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Polynomial Theorems and Roots

Fundamental Theorem of Algebra and finding roots

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Polynomial Theorems

Fundamental Theorem of Algebra

Every polynomial of degree nn has exactly nn roots (counting multiplicity), including complex roots.

Example: A degree 3 polynomial has 3 roots (some may be repeated or complex).

Rational Root Theorem

If pq\frac{p}{q} is a rational root of anxn+...+a0=0a_nx^n + ... + a_0 = 0, then:

  • pp divides a0a_0 (constant term)
  • qq divides ana_n (leading coefficient)

Example: For 2x3+5x2−4x−3=02x^3 + 5x^2 - 4x - 3 = 0

Possible rational roots: ±1,±3,±12,±32\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}

Descartes' Rule of Signs

Count sign changes in P(x)P(x) to find:

  • Positive roots: Number of sign changes (or less by an even number)
  • Negative roots: Number of sign changes in P(−x)P(-x)

Complex Conjugate Theorem

If a+bia + bi is a root of a polynomial with real coefficients, then a−bia - bi is also a root.

Example: If 2+3i2 + 3i is a root, then 2−3i2 - 3i must also be a root.

📚 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: x3−5x2+2x−6=0x^3 - 5x^2 + 2x - 6 = 0

💡 Show Solution

Use the Rational Root Theorem:

Factors of constant term (−6-6): ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6 Factors of leading coefficient (11): ±1\pm 1

Possible rational roots: ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6

Answer: ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6

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 22, −3-3, and 1+i1 + i. What is the minimum degree?

💡 Show Solution

We have roots: 22, −3-3, and 1+i1 + i

By the Complex Conjugate Theorem, if 1+i1 + i is a root, then 1−i1 - i must also be a root (assuming real coefficients).

Total roots: 22, −3-3, 1+i1 + i, 1−i1 - i

That's 4 roots, so minimum degree is 4.

Answer: Degree 4

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 33 and 2−i2 - i

💡 Show Solution

Since coefficients are real and 2−i2 - i is a root, then 2+i2 + i must also be a root.

Roots: 33, 2−i2 - i, 2+i2 + i

Step 1: Write factors (x−3)(x−(2−i))(x−(2+i))(x - 3)(x - (2 - i))(x - (2 + i))

Step 2: Multiply the complex factors first (x−2+i)(x−2−i)(x - 2 + i)(x - 2 - i) =[(x−2)+i][(x−2)−i]= [(x - 2) + i][(x - 2) - i] =(x−2)2−i2= (x - 2)^2 - i^2 =(x−2)2+1= (x - 2)^2 + 1 =x2−4x+4+1= x^2 - 4x + 4 + 1 =x2−4x+5= x^2 - 4x + 5

Step 3: Multiply by (x−3)(x - 3) (x−3)(x2−4x+5)(x - 3)(x^2 - 4x + 5) =x3−4x2+5x−3x2+12x−15= x^3 - 4x^2 + 5x - 3x^2 + 12x - 15 =x3−7x2+17x−15= x^3 - 7x^2 + 17x - 15

Answer: P(x)=x3−7x2+17x−15P(x) = x^3 - 7x^2 + 17x - 15

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

Explain using:

⚠️ Common Mistakes: Polynomial Theorems and Roots

Avoid these 3 frequent errors

🌍 Real-World Applications: Polynomial Theorems and Roots

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve x2−5x+6=0x^2 - 5x + 6 = 0.

2Factor the quadratic
3Set each factor equal to zero
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Graphing Polynomial Functions
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Fundamental Theorem of Algebra and finding roots
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