Solution

Step 1: Identify degree and leading coefficient

Degree: 5 (odd)

Leading coefficient: -2 (negative)

Step 2: Apply Leading Term Test

Odd degree + Negative leading coefficient:

As x → -∞, f(x) → +∞

As x → +∞, f(x) → -∞

Interpretation: The graph rises on the left and falls on the right.

Solution

Zero	Multiplicity	Graph Behavior
x = 0	1 (odd)	Crosses x-axis
x = -3	2 (even)	Touches x-axis (bounces)
x = 2	3 (odd)	Crosses x-axis (flattens near crossing)

Additional information:

Total degree: 1 + 2 + 3 = 6 (even)

Leading coefficient: 1 (positive)

End behavior: Both ends point upward

Solution

Step 1: Use Rational Root Theorem

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

Step 2: Test x = 1

f(1) = 1 - 2 - 5 + 6 = 0 Zero found!

Step 3: Use synthetic division with x = 1

    1 |  1   -2   -5    6
      |       1   -1   -6
      |___________________
         1   -1   -6    0

Quotient: x² - x - 6

Step 4: Factor the quotient

x² - x - 6 = (x - 3)(x + 2)

Step 5: Complete factorization

f(x) = (x - 1)(x - 3)(x + 2)

All zeros: x = -2, 1, 3

Solution

Step 1: Factor to find zeros

f(x) = -x³ + 4x = -x(x² - 4) = -x(x - 2)(x + 2)

Zeros: x = -2, 0, 2 (all multiplicity 1, so graph crosses at each)

Step 2: Find y-intercept

f(0) = 0, so y-intercept is (0, 0)

Step 3: Determine end behavior

Degree: 3 (odd), Leading coefficient: -1 (negative)

As x → -∞, f(x) → +∞

As x → +∞, f(x) → -∞

Step 4: Test points

f(-3) = -(-27) + 12 = 39 (positive)

f(1) = -1 + 4 = 3 (positive)

f(3) = -27 + 12 = -15 (negative)

Graph characteristics:

• Comes from upper left (positive infinity)
• Crosses at x = -2
• Goes up to a local maximum between -2 and 0
• Crosses at x = 0
• Goes down to a local minimum between 0 and 2
• Crosses at x = 2
• Continues down to negative infinity

Solution

Step 1: Write in factored form

f(x) = 3(x + 1)²(x - 2)²

Step 2: Verify degree

Degree = 2 + 2 = 4

Step 3: Determine end behavior

Degree: 4 (even)

Leading coefficient: 3 (positive)

As x → -∞, f(x) → +∞

As x → +∞, f(x) → +∞

Both ends point upward.

Additional notes:

Graph touches (bounces) at both x = -1 and x = 2 (even multiplicities)

The graph never crosses the x-axis; it touches at two points and is positive everywhere else

Solution

Step 1: Recognize as sum of cubes

8x³ + 27 = (2x)³ + 3³

Step 2: Apply sum of cubes formula

a³ + b³ = (a + b)(a² - ab + b²)

where a = 2x and b = 3

Step 3: Substitute

(2x + 3)[(2x)² - (2x)(3) + 3²]

(2x + 3)(4x² - 6x + 9)

Answer: (2x + 3)(4x² - 6x + 9)

Solution

Step 1: Group terms in pairs

(3x³ - 6x²) + (4x - 8)

Step 2: Factor GCF from each group

3x²(x - 2) + 4(x - 2)

Step 3: Factor out common binomial

(x - 2)(3x² + 4)

Answer: (x - 2)(3x² + 4)

Note: 3x² + 4 cannot be factored further over the real numbers

Solution

Step 1: Recognize as difference of cubes

64x³ - 1 = (4x)³ - 1³

Step 2: Apply difference of cubes formula

a³ - b³ = (a - b)(a² + ab + b²)

where a = 4x and b = 1

Step 3: Substitute

(4x - 1)[(4x)² + (4x)(1) + 1²]

(4x - 1)(16x² + 4x + 1)

Answer: (4x - 1)(16x² + 4x + 1)

Solution

Step 1: List potential rational zeros

p (factors of -6): ±1, ±2, ±3, ±6

q (factors of 2): ±1, ±2

Potential zeros: ±1, ±2, ±3, ±6, ±1/2, ±3/2

Step 2: Test x = -2

f(-2) = 2(-8) - 4 - 13(-2) - 6 = -16 - 4 + 26 - 6 = 0

Step 3: Synthetic division with x = -2

   -2 |  2   -1  -13   -6
      |      -4   10    6
      |___________________
         2   -5   -3    0

Quotient: 2x² - 5x - 3

Step 4: Factor the quotient

2x² - 5x - 3 = (2x + 1)(x - 3)

Step 5: Complete factorization

f(x) = (x + 2)(2x + 1)(x - 3)

All zeros: x = -2, x = -1/2, x = 3

Solution

Step 1: Factor out GCF

2x⁴ - 162 = 2(x⁴ - 81)

Step 2: Recognize difference of squares

x⁴ - 81 = (x²)² - 9² = (x² - 9)(x² + 9)

Step 3: Factor x² - 9 further (difference of squares)

x² - 9 = (x - 3)(x + 3)

Step 4: Complete factorization

Answer: 2(x - 3)(x + 3)(x² + 9)

Note: x² + 9 cannot be factored over the real numbers

Solution

Step 1: Factor numerator and denominator

Numerator: 3x² - 12 = 3(x² - 4) = 3(x - 2)(x + 2)

Denominator: x² + x - 6 = (x + 3)(x - 2)

Step 2: Cancel common factors

f(x) = 3(x + 2)/(x + 3), x ≠ 2

Step 3: Find vertical asymptotes

Set remaining denominator = 0: x + 3 = 0

Vertical asymptote: x = -3

Step 4: Find horizontal asymptote

After cancellation: degree of numerator = 1, degree of denominator = 1

Leading coefficients: 3/1

Horizontal asymptote: y = 3

Note: x = 2 creates a hole, not a vertical asymptote

Solution

Step 1: Factor completely

Numerator: x² + 5x + 6 = (x + 2)(x + 3)

Denominator: x² - 4 = (x - 2)(x + 2)

Step 2: Identify common factors

Common factor: (x + 2)

Step 3: Cancel common factor

g(x) = (x + 3)/(x - 2), x ≠ -2

Step 4: Find x-coordinate of hole

x + 2 = 0 → x = -2

Step 5: Find y-coordinate using simplified function

y = (-2 + 3)/(-2 - 2) = 1/(-4) = -1/4

Hole at: (-2, -1/4)

Vertical asymptote at: x = 2

Solution

Step 1: Set denominator equal to zero

x² - 2x - 15 = 0

Step 2: Factor

(x - 5)(x + 3) = 0

Step 3: Solve

x = 5 or x = -3

Step 4: State domain

Domain: All real numbers except x = -3 and x = 5

Interval notation: (-∞, -3) ∪ (-3, 5) ∪ (5, ∞)

Set notation: {x | x ≠ -3, x ≠ 5}

Solution

Step 1: Factor

Numerator: 2x² - 8 = 2(x² - 4) = 2(x - 2)(x + 2)

Denominator: x² + 3x + 2 = (x + 1)(x + 2)

Step 2: Simplify

f(x) = 2(x - 2)/(x + 1), x ≠ -2

Step 3: Find hole

Hole at x = -2, y = 2(-2 - 2)/(-2 + 1) = 2(-4)/(-1) = 8

Hole: (-2, 8)

Step 4: Find vertical asymptote

x + 1 = 0 → x = -1

Vertical asymptote: x = -1

Step 5: Find horizontal asymptote

After simplification: degree 1/degree 1, coefficients 2/1

Horizontal asymptote: y = 2

Step 6: Find x-intercept (set numerator = 0)

2(x - 2) = 0 → x = 2

X-intercept: (2, 0)

Step 7: Find y-intercept (evaluate f(0))

f(0) = 2(0 - 2)/(0 + 1) = -4/1 = -4

Y-intercept: (0, -4)

Solution

Step 1: Compare degrees

Degree of numerator: 3

Degree of denominator: 3

Degrees are equal

Step 2: Find ratio of leading coefficients

Leading coefficient of numerator: 5

Leading coefficient of denominator: 3

Ratio: 5/3

Horizontal asymptote: y = 5/3

Interpretation: As x → ±∞, f(x) → 5/3

Solution

Step 1: Factor

x² - 7x + 10 = (x - 2)(x - 5)

Step 2: Find critical values

x = 2 and x = 5

Step 3: Test intervals

Interval	Test Point	Sign
(-∞, 2)	x = 0	(-)(-) = +
(2, 5)	x = 3	(+)(-) = -
(5, ∞)	x = 6	(+)(+) = +

Step 4: Select negative interval

We want f(x) < 0 (negative)

Solution: (2, 5)

Or: 2 < x < 5

Solution

Step 1: Factor

x³ + 2x² - 3x = x(x² + 2x - 3) = x(x + 3)(x - 1)

Step 2: Find zeros

x = -3, 0, 1

Step 3: Create sign chart

Interval:  (-∞,-3) | (-3,0) | (0,1) | (1,∞)
Sign:         -    |   +    |   -   |   +

Step 4: We want f(x) ≥ 0 (positive or zero)

Positive intervals: (-3, 0) and (1, ∞)

Include boundaries (≥): x = -3, 0, 1

Solution: [-3, 0] ∪ [1, ∞)

Solution

Step 1: Find critical values

Numerator = 0: x = 4

Denominator = 0: x = -1

Step 2: Test intervals

Interval	Test Point	Sign
(-∞, -1)	x = -2	(-)/(-) = +
(-1, 4)	x = 0	(-)/(+) = -
(4, ∞)	x = 5	(+)/(+) = +

Step 3: Select positive intervals

We want f(x) > 0 (positive)

Do not include x = -1 (makes denominator zero)

Do not include x = 4 (strict inequality >)

Solution: (-∞, -1) ∪ (4, ∞)

Solution

Step 1: Factor numerator

(x² - 9)/(x - 1) = (x - 3)(x + 3)/(x - 1)

Step 2: Find critical values

From numerator: x = -3, 3

From denominator: x = 1

All critical values: -3, 1, 3

Step 3: Create sign chart

Interval:  (-∞,-3) | (-3,1) | (1,3) | (3,∞)
Sign:         -    |   +    |   -   |   +

Step 4: We want f(x) ≤ 0 (negative or zero)

Negative intervals: (-∞, -3) and (1, 3)

Include x = -3 and x = 3 (≤ and denominator ≠ 0)

Never include x = 1 (denominator = 0)

Solution: (-∞, -3] ∪ (1, 3]

Solution

Step 1: Identify zeros and multiplicities

x = 0 (multiplicity 1 - odd)

x = 2 (multiplicity 2 - even)

x = -1 (multiplicity 1 - odd)

Step 2: Key insight

Sign changes at odd multiplicity zeros (0 and -1)

Sign does NOT change at even multiplicity zero (2)

Step 3: Create sign chart

Interval:  (-∞,-1) | (-1,0) | (0,2) | (2,∞)
Factor x:      -   |   -    |   +   |   +
Factor(x-2)²:  +   |   +    |   +   |   +
Factor(x+1):   -   |   +    |   +   |   +
Product:       +   |   -    |   +   |   +

Step 4: Select positive intervals

We want f(x) > 0 (positive)

Positive intervals: (-∞, -1), (0, 2), and (2, ∞)

Note: (0, 2) and (2, ∞) can be combined as (0, ∞)

Solution: (-∞, -1) ∪ (0, ∞)

Or: x < -1 or x > 0