The Coefficients Already Know

A polynomial's degree, the highest power of its variable, gives it a name. 5t - 9 has degree 1, so it is a linear polynomial. 3v^2 + v - 6 has degree 2, a quadratic polynomial. -2z^3 + z - 7 has degree 3, a cubic polynomial. Not every expression built from a variable counts as a polynomial, though: 1/(x+1) and sqrt(x) + 2 are not polynomials, because a polynomial only allows whole-number, non-negative powers of its variable, never a variable in a denominator or under a root. The general quadratic is written ax^2 + bx + c, with a not equal to 0 (otherwise it would collapse to a linear polynomial), and the general cubic is ax^3 + bx^2 + cx + d, again with a not equal to 0.

If p(x) is a polynomial and k is a real number, then p(k) means substituting k for x throughout the expression. A number k is called a zero of p(x) if p(k) = 0. Consider p(x) = x^2 - x - 6. Substituting x = -2 gives p(-2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0, so -2 is a zero. Substituting x = 3 gives p(3) = 9 - 3 - 6 = 0, so 3 is a zero too. Both -2 and 3 are zeroes of x^2 - x - 6, and, as will soon become clear, the coefficients of this very polynomial already contain a shortcut to facts about these two numbers, without needing to find them by substitution at all.

Every zero of a polynomial p(x) is exactly the x-coordinate of a point where the graph of y = p(x) crosses the x-axis, since y = 0 there. A linear polynomial's graph is a straight line, and a straight line (that isn't horizontal) crosses the x-axis at exactly one point, so a linear polynomial always has exactly one zero. A quadratic polynomial's graph is a parabola, which can meet the x-axis in three different ways: at two distinct points (two zeroes), touching it at just one point (one repeated zero), or missing it completely (no real zero at all). A cubic polynomial's graph can cross the x-axis up to three times. In general, a polynomial of degree n has AT MOST n zeroes, never more, because its graph cannot cross the x-axis more than n times.

Take the quadratic p(x) = 4x^2 - 4x - 3. Splitting the middle term needs two numbers multiplying to 4 x (-3) = -12 and adding to -4: those numbers are -6 and 2. So 4x^2 - 6x + 2x - 3 = 2x(2x - 3) + 1(2x - 3) = (2x + 1)(2x - 3), giving zeroes x = -1/2 and x = 3/2. Now compare these zeroes with the original coefficients a=4, b=-4, c=-3. The sum of the zeroes is -1/2 + 3/2 = 1, and -b/a = -(-4)/4 = 1: they match. The product of the zeroes is (-1/2)(3/2) = -3/4, and c/a = -3/4: they match too. This is not a coincidence for this one polynomial; it holds for every quadratic.

Why does this always work? If alpha and beta are the zeroes of ax^2 + bx + c, then (x - alpha) and (x - beta) must be its factors, so ax^2 + bx + c = k(x - alpha)(x - beta) for some constant k. Expanding the right side: k(x - alpha)(x - beta) = kx^2 - k(alpha + beta)x + k(alpha)(beta). Comparing the coefficients of x^2, x, and the constant term on both sides gives a = k, b = -k(alpha + beta), and c = k(alpha)(beta). Substituting k = a into the second and third equations and rearranging gives alpha + beta = -b/a and (alpha)(beta) = c/a exactly. This is a proof, not just a pattern spotted in examples: it holds for every quadratic with real coefficients, whatever alpha and beta turn out to be.

The relationship also runs in reverse: given only the sum and product of two zeroes, a matching quadratic can be built without ever knowing the zeroes as separate numbers. If the sum is 5 and the product is -6, the polynomial x^2 - 5x - 6 fits, since -b/a = -(-5)/1 = 5 and c/a = -6/1 = -6. Checking by factorising: x^2 - 5x - 6 needs two numbers adding to -5 and multiplying to -6, namely -6 and 1, giving (x - 6)(x + 1) = 0, so the zeroes are 6 and -1. Indeed 6 + (-1) = 5 and 6 x (-1) = -6, confirming the construction. Any real multiple of this polynomial, such as 2x^2 - 10x - 12, has the exact same zeroes and therefore the exact same sum and product too.

Does a similar relationship exist for a cubic? Consider p(x) = x^3 - 6x^2 + 11x - 6, which has zeroes 1, 2 and 3 (checked directly: (x-1)(x-2)(x-3) expands to exactly this polynomial). For a cubic ax^3 + bx^2 + cx + d with zeroes alpha, beta and gamma, there are now THREE relationships: the sum alpha+beta+gamma = -b/a, the sum of pairwise products (alpha)(beta) + (beta)(gamma) + (alpha)(gamma) = c/a, and the full product (alpha)(beta)(gamma) = -d/a. Checking against 1, 2, 3: the sum is 1+2+3=6, matching -b/a = -(-6)/1 = 6. The pairwise-product sum is (1)(2)+(2)(3)+(1)(3) = 2+6+3 = 11, matching c/a = 11/1 = 11. The full product is 1x2x3=6, matching -d/a = -(-6)/1 = 6. Interesting as this is, CBSE's own Class 10 syllabus flags the cubic relationships as beyond what is examined directly - worth knowing exists, without needing to memorise for a test.