Tenth Grade Mathematics: Quadratic and Cubic Polynomial Formulation

While the preceding chapter focused on analyzing the zeros and coefficients of an existing polynomial, practical mathematics frequently demands the reverse operation: synthesizing an unknown polynomial when given its zeros or conditions on their sums and products. Whether designing roller-coaster curves, engineering trajectory paths, or establishing kinematic formulas, polynomial formulation provides the formal procedure for converting specific root constraints into fully developed algebraic equations.

Formulation of a Quadratic Polynomial

Suppose we are required to construct a quadratic polynomial whose zeros are alpha and beta. By the Factor Theorem, the polynomial must have (x - alpha) and (x - beta) as linear factors.

General Quadratic Formulation:
P(x) = k * (x - alpha) * (x - beta)
     = k * [x^2 - (alpha + beta)*x + alpha * beta]

where k is any non-zero real constant. In standard notation, letting S = alpha + beta (Sum of zeros) and P = alpha * beta (Product of zeros), the family of polynomials is given by:

P(x) = k * [x^2 - S*x + P],   where k != 0

Geometric Significance of the Parameter k:
All polynomials of the form k * (x^2 - S*x + P) share the exact same zeros
along the x-axis, but k scales the vertical steepness and opening direction:
                 y
                 |        k = 2 (steeper)
                 |   \   | k = 1 (standard)
                 |    \  |  /
           ------+-----+---+----+--- x
               alpha   |  beta
                 |     / | \   k = -1 (opens downwards)

If no additional constraint (such as a specific point through which the curve must pass) is provided, we commonly select k = 1 or choose k to eliminate fractional denominators, expressing the polynomial with integer coefficients.

Step-by-Step Construction Examples

Example 1: Given Sum S = -3 and Product P = 2.
Formulation: P(x) = k * [x^2 - S*x + P]
             P(x) = k * [x^2 - (-3)*x + 2]
             P(x) = k * (x^2 + 3*x + 2)
Choosing k = 1 gives the quadratic polynomial: x^2 + 3*x + 2.

Example 2: Given zeros alpha = 1/4 and beta = -1.
Sum S = 1/4 + (-1) = -3/4.
Product P = (1/4) * (-1) = -1/4.
Formulation: P(x) = k * [x^2 - (-3/4)*x + (-1/4)]
             P(x) = k * [x^2 + (3/4)*x - 1/4]
To clear denominators, choose k = 4:
             P(x) = 4 * [x^2 + (3/4)*x - 1/4] = 4*x^2 + 3*x - 1.

Formulation of a Cubic Polynomial

When three zeros—alpha, beta, and gamma—are specified, the polynomial is constructed by taking the product of three linear factors:

P(x) = k * (x - alpha) * (x - beta) * (x - gamma)

Expanding this product produces the general cubic formulation in terms of its three fundamental symmetric quantities:

General Cubic Formulation:
P(x) = k * [x^3 - S1*x^2 + S2*x - S3]

where:
S1 = alpha + beta + gamma                       (Sum of zeros)
S2 = alpha*beta + beta*gamma + gamma*alpha       (Sum of pairwise products)
S3 = alpha * beta * gamma                       (Product of all three zeros)
k  = any non-zero real constant

Example: Formulate a cubic polynomial whose zeros are 2, -3, and 5.
Step 1: Compute S1:  2 + (-3) + 5 = 4
Step 2: Compute S2:  2*(-3) + (-3)*5 + 5*2 = -6 - 15 + 10 = -11
Step 3: Compute S3:  2 * (-3) * 5 = -30
Step 4: Substitute into formulation:
        P(x) = k * [x^3 - (4)*x^2 + (-11)*x - (-30)]
        P(x) = k * [x^3 - 4*x^2 - 11*x + 30]
Choosing k = 1 gives: P(x) = x^3 - 4*x^2 - 11*x + 30.

Constructing Transformed Polynomials

A frequent challenge in algebra is to formulate a new polynomial whose roots are related to the roots of a known polynomial by a specific transformation (for example, roots that are reciprocal, doubled, or shifted).

Problem Pattern:
Given that alpha and beta are zeros of a*x^2 + b*x + c, formulate a quadratic
polynomial whose zeros are 2*alpha and 2*beta.

Methodology:
1. Find original S = alpha + beta = -b/a, and P = alpha*beta = c/a.
2. Find new sum S' = 2*alpha + 2*beta = 2*(alpha + beta) = 2*S = -2*b/a.
3. Find new product P' = (2*alpha) * (2*beta) = 4*(alpha*beta) = 4*P = 4*c/a.
4. Construct: Q(x) = x^2 - S'*x + P' = x^2 - (-2*b/a)*x + 4*c/a
                   = a*x^2 + 2*b*x + 4*c.

Chapter Exercises

Exercise 1: Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively: (a) 1/4, -1; (b) sqrt(2), 1/3; and (c) 0, sqrt(5).

Exercise 2: If alpha and beta are the zeros of the polynomial 2x^2 - 4x + 5, form a quadratic polynomial whose zeros are (2alpha + 1) and (2beta + 1).

Exercise 3: Formulate a cubic polynomial with integer coefficients whose zeros are 1/2, 1, and -2.

Solutions and Explanations

Solution 1: We use the formulation P(x) = k * [x^2 - Sx + P]. For (a), S = 1/4 and P = -1. P(x) = k * [x^2 - (1/4)x - 1]. Choosing k = 4 gives 4x^2 - x - 4. For (b), S = sqrt(2) and P = 1/3. P(x) = k * [x^2 - sqrt(2)x + 1/3]. Choosing k = 3 yields 3x^2 - 3sqrt(2)x + 1. For (c), S = 0 and P = sqrt(5). P(x) = k * [x^2 - 0x + sqrt(5)] = k * [x^2 + sqrt(5)]. Choosing k = 1 gives x^2 + sqrt(5).

Solution 2: For 2x^2 - 4x + 5, coefficients are a = 2, b = -4, c = 5. The original sum is alpha + beta = -(-4)/2 = 2, and the original product is alpha * beta = 5/2. The new zeros are u = 2alpha + 1 and v = 2beta + 1. The new sum is S' = u + v = 2alpha + 1 + 2beta + 1 = 2(alpha + beta) + 2 = 2(2) + 2 = 6. The new product is P' = u * v = (2alpha + 1)(2beta + 1) = 4alphabeta + 2(alpha + beta) + 1 = 4(5/2) + 2(2) + 1 = 10 + 4 + 1 = 15. The new polynomial is Q(x) = x^2 - S'x + P' = x^2 - 6x + 15.

Solution 3: The zeros are alpha = 1/2, beta = 1, gamma = -2. Compute S1: 1/2 + 1 - 2 = 3/2 - 2 = -1/2. Compute S2: (1/2)(1) + (1)(-2) + (-2)(1/2) = 1/2 - 2 - 1 = -5/2. Compute S3: (1/2) * (1) * (-2) = -1. Substitute into cubic formula: P(x) = k * [x^3 - (-1/2)x^2 + (-5/2)x - (-1)] = k * [x^3 + (1/2)x^2 - (5/2)x + 1]. To ensure integer coefficients, choose k = 2: P(x) = 2x^3 + x^2 - 5*x + 2.