use the Factor Theorem: (x - a) is a factor if and only if P(a) = 0

Step 2: For (x + 3), we have x - (-3), so a = -3: Check if P(-3) = 0

Step 3: Evaluate P(-3): P(-3) = (-3)³ + 4(-3)² - 3(-3) - 18 P(-3) = -27 + 4(9) + 9 - 18 P(-3) = -27 + 36 + 9 - 18 P(-3) = 0

Step 4: Conclusion: Since P(-3) = 0, (x + 3) IS a factor

Answer: Yes, (x + 3) is a factor of P(x)

4Problem 4medium

❓ Question:

A polynomial has roots $2$ , $-3$ , and $1 + i$ . What is the minimum degree?

💡 Show Solution

5Problem 5medium

❓ Question:

Find all zeros of P(x) = x³ - 6x² + 11x - 6 given that x = 1 is a zero.

💡 Show Solution

Step 1: Since x = 1 is a zero, (x - 1) is a factor: Use synthetic or long division to find the other factor

Step 2: Synthetic division by (x - 1): 1 | 1 -6 11 -6 | 1 -5 6 ___________________ 1 -5 6 0

Result: x² - 5x + 6

Step 3: Factor the quadratic: x² - 5x + 6 = (x - 2)(x - 3)

Step 4: Write complete factorization: P(x) = (x - 1)(x - 2)(x - 3)

Step 5: Find all zeros: x - 1 = 0 → x = 1 x - 2 = 0 → x = 2 x - 3 = 0 → x = 3

Answer: The zeros are x = 1, 2, 3

6Problem 6medium

❓ Question:

Use the Rational Root Theorem to find all possible rational roots of P(x) = 2x³ - 3x² - 11x + 6.

💡 Show Solution

Step 1: Recall the Rational Root Theorem: Possible rational roots = ±(factors of constant term)/(factors of leading coefficient)

Step 2: Find factors of constant term (6): ±1, ±2, ±3, ±6

Step 3: Find factors of leading coefficient (2): ±1, ±2

Step 4: List all possible rational roots: ±1/1, ±2/1, ±3/1, ±6/1, ±1/2, ±2/2, ±3/2, ±6/2

Step 5: Simplify and remove duplicates: ±1, ±2, ±3, ±6, ±1/2, ±3/2

Step 6: Test to find actual roots (optional): P(1/2) = 2(1/8) - 3(1/4) - 11(1/2) + 6 = 1/4 - 3/4 - 11/2 + 6 = 0 ✓ So x = 1/2 is a root

Answer: Possible rational roots are ±1, ±2, ±3, ±6, ±1/2, ±3/2

7Problem 7hard

❓ Question:

Find a polynomial with real coefficients that has roots $3$ and $2 - i$

💡 Show Solution

Since coefficients are real and $2 - i$ is a root, then must also be a root.

8Problem 8hard

❓ Question:

Find a polynomial with integer coefficients that has zeros at x = 2, x = -3, and x = 1/2.

💡 Show Solution

Step 1: Write factors for each zero: x = 2 → factor (x - 2) x = -3 → factor (x + 3) x = 1/2 → factor (x - 1/2) or (2x - 1)

Step 2: To avoid fractions, use (2x - 1): P(x) = (x - 2)(x + 3)(2x - 1)

Step 3: Multiply first two factors: (x - 2)(x + 3) = x² + 3x - 2x - 6 = x² + x - 6

Step 4: Multiply by the third factor: (x² + x - 6)(2x - 1)

Step 5: Distribute 2x: 2x(x²) + 2x(x) + 2x(-6) = 2x³ + 2x² - 12x

Step 6: Distribute -1: -1(x²) - 1(x) - 1(-6) = -x² - x + 6

Step 7: Combine: 2x³ + 2x² - 12x - x² - x + 6 = 2x³ + x² - 13x + 6

Step 8: Verify zeros: P(2) = 16 + 4 - 26 + 6 = 0 ✓ P(-3) = -54 + 9 + 39 + 6 = 0 ✓ P(1/2) = 2(1/8) + 1/4 - 13/2 + 6 = 1/4 + 1/4 - 13/2 + 6 = 0 ✓

Answer: P(x) = 2x³ + x² - 13x + 6

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Polynomial Theorems and Roots

Try the Interactive Version!

Polynomial Theorems

Fundamental Theorem of Algebra

Rational Root Theorem

Descartes' Rule of Signs

Complex Conjugate Theorem

📚 Practice Problems

1Problem 1easy

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2Problem 2easy

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3Problem 3easy

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4Problem 4medium

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5Problem 5medium

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6Problem 6medium

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7Problem 7hard

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8Problem 8hard

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⚠️ Common Mistakes: Polynomial Theorems and Roots

🌍 Real-World Applications: Polynomial Theorems and Roots

📝 Worked Example: Solving a Quadratic by Factoring

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