Ch 3Theory of Equations — Class 12 Mathematics, concept-first.

Solving polynomial equations is one of the oldest pursuits in mathematics — Sumerian and Babylonian scribes were already working with them around 2000 BCE, and mathematicians across Egypt, Greece, Ind…

Before building new tools, this section pins down the vocabulary this whole chapter is built on: what exactly a polynomial, a polynomial equation and a root are, and revisits the one polynomial equati…

For a non-negative integer , a polynomial of degree in one variable is an expression where the coefficients () are constants and the leading coefficient . The variable may be real or complex.

For the quadratic equation (), the quantity is the discriminant. The two roots are and , usually combined as

Vieta's Formulae, named for the French mathematician François Viète, relate the coefficients of a polynomial equation to sums and products of its roots — without ever needing to know the individual ro…

Let be the roots of . Since determine the polynomial up to the factor , Comparing coefficients of like powers on both sides gives Vieta's formula for a quadratic:

Everything learned for the quadratic case extends to a polynomial of any degree. Before writing the general Vieta relations, though, we need to pin down how many roots a degree- equation can even have…

If is a root of , then is a factor of , so . If are both roots, is a factor, so ; and in general, if has (distinct) roots, then .

Cubic case. Consider . By the Fundamental Theorem of Algebra it has exactly three roots (with multiplicity), so Comparing coefficients (valid since ): For a monic cubic () this reads especially cleanl…

Building an equation from given roots, directly. One way to write a degree- equation with roots is to just multiply out the factors: .

Vieta's formulae link coefficients to combinations of roots. This section asks a different question: what does the type of the coefficients (real, or more restrictively rational, or more restrictively…

For a quadratic with real coefficients, if is a root then is also a root — this section proves the same is true for any degree.

Restricting further to rational coefficients produces an analogous surd-conjugate result. For with rational and : when the (repeated) root is real and rational.

Restricting all the way to integer coefficients pins the discriminant test down further: for with integers, is automatically an integer, so So an integer-coefficient quadratic has rational roots exact…

Certain geometric facts are most cleanly proved using polynomial equations — reducing a geometry question ("how many points can these two curves share?") to an algebra question ("how many roots can th…

Even without an exact formula, a handful of general facts help locate the real roots of a higher-degree polynomial equation :

Not every higher-degree equation submits to Vieta's formula or the Rational Root Theorem cleanly. But very often extra information is available — either given outright ("one root is ") or spottable ju…

If is a known imaginary root of a real-coefficient quartic, the Complex Conjugate Root Theorem (§3.4.1) hands us for free — so and are both factors, and hence so is their product: Dividing the origina…

If has degree and involves only even powers of (every odd-power coefficient is ), substitute : this turns into a genuine degree- equation in .

The sum of the coefficients of is nothing other than (substitute everywhere — every power of is , so what remains is exactly the sum of the coefficients). So:

A companion test to §3.7.3: suppose the sum of the odd-power coefficients of equals the sum of the even-power coefficients.

Being told the roots of a cubic are in a specific progression hands over enough extra structure to solve it, by combining the assumed form with Vieta's relations (§3.3.2.2: , , ).

Quartic equations of the shape () can sometimes be rewritten so the four linear factors pair up into two quadratics that share the same leading and linear terms — say .

So far every technique in §3.7 needed some extra fact about the roots. This section covers the two main tools for when the equation is given completely bare, with nothing else known.

This is the systematic version of the guessing in §3.8: instead of testing arbitrary numbers, list every divisor of the constant term as a candidate numerator, every divisor of the leading coefficient…

Some equations have a special coefficient symmetry that lets a substitution do all the work, without needing to guess any individual root.

Some equations aren't polynomial equations at all — yet a well-chosen substitution converts them into one that genuinely is, which can then be solved by every tool above.

Every technique so far either needed extra information about the roots, or actually solved the equation.

Worked illustration. For (the term is absent, i.e. it has coefficient ), the sign of each nonzero coefficient in order (from down to the constant) is .

Descartes' Rule gives an upper bound, not an exact count — the next two sub-sections work through fully-solved polynomials to see when that bound is attained exactly and when it isn't, then combine th…

Worked illustration 1 (bound attained exactly). has roots . 's coefficient signs (nonzero ones) are : 2 sign changes, so at most positive roots. has signs : 1 sign change, so at most negative root.

Turning the two real-root bounds into a non-real-root bound. Let be the number of sign changes in (so at most positive roots) and the number of sign changes in (so at most negative roots).

In this chapter we studied:

Sample papers and previous-year board questions for this subject.