TN 12th Maths — Learn, Visualize, Practice

1. Polynomials, roots and multiplicity

A polynomial of degree $n$ in one variable is $P(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ , with the leading coefficient $a_n \ne 0$ . A number $c$ with $P(c) = 0$ is called a zero of the polynomial, or a root of the equation $P(x) = 0$ . If the leading coefficient is $1$ , the polynomial is called monic.

DefinitionFundamental Theorem of Algebra

Every polynomial equation $P(x) = 0$ of degree $n \ge 1$ has at least one root in $\mathbb{C}$ . Applying this repeatedly, a degree- $n$ equation has exactly $n$ roots in $\mathbb{C}$ when the roots are counted with their multiplicities.

So a degree- $n$ equation can have at most $n$ roots. It cannot have more, even after counting repeats — that upper bound is one of the most-used facts in the whole chapter.

DefinitionMultiplicity

$a$ is a root of multiplicity $k$ if $(x-a)^k$ divides $P(x)$ but $(x-a)^{k+1}$ does not. A root of multiplicity $1$ is called a simple root.

Example

Find the roots of

x^2 - 6x + 9 = 0

and state each multiplicity.

Example

How many roots (with multiplicity) does

x^3 - 7x^2 + 15x - 9 = 0

have, given

x=3

is a double root?

2. Quadratic equations and the discriminant

For $ax^2 + bx + c = 0$ the discriminant is $\Delta = b^2 - 4ac$ , and the roots are $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . When $a, b, c$ are real, $\Delta$ decides the nature of the roots before you solve.

Nature of the roots (real coefficients)

$\Delta > 0$ : two distinct real roots.
$\Delta = 0$ : equal real (repeated) roots.
$\Delta < 0$ : no real roots — a conjugate pair of imaginary roots.

Example

If

p

is real, discuss the nature of the roots of

4x^2 + 4px + (p+2) = 0

.

Example

Find

k

so that

x^2 + (2k+2)x + 9 = 0

has equal roots.

3. Vieta's formulae — roots ↔ coefficients

The roots of a polynomial are locked to its coefficients. You never need to solve the equation to know the sum or product of its roots — Vieta's formulae read them straight off. They come from expanding $a(x-\alpha)(x-\beta)\cdots$ and matching coefficients.

Quadratic ax² + bx + c = 0, roots α, β

\alpha+\beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}

Cubic ax³ + bx² + cx + d = 0, roots α, β, γ

\alpha+\beta+\gamma = -\dfrac{b}{a},\quad \alpha\beta+\beta\gamma+\gamma\alpha = \dfrac{c}{a},\quad \alpha\beta\gamma = -\dfrac{d}{a}

Quartic ax⁴ + bx³ + cx² + dx + e = 0, roots α, β, γ, δ

\textstyle\sum\alpha = -\dfrac{b}{a},\ \ \sum\alpha\beta = \dfrac{c}{a},\ \ \sum\alpha\beta\gamma = -\dfrac{d}{a},\ \ \alpha\beta\gamma\delta = \dfrac{e}{a}

Key idea: For a monic polynomial the pattern is simply: sum of roots $= -(\text{coeff of } x^{n-1})$, product of roots $= (-1)^n(\text{constant term})$, with alternating signs for the symmetric sums in between.

Example

The roots of

x^3 - 6x^2 + 11x - 6 = 0

are

\alpha, \beta, \gamma

. Find

\alpha+\beta+\gamma

,

\alpha\beta+\beta\gamma+\gamma\alpha

and

\alpha\beta\gamma

.

Example

If

\alpha, \beta, \gamma

are the roots of

x^3 + px^2 + qx + r = 0

, find

\sum\dfrac{1}{\beta\gamma}

in terms of the coefficients.

4. Symmetric functions of the roots

Many problems ask for a symmetric expression in the roots — $\sum\alpha^2$ , $\sum\alpha^2\beta$ , $\sum\dfrac{1}{\alpha}$ and so on. The trick is to rewrite each one using only $\sum\alpha$ , $\sum\alpha\beta$ , $\alpha\beta\gamma$ , which Vieta hands you directly. The single most useful identity is:

\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)

Example

Find the sum of the squares of the roots of

ax^4 + bx^3 + cx^2 + dx + e = 0

.

Example

The roots of

2x^2 - 7x + 13 = 0

are

\alpha, \beta

. Build a quadratic whose roots are

\alpha^2

and

\beta^2

.

Note: Handle repeated roots carefully in these sums. If a cubic has roots $1, 2, 2$ then $\sum\alpha\beta = (1)(2) + (2)(2) + (2)(1) = 8$, counting each pair once.

5. Forming an equation from its roots

Vieta's formulae run in reverse: given the roots, you can write down the polynomial without multiplying out the factors. For a quadratic with roots $\alpha, \beta$ :

x^2 - (\alpha+\beta)\,x + \alpha\beta = 0 \quad\text{i.e.}\quad x^2 - (\text{sum})x + (\text{product}) = 0

For a cubic with roots $\alpha, \beta, \gamma$ the pattern continues with alternating signs:

x^3 - (\textstyle\sum\alpha)x^2 + (\textstyle\sum\alpha\beta)x - \alpha\beta\gamma = 0

Example

Form the quadratic whose roots are

3

and

-2

.

Example

Construct a cubic equation with roots

1, 2, 3

.

6. Transforming the roots of an equation

A powerful idea: build a new equation whose roots are simple functions of the old roots — without ever finding those roots. The whole game is to express the new sum and product using the old $\sum\alpha$ and $\alpha\beta$ .

Common transformations of ax² + bx + c = 0

Roots increased by $h$ (new roots $\alpha+h,\ \beta+h$ ): replace $x$ by $x-h$ .
Roots multiplied by $k$ (new roots $k\alpha,\ k\beta$ ): replace $x$ by $\dfrac{x}{k}$ .
Reciprocal roots ( $\dfrac{1}{\alpha},\ \dfrac{1}{\beta}$ ): reverse the coefficients — $cx^2 + bx + a = 0$ .

Example

If

\alpha, \beta

are the roots of

17x^2 + 43x - 73 = 0

, form a quadratic whose roots are

\alpha+2

and

\beta+2

.

Example

If

\alpha, \beta, \gamma

are the roots of

x^3 + 2x^2 + 3x + 4 = 0

, form a cubic whose roots are

\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}

.

7. Imaginary and irrational (surd) roots

When the coefficients are restricted, non-real and irrational roots are forced to come in pairs. These theorems let you write down a second root for free — the key to reducing a high-degree problem.

DefinitionComplex Conjugate Root Theorem

If a polynomial has real coefficients and $a + bi$ ( $b \ne 0$ ) is a root, then its conjugate $a - bi$ is also a root.

DefinitionConjugate Surd Root Theorem

If a polynomial has rational coefficients and $p + \sqrt{q}$ (with $\sqrt{q}$ irrational) is a root, then $p - \sqrt{q}$ is also a root.

Note: An odd-degree polynomial with real coefficients must have at least one real root: imaginary roots pair up, and pairs cannot account for an odd total.

Example

Find the monic polynomial equation of minimum degree with real coefficients having

2 - 3i

as a root.

Example

Find a polynomial equation of minimum degree with rational coefficients having

2 - \sqrt{3}

as a root.

Watch out: 'Rational coefficients' is essential in the surd theorem. $x - (2+\sqrt3) = 0$ has $2+\sqrt3$ as a root but not $2-\sqrt3$ — because its coefficients are not all rational.

8. Using a known root to solve higher-degree equations

Once a conjugate pair is known, their factors multiply to a real (or rational) quadratic factor. Divide it out and the leftover quotient is easy to solve.

An imaginary pair $\alpha \pm i\beta$ gives the factor $x^2 - 2\alpha x + (\alpha^2+\beta^2)$ .
A surd pair $p \pm \sqrt{q}$ gives the factor $x^2 - 2px + (p^2 - q)$ .

Example

If

2 + i

and

3 - \sqrt{2}

are roots of

x^6 - 13x^5 + 62x^4 - 126x^3 + 65x^2 + 127x - 140 = 0

, find all roots.

9. Special polynomial equations

Certain patterns in the coefficients hand you a root immediately, or collapse the degree. Spotting them is often the whole trick to a higher-degree question.

Sum of all coefficients $= 0$ : then $P(1) = 0$ , so $x = 1$ is a root.
Sum of odd-power coefficients equals sum of even-power coefficients: then $P(-1) = 0$ , so $x = -1$ is a root.
Only even powers of $x$ appear: substitute $y = x^2$ to halve the degree, solve, then take square roots.

Example

Solve

x^3 - 3x^2 - 33x + 35 = 0

.

Example

Solve

2x^3 + 11x^2 - 9x - 18 = 0

.

Example

Solve the even-power equation

x^4 - 9x^2 + 20 = 0

.

10. Reciprocal equations

DefinitionReciprocal equation

An equation whose coefficients read the same forwards and backwards (Type I: $a_r = a_{n-r}$ ) or the same in magnitude but opposite in sign (Type II: $a_r = -a_{n-r}$ ). Its roots occur in reciprocal pairs $r$ and $\dfrac{1}{r}$ .

Standard results

Odd degree, Type I: $x = -1$ is always a root.
Odd degree, Type II: $x = 1$ is always a root.
Even degree, Type II: $x = 1$ and $x = -1$ are roots (the middle term is $0$ ).
Even degree: divide through by $x^{n/2}$ and set $y = x + \dfrac{1}{x}$ (or $x - \dfrac{1}{x}$ ) to halve the degree.

Example

Solve

7x^3 - 43x^2 = 43x - 7

.

Example

Solve

x^4 - 10x^3 + 26x^2 - 10x + 1 = 0

.

11. Rational Root Theorem

When the coefficients are integers, this theorem turns root-guessing into checking a short finite list.

DefinitionRational Root Theorem

For a polynomial with integer coefficients, any rational root $\dfrac{p}{q}$ in lowest terms has $p$ dividing the constant term $a_0$ and $q$ dividing the leading coefficient $a_n$ .

Note: A monic integer polynomial ($a_n = 1$) can only have integer rational roots, and each must divide the constant term. The theorem only lists candidates — it does not guarantee any of them is actually a root; you must test them.

Example

Find the roots of

2x^3 + 3x^2 + 2x + 3 = 0

.

Example

List all candidate rational roots of

2x^3 + x - 6 = 0

.

12. Descartes' Rule of Signs

Before solving, you can bound how many positive and negative real roots are even possible, just by counting sign changes in the coefficients.

DefinitionDescartes' Rule of Signs

If $s$ is the number of sign changes in $P(x)$ and $p$ is the number of positive real roots, then $s - p$ is a non-negative even integer. Apply the same rule to $P(-x)$ to bound the negative real roots.

Key idea: The number of positive roots is $s$, or less than $s$ by an even number. A lower bound on imaginary roots is $n - (m + k)$, where $m, k$ are the sign changes in $P(x)$ and $P(-x)$.

Example

Discuss the positive and negative real roots of

P(x) = x^3 - 2x^2 + x - 5

.

Example

Show that

P(x) = 9x^9 + 2x^5 - 7x^4 - 29x^2 + 9

has at least six imaginary roots.

Note: Descartes' rule gives only upper bounds (and a parity condition) — never the exact roots. Combine it with the sum-of-coefficients trick to pin down exact counts in special cases.