Polynomial Theorems and Roots

Fundamental Theorem of Algebra and finding roots

Polynomial Theorems

Fundamental Theorem of Algebra

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

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

Rational Root Theorem

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

$p$ divides $a_0$ (constant term)
$q$ divides $a_n$ (leading coefficient)

Example: For $2x^3 + 5x^2 - 4x - 3 = 0$

Possible rational roots: $\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}$

Descartes' Rule of Signs

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

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

Complex Conjugate Theorem

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

Example: If $2 + 3i$ is a root, then $2 - 3i$ 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: $x^3 - 5x^2 + 2x - 6 = 0$

💡 Show Solution

Use the Rational Root Theorem:

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

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

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

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

💡 Show Solution

We have roots: $2$ , $-3$ , and $1 + i$

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

Total roots: $2$ , $-3$ , $1 + i$ , $1 - i$

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 $3$ and $2 - i$

💡 Show Solution

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

Roots: $3$ , $2 - i$ , $2 + i$

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

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

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

Answer: $P(x) = x^3 - 7x^2 + 17x - 15$

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

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