Solution:

The number of zeros is the number of times the graph of p(x) intersects the x-axis.

• Given: p(x) = x² – 4x + 3 (quadratic, degree 2).
• Factorize: x² – 4x + 3 = (x – 1)(x – 3).
• Roots: x = 1, x = 3.
• Graph intersects x-axis at 2 points.

Answer: Number of zeros = 2.

Solution:

The number of zeros is the number of x-axis intersections.

• Given: p(x) = x – 5 (linear, degree 1).
• Root: x = 5.
• Graph is a line intersecting x-axis at 1 point.

Answer: Number of zeros = 1.

Solution:

The number of zeros is the number of x-axis intersections.

• Given: p(x) = 2x² + 8 (quadratic, degree 2).
• Discriminant: b² – 4ac = 0 – 4(2)(8) = –64 < 0.
• No real roots, so the graph does not intersect x-axis.

Answer: Number of zeros = 0.

Solution:

The number of zeros is the number of x-axis intersections.

• Given: p(x) = x³ – 2x² (cubic, degree 3).
• Factorize: x²(x – 2).
• Roots: x = 0, x = 2.
• Graph intersects x-axis at 2 points (double root at x = 0).

Answer: Number of zeros = 2.

Solution:

The number of zeros is the number of x-axis intersections.

• Given: p(x) = x² – 6x + 9 (quadratic, degree 2).
• Factorize: (x – 3)².
• Root: x = 3 (double root).
• Graph touches x-axis at 1 point.

Answer: Number of zeros = 1.

Solution:

The number of zeros is the number of x-axis intersections.

• Given: p(x) = 3 (constant, degree 0).
• No variable, so no intersection with x-axis.

Answer: Number of zeros = 0.

Solution:

For a quadratic polynomial ax² + bx + c, zeros are values where p(x) = 0. Sum of zeros = –b/a, product of zeros = c/a.

• Given: p(x) = x² – 7x + 12 (a = 1, b = –7, c = 12).
• Factorize: x² – 7x + 12 = x² – 4x – 3x + 12 = (x – 4)(x – 3).
• Zeros: x = 3, x = 4.
• Sum of zeros: 3 + 4 = 7 = –(–7)/1 = –b/a.
• Product of zeros: 3 × 4 = 12 = 12/1 = c/a.
• Relationships verified.

Answer: Zeros are 3 and 4. Sum = 7 = –b/a, Product = 12 = c/a.

Solution:

A quadratic polynomial with zeros α and β is x² – (α + β)x + αβ.

• Zeros: α = 5, β = –2.
• Sum of zeros: α + β = 5 + (–2) = 3.
• Product of zeros: αβ = 5 × (–2) = –10.
• Polynomial: x² – (3)x + (–10) = x² – 3x – 10.

Answer: The quadratic polynomial is x² – 3x – 10.

Solution:

The division algorithm states: Dividend = Divisor × Quotient + Remainder.

• Divide 2x² + 7x + 3 by x + 2 using polynomial long division.
• Step 1: 2x² ÷ x = 2x. Multiply: 2x(x + 2) = 2x² + 4x.
• Subtract: (2x² + 7x + 3) – (2x² + 4x) = 3x + 3.
• Step 2: 3x ÷ x = 3. Multiply: 3(x + 2) = 3x + 6.
• Subtract: (3x + 3) – (3x + 6) = –3.
• Quotient = 2x + 3, Remainder = –3.
• Verify: 2x² + 7x + 3 = (x + 2)(2x + 3) + (–3).
• Right side: (x + 2)(2x + 3) = 2x² + 3x + 4x + 6 = 2x² + 7x + 6. Add –3: 2x² + 7x + 6 – 3 = 2x² + 7x + 3, which matches the dividend.

Answer: Quotient = 2x + 3, Remainder = –3. Division algorithm verified.

Solution:

Use polynomial long division.

• Divide x³ – 5x² + 6x + 2 by x – 1.
• Step 1: x³ ÷ x = x². Multiply: x²(x – 1) = x³ – x².
• Subtract: (x³ – 5x² + 6x + 2) – (x³ – x²) = –4x² + 6x + 2.
• Step 2: –4x² ÷ x = –4x. Multiply: –4x(x – 1) = –4x² + 4x.
• Subtract: (–4x² + 6x + 2) – (–4x² + 4x) = 2x + 2.
• Step 3: 2x ÷ x = 2. Multiply: 2(x – 1) = 2x – 2.
• Subtract: (2x + 2) – (2x – 2) = 4.
• Quotient = x² – 4x + 2, Remainder = 4.

Answer: Quotient = x² – 4x + 2, Remainder = 4.

Solution:

By the Factor Theorem, x – a is a factor of p(x) if p(a) = 0. Here, a = 3.

• Given: p(x) = x³ – 7x² + 14x – 6.
• Calculate p(3): 3³ – 7(3²) + 14(3) – 6 = 27 – 63 + 42 – 6 = 27 – 63 + 36 = 0.
• Since p(3) = 0, x – 3 is a factor.

Answer: x – 3 is a factor.

Solution:

By the Remainder Theorem, the remainder when p(x) is divided by x – a is p(a).

• Given: p(x) = x³ – 4x² + 5x – 2, a = 2.
• Calculate p(2): 2³ – 4(2²) + 5(2) – 2 = 8 – 16 + 10 – 2 = 0.

Answer: The remainder is 0.

Solution:

By the Remainder Theorem, the remainder when p(x) is divided by x – a is p(a). Here, x + 1 = x – (–1), so a = –1.

• Given: p(x) = x⁴ – 3x² + 2x + 1.
• Calculate p(–1): (–1)⁴ – 3(–1)² + 2(–1) + 1 = 1 – 3 – 2 + 1 = –3.

Answer: The remainder is –3.

Solution:

Verify zeros and use the sum of zeros for a cubic polynomial ax³ + bx² + cx + d, where sum = –b/a.

• Given: p(x) = x³ – 3x² + 2x (a = 1, b = –3).
• Verify x = 1: p(1) = 1³ – 3(1²) + 2(1) = 1 – 3 + 2 = 0. So, x = 1 is a zero.
• Verify x = 2: p(2) = 2³ – 3(2²) + 2(2) = 8 – 12 + 4 = 0. So, x = 2 is a zero.
• Sum of zeros: 1 + 2 + third zero = –(–3)/1 = 3.
• Third zero = 3 – (1 + 2) = 0.
• Verify x = 0: p(0) = 0 – 0 + 0 = 0. Confirmed.

Answer: Zeros are 1, 2, and 0.

Solution:

For a cubic polynomial ax³ + bx² + cx + d, sum of zeros = –b/a.

• Given: p(x) = x³ – 6x² + 11x – 6 (a = 1, b = –6).
• Zeros: 2, 3, and third zero.
• Sum of zeros: 2 + 3 + third zero = –(–6)/1 = 6.
• Third zero = 6 – (2 + 3) = 1.
• Verify: p(1) = 1³ – 6(1²) + 11(1) – 6 = 1 – 6 + 11 – 6 = 0. Confirmed.

Answer: The third zero is 1.

Solution:

A quadratic polynomial with zeros α and β is x² – (α + β)x + αβ.

• Zeros: α = 3 + √2, β = 3 – √2.
• Sum of zeros: α + β = (3 + √2) + (3 – √2) = 6.
• Product of zeros: αβ = (3 + √2)(3 – √2) = 9 – 2 = 7.
• Polynomial: x² – (6)x + 7 = x² – 6x + 7.

Answer: The quadratic polynomial is x² – 6x + 7.

Solution:

A quadratic polynomial with zeros α and β is x² – (α + β)x + αβ.

• Zeros: α = –3/4, β = 2/5.
• Sum of zeros: α + β = –3/4 + 2/5 = (–15 + 8)/20 = –7/20.
• Product of zeros: αβ = (–3/4) × (2/5) = –6/20 = –3/10.
• Polynomial: x² – (–7/20)x + (–3/10) = x² + (7/20)x – 3/10.
• To avoid fractions, multiply by 20: 20x² + 7x – 6.

Answer: The quadratic polynomial is 20x² + 7x – 6.

Solution:

For a quadratic polynomial ax² + bx + c, sum of zeros = –b/a, product of zeros = c/a.

• Given: p(x) = x² – kx + 6 (a = 1, b = –k, c = 6).
• Product of zeros = 6/1 = 6.
• Let zeros be 2m and 3m (ratio 2:3).
• Product: 2m × 3m = 6m² = 6 → m² = 1 → m = ±1.
• Case 1: m = 1. Zeros: 2, 3. Sum = 2 + 3 = 5 = –(–k)/1 = k.
• Case 2: m = –1. Zeros: –2, –3. Sum = –2 – 3 = –5 = k.
• Verify: For k = 5, zeros are 2, 3 (satisfies). For k = –5, zeros are –2, –3 (satisfies).

Answer: k = 5 or k = –5.