Find the quadratic polynomial whose zeroes are

General
1

find the quadratic polynomial whose zeroes are

2

anonymous6 said find the quadratic polynomial whose zeroes are 3 and -4.

Quadratic Polynomial Formation

To find a quadratic polynomial given its zeroes, we can use the fact that if x_1 and x_2 are the roots (or zeroes) of a polynomial, then the polynomial can be expressed as:

$$ p(x) = a(x - x_1)(x - x_2) $$

where a is a non-zero constant that determines the “stretch” or “compression” of the polynomial, and (x - x_1)(x - x_2) represents the factors derived from the roots.

For the zeroes 3 and -4, we can substitute these values into the form above:

  1. Identify Zeroes: x_1 = 3, x_2 = -4.

  2. Substitute into General Form:
    $$ p(x) = a(x - 3)(x + 4) $$

  3. Expand the Expression: Use the distributive property to expand the expression:

    p(x) = a[(x - 3)(x + 4)] = a[x(x + 4) - 3(x + 4)]

  4. Continue Expansion:

    = a[x^2 + 4x - 3x - 12]

  5. Simplify Further:

    = a[x^2 + (4x - 3x) - 12] = a[x^2 + x - 12]

  6. Selecting ‘a’: Typically, a is chosen to be 1 for the simplest form of the polynomial unless specified otherwise. So, we have:

    p(x) = 1(x^2 + x - 12) = x^2 + x - 12

Now we have the quadratic polynomial whose zeroes are 3 and -4: x^2 + x - 12.

Verifying the Polynomial with Zero Factor Property

To confirm that x^2 + x - 12 is indeed the correct polynomial with the given zeroes, let’s check it with the Zero Factor Property:

  • Substitute x_1 = 3:
    $$ 3^2 + 3 - 12 = 9 + 3 - 12 = 0 $$

  • Substitute x_2 = -4:
    $$ (-4)^2 + (-4) - 12 = 16 - 4 - 12 = 0 $$

Both substitutions result in zero, confirming that 3 and -4 are roots of the polynomial.

Properties of Quadratic Polynomials

Quadratic polynomials can be examined using their standard form ax^2 + bx + c, where:

  1. Sum of Zeroes: The sum of the zeroes can be found using the formula:
    $$ x_1 + x_2 = -\frac{b}{a} $$

    Here, a = 1, b = 1, so:
    $$ 3 + (-4) = -1 = -\frac{1}{1} $$

  2. Product of Zeroes: The product of the zeroes is given by:
    $$ x_1 \cdot x_2 = \frac{c}{a} $$

    Given c = -12, a = 1, we have:
    $$ 3 \cdot (-4) = -12 = \frac{-12}{1} $$

Both properties validate the original polynomial construction.

Graphical Representation of Polynomials

When graphing x^2 + x - 12, it can be useful to note:

  • Vertex: This forms the peak or trough of the parabola and can be found using:
    $$ x = -\frac{b}{2a} $$

    For our polynomial:
    $$ x = -\frac{1}{2 \times 1} = -\frac{1}{2} $$

  • Axis of Symmetry: The line x = -\frac{1}{2} acts as the axis of symmetry for the parabola.

  • Intercepts: The y-intercept is the constant term c, which is -12.

  • Opening Direction: Since the leading coefficient a = 1 is positive, the parabola opens upwards.

  • Zeroes on Graph: The points (3,0) and (-4,0) are where the graph intersects the x-axis, coinciding with the polynomial’s zeroes.

Exploration Through Examples

  1. Example 1: Find the quadratic polynomial with zeroes 5 and -2.

    Solution:**
    Substitute x_1 = 5 and x_2 = -2 into the formula:
    $$ p(x) = a(x - 5)(x + 2) $$
    Simplify to:
    $$ p(x) = a[x^2 - 3x - 10] $$

    Commonly a = 1, so:
    $$ p(x) = x^2 - 3x - 10 $$

    Confirm using zero factor property for both zeroes: 5 and -2.

  2. Example 2: Find a quadratic polynomial with zeroes \frac{1}{2} and -3.

    Solution:
    Substitute x_1 = \frac{1}{2} and x_2 = -3:
    $$ p(x) = a\left(x - \frac{1}{2}\right)(x + 3) $$

    $$ p(x) = a\left(x^2 + \frac{3x}{2} - \frac{x}{2} - \frac{3}{2}\right) $$

    Simplify:
    $$ p(x) = a[x^2 + \frac{2x}{2} - \frac{3}{2}] $$
    $$ p(x) = a[x^2 + x - \frac{3}{2}] $$

    Letting a = 2 to remove the fraction, we have:
    $$ p(x) = 2x^2 + 2x - 3 $$

Each example reinforces the process of constructing a quadratic polynomial from its known zeroes, ensuring that we use the methods consistently, incorporating verification steps for accuracy.

@user