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Fundamental Theorem of Algebra and finding roots
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Every polynomial of degree has exactly roots (counting multiplicity), including complex roots.
Example: A degree 3 polynomial has 3 roots (some may be repeated or complex).
If is a rational root of , then:
Use the Remainder Theorem to find the remainder when P(x) = 2x³ - 5x² + 3x - 7 is divided by (x - 2).
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
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Example: For
Possible rational roots:
Count sign changes in to find:
If is a root of a polynomial with real coefficients, then is also a root.
Example: If is a root, then must also be a root.
List all possible rational roots:
Use the Rational Root Theorem:
Factors of constant term (): Factors of leading coefficient ():
Possible rational roots:
Answer:
Determine if (x + 3) is a factor of P(x) = x³ + 4x² - 3x - 18.
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)
A polynomial has roots , , and . What is the minimum degree?
We have roots: , , and
By the Complex Conjugate Theorem, if is a root, then must also be a root (assuming real coefficients).
Find all zeros of P(x) = x³ - 6x² + 11x - 6 given that x = 1 is a zero.
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
Use the Rational Root Theorem to find all possible rational roots of P(x) = 2x³ - 3x² - 11x + 6.
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
Find a polynomial with real coefficients that has roots and
Since coefficients are real and is a root, then must also be a root.
Roots: , ,
Step 1: Write factors
Step 2: Multiply the complex factors first
Step 3: Multiply by
Answer:
Find a polynomial with integer coefficients that has zeros at x = 2, x = -3, and x = 1/2.
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
Total roots: , , ,
That's 4 roots, so minimum degree is 4.
Answer: Degree 4