1. Building Polynomials with Specified Zeros Step 1: Learning Polynomial Building - Provided zeros: -3 and 5 - Simple polynomial form: (x + 3)(x - 5) - Expanding: x² - 2x - 15 Step 2: Freedom Degree Polynomials may be formed by multiplying the simple form by any non-zero constant. Possible Polynomials:Read more

    Building Polynomials with Specified Zeros

    Step 1: Learning Polynomial Building
    – Provided zeros: -3 and 5
    – Simple polynomial form: (x + 3)(x – 5)
    – Expanding: x² – 2x – 15

    Step 2: Freedom Degree
    Polynomials may be formed by multiplying the simple form by any non-zero constant.

    Possible Polynomials:
    1. x² – 2x – 15
    2. 2x² – 4x – 30
    3. 3x² – 6x – 45

    General Form:
    For any non-zero constant k:
    k(x² – 2x – 15)

    Mathematical Insight:
    – The k is the parameter for infinite scaling of the polynomial
    – Each scale produces a different polynomial with the same zeros
    – These are actually the same polynomial to a constant factor

    Step 3: Counting Polynomials
    – We can make polynomials with these zeros by multiplying the fundamental form
    by any real number
    – This produces an infinite number of polynomials

    Conclusion
    There are MORE THAN 3 polynomials with zeros -3 and 5.

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  2. Building a Quadratic Polynomial with Given Zeros Step 1: Defining Zeros - Given zeros: 4 and -3 - The polynomial shall thus be in the form: (x - 4)(x + 3) Step 2: Algebraic Expansion (x - 4)(x + 3) = x² + 3x - 4x - 12 = x² - x - 12 Step 3: Checking Zero Properties Let's test if the zeros hold: - WheRead more

    Building a Quadratic Polynomial with Given Zeros

    Step 1: Defining Zeros
    – Given zeros: 4 and -3
    – The polynomial shall thus be in the form: (x – 4)(x + 3)

    Step 2: Algebraic Expansion
    (x – 4)(x + 3) = x² + 3x – 4x – 12
    = x² – x – 12

    Step 3: Checking Zero Properties
    Let’s test if the zeros hold:
    – When x = 4:
    4² – 4 – 12 = 16 – 4 – 12 = 0 OK
    – When x = -3:
    (-3)² – (-3) – 12 = 9 + 3 – 12 = 0 OK

    Step 4: Coefficient Analysis
    – Coefficient of x²: 1
    – Coefficient of x: -1
    – Constant term: -12

    Mathematical Insights:
    – The zeros of a quadratic define its form
    – The general form illustrates how the zeros are connected to the coefficients of the polynomial
    – Vieta’s formulas support the connection between coefficients and zeros

    Conclusion:
    The quadratic polynomial whose zeros are 4 and -3 is x² – x – 12.

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  3. Solving for the Unknown Zero of a Quadratic Polynomial Step 1: Given Information - Polynomial: 2x² + 7x + 3 - One known zero: -3/2 Step 2: Vieta's Formulas for Quadratic Polynomials Let the two zeros be p and q: - p + q = -b/a - p * q = c/a Where in 2x² + 7x + 3: - a = 2 - b = 7 - c = 3 Step 3: UsinRead more

    Solving for the Unknown Zero of a Quadratic Polynomial

    Step 1: Given Information
    – Polynomial: 2x² + 7x + 3
    – One known zero: -3/2

    Step 2: Vieta’s Formulas for Quadratic Polynomials
    Let the two zeros be p and q:
    – p + q = -b/a
    – p * q = c/a

    Where in 2x² + 7x + 3:
    – a = 2
    – b = 7
    – c = 3

    Step 3: Using the Known Zero
    If p = -3/2, then we need to find q

    Sum of zeros formula:
    p + q = -b/a
    -3/2 + q = -7/2

    Step 4: Solving for the Unknown Zero
    q = -7/2 – (-3/2)
    = -7/2 + 3/2
    = -4/2
    = -2

    Verification:
    – First zero: p = -3/2
    – Second zero: q = -2
    – Check: (-3/2) + (-2) = -7/2 ✓
    – Check: (-3/2) * (-2) = 3/a ✓

    Mathematical Insight:
    Vieta’s formulas provide a powerful method to find
    polynomial zeros without fully solving the equation.

    Conclusion:
    The other zero of the polynomial is -2 (or 1 in the given options).

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  4. Determining the Remainder of Polynomial Division Step 1: Remainder Theorem Basics - The remainder theorem says that when a polynomial f(x) is divided by (x - a), the remainder is f(a) - This implies we can determine the remainder by evaluating the polynomial at x = 2 Step 2: Evaluating a PolynomialRead more

    Determining the Remainder of Polynomial Division

    Step 1: Remainder Theorem Basics
    – The remainder theorem says that when a polynomial f(x) is divided by (x – a),
    the remainder is f(a)
    – This implies we can determine the remainder by evaluating the polynomial at x = 2

    Step 2: Evaluating a Polynomial
    Polynomial: f(x) = x³ – 6x² + 11x – 6
    Substituting x = 2:

    f(2) = 2³ – 6(2)² + 11(2) – 6
    = 8 – 6(4) + 22 – 6
    = 8 – 24 + 22 – 6
    = 0

    Mathematical Insight:
    – By simply putting the root of the divisor (2) in the polynomial
    – We can easily find the remainder without long division
    – It is a very strong method that makes polynomial remainder computation easier

    Step 3: Confirmation
    – The remainder is 0
    – It indicates (x – 2) evenly divides the polynomial
    – There is no remaind value when divided

    Conclusion:
    The remainder upon division of x³ – 6x² + 11x – 6 by (x – 2) is zero.

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  5. Constructing a Quadratic Polynomial with Specific Zero Properties Step 1: Understanding Vieta's Formulas For a quadratic polynomial ax² + bx + c with zeros p and q: - Sum of zeros: p + q = -b/a - Product of zeros: p * q = c/a Given Conditions: - Sum of zeros = -4 - Product of zeros = 3 Step 2: AnalyRead more

    Constructing a Quadratic Polynomial with Specific Zero Properties

    Step 1: Understanding Vieta’s Formulas
    For a quadratic polynomial ax² + bx + c with zeros p and q:
    – Sum of zeros: p + q = -b/a
    – Product of zeros: p * q = c/a

    Given Conditions:
    – Sum of zeros = -4
    – Product of zeros = 3

    Step 2: Analyzing the Coefficients
    Let’s consider a standard quadratic form: x² + 4x + c

    Checking Sum of Zeros:
    – p + q = -4
    – This means the coefficient of x must be -4

    Checking Product of Zeros:
    – p * q = 3
    – This means the constant term must be 3

    Step 3: Verification
    The polynomial becomes: x² – 4x + 3

    Mathematical Verification:
    Let’s find the zeros using the quadratic formula:
    x = [4 ± √(16 – 4(1)(3))] / 2(1)
    = [4 ± √(16 – 12)] / 2
    = [4 ± √4] / 2
    = [4 ± 2] / 2

    Zeros are:
    – p = (4 + 2)/2 = 3
    – q = (4 – 2)/2 = 1

    Checking Conditions:
    – Sum of zeros: 3 + 1 = 4 ✓
    – Product of zeros: 3 * 1 = 3 ✓

    Key Insights:
    – Vieta’s formulas provide a powerful way to relate
    zeros to polynomial coefficients
    – We can construct polynomials by understanding
    the relationships between zeros and coefficients

    Conclusion:
    The polynomial that satisfies the given conditions is x² – 4x + 3.

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