Forming Quadratic Polynomial from Zeroes

If alpha and beta are the zeroes of a quadratic polynomial, then the polynomial can be expressed as:

p(x) = k[x^2 - (alpha + beta)x + alpha * beta]

where k is any non-zero real number.

In other words: p(x) = k[x^2 - (sum of zeroes)x + (product of zeroes)]

This formula comes from the fact that if alpha and beta are zeroes, then (x - alpha) and (x - beta) are factors. So:

p(x) = k(x - alpha)(x - beta) = k[x^2 - (alpha + beta)x + alpha*beta]

The constant k allows for infinitely many polynomials with the same zeroes. For example, x^2 - 5x + 6, 2x^2 - 10x + 12, and -3x^2 + 15x - 18 all have the same zeroes (2 and 3) but differ by the value of k (which is 1, 2, and -3 respectively).

Convention: In Class 10 problems, unless otherwise specified, we usually take k = 1 (if the result has integer coefficients) or choose k to eliminate fractions in the coefficients. The simplest polynomial (with smallest positive integer coefficients and leading coefficient positive) is preferred.

Important note: Two polynomials have the same zeroes if and only if one is a non-zero scalar multiple of the other. The zeroes determine the polynomial up to a constant factor.

The formula for forming a quadratic polynomial from its zeroes is derived from the Factor Theorem and the relationship between zeroes and coefficients.

Starting point: If alpha is a zero of p(x), then p(alpha) = 0, which means (x - alpha) is a factor of p(x) (Factor Theorem).

Step 1: If alpha and beta are the two zeroes of a quadratic polynomial, then both (x - alpha) and (x - beta) are factors. Since the polynomial has degree 2, it can have at most two linear factors. Therefore:

p(x) = a(x - alpha)(x - beta)

where 'a' is the leading coefficient (a non-zero constant).

Step 2: Expand the product:

p(x) = a[x^2 - beta*x - alpha*x + alpha*beta]

= a[x^2 - (alpha + beta)x + alpha*beta]

Step 3: Comparing with the standard form ax^2 + bx + c:

Coefficient of x = -a(alpha + beta), so b = -a(alpha + beta), giving alpha + beta = -b/a.

Constant term = a * alpha * beta, so c = a * alpha * beta, giving alpha * beta = c/a.

These are precisely the sum and product of zeroes formulas we already know, confirming the consistency of the derivation.

Step 4: Writing the formula with k instead of a (to emphasise that k is a free parameter):

p(x) = k[x^2 - (sum of zeroes)x + (product of zeroes)]

This is the general formula for forming a quadratic polynomial from its zeroes.

Why k can be any non-zero constant: Multiplying a polynomial by a non-zero constant does not change its zeroes. If p(alpha) = 0, then k * p(alpha) = k * 0 = 0. So k * p(x) has the same zeroes as p(x).

Different types of zeroes lead to different forms of the resulting polynomial:

Type 1: Integer Zeroes

Example: Zeroes are 4 and -3. Sum = 1, Product = -12. Polynomial: x^2 - x - 12. This is the simplest case — direct substitution gives integer coefficients.

Type 2: Fractional Zeroes

Example: Zeroes are 1/2 and -3/4. Sum = 1/2 + (-3/4) = -1/4. Product = (1/2)(-3/4) = -3/8. Polynomial: k[x^2 - (-1/4)x + (-3/8)] = k[x^2 + x/4 - 3/8]. Taking k = 8: 8x^2 + 2x - 3.

Type 3: Irrational Conjugate Zeroes

Example: Zeroes are 2 + sqrt(3) and 2 - sqrt(3). Sum = 4. Product = (2)^2 - (sqrt(3))^2 = 4 - 3 = 1. Polynomial: x^2 - 4x + 1. Note that irrational zeroes of a polynomial with rational coefficients always come in conjugate pairs.

Type 4: One Zero is Zero

Example: Zeroes are 0 and 5. Sum = 5, Product = 0. Polynomial: x^2 - 5x. The constant term is 0, so the polynomial has no constant term.

Type 5: Equal Zeroes

Example: Both zeroes are 3. Sum = 6, Product = 9. Polynomial: x^2 - 6x + 9 = (x - 3)^2. This is a perfect square trinomial.

Type 6: Zeroes with a Given Relationship

Sometimes the zeroes are not given directly but a relationship is stated (e.g., one zero is double the other, or zeroes differ by 3). The sum and product formulas are used to set up and solve equations to find the actual zeroes or the coefficients.

Method 1: Direct Substitution into the Formula

Step 1: Find sum S = alpha + beta. Step 2: Find product P = alpha * beta. Step 3: Write p(x) = k(x^2 - Sx + P). Step 4: Choose k for simplest integer coefficients.

Method 2: Factor Form Expansion

Step 1: Write p(x) = k(x - alpha)(x - beta). Step 2: Expand using FOIL or distributive property. Step 3: Simplify.

This method is sometimes easier when the zeroes are simple numbers.

Method 3: Using Given Conditions

When zeroes are not directly given but described by conditions (e.g., sum = 5, product = 6), directly substitute into the formula without finding individual zeroes.

Method 4: For Cubic Polynomials

Step 1: Find S1 = alpha + beta + gamma (sum). Step 2: Find S2 = alpha*beta + beta*gamma + gamma*alpha (sum of products in pairs). Step 3: Find P = alpha*beta*gamma (product). Step 4: Write p(x) = k(x^3 - S1*x^2 + S2*x - P).

Forming polynomials from their zeroes is a fundamental technique with applications across mathematics and science.

Constructing Equations from Solutions: In many real-world problems, the solutions (zeroes) are known from experiments or observations, and the governing equation needs to be reconstructed. For example, if a ball hits the ground at t = 0 and t = 4 seconds, the height function h(t) can be modelled as h(t) = k * t * (t - 4).

Curve Fitting: In data science and statistics, polynomial curves are fitted to data points. If the data crosses zero at known values, the polynomial can be constructed using these zeroes as a starting point.

Signal Processing: In electronics and signal processing, filters are designed using polynomials. The zeroes of the polynomial correspond to frequencies that are blocked by the filter. Engineers construct the filter polynomial from the desired zero frequencies.

Competitive Examinations: JEE, NTSE, and Olympiad exams frequently test the ability to form polynomials from conditions on zeroes (sum, product, difference, ratio, etc.). This skill is also foundational for higher mathematics topics like polynomial rings and field theory.

Error Checking: After solving a quadratic equation, you can reconstruct the polynomial from the solutions you found and compare it with the original polynomial to verify your answers.