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Advanced Algebra โ€ข Complete Guide

Table of Contents

Foundations &
Classical Algebra

A structured mathematical guide covering equations, roots, reciprocal equations, Descartes’ Rule of Signs, cubic equations and classical methods for solving quartic equations.

Designed & presented by Grandmaster Bikram Sutradhar

What You Will Master

  • โœ“ Transformation of Equations
  • โœ“ Reciprocal Equations
  • โœ“ Descartes’ Rule of Signs
  • โœ“ Theory of Roots
  • โœ“ Cardano’s Cubic Method
  • โœ“ Ferrari’s Quartic Method
  • โœ“ Descartes’ Quartic Method
  • โœ“ Exam-Oriented Problem Solving
0
Concepts
0
MCQs
0
Examples
0
Core Topics
Course Map

Complete Algebra Roadmap

Study the topics in the following sequence.

01

Transformation

Learn how equations change under substitutions and transformations.

02

Reciprocal Equations

Understand reciprocal or palindromic polynomial equations.

03

Descartes’ Rule

Determine possible numbers of positive and negative real roots.

04

Theory of Roots

Master relations between coefficients and roots.

05

Cardano’s Method

Classical solution of cubic equations.

06

Ferrari’s Method

Classical method for solving quartic equations.

07

Descartes’ Quartic Method

A classical alternative approach to quartic equations.

Master Notes

Foundations & Classical Algebra

1. Transformation of Equations

Transformation of an equation means changing the form of an equation without losing the relationship between its roots and the transformed variable.

Important Types

  • Changing the variable.
  • Changing the roots by adding or subtracting a constant.
  • Multiplying or dividing the roots by a constant.
  • Replacing x by a function of another variable.
  • Reciprocal transformation.

Root Translation

Suppose

f(x) = 0

and put

x = y + a

Then the equation in y has roots related to the original roots by

y = x – a
Example:

Transform
xยฒ – 6x + 5 = 0
Put
x = y + 3
Then
(y+3)ยฒ – 6(y+3) + 5 = 0
Therefore
yยฒ – 4 = 0
Hence
y = ยฑ2
and therefore
x = y + 3 = 5, 1
Exam Tip: When an equation contains a large xยฒ and x coefficient, the substitution x = y – b/(2a) can remove the quadratic term from a cubic after suitable normalization.

2. Reciprocal Equations

A reciprocal equation is a polynomial equation whose coefficients show symmetry when written in reverse order.

Example

xโด + 3xยณ + 5xยฒ + 3x + 1 = 0
Notice that the coefficients are
1, 3, 5, 3, 1
which read the same forwards and backwards.For x โ‰  0, divide by xยฒ:
xยฒ + 3x + 5 + 3/x + 1/xยฒ = 0
Group the reciprocal terms:
(xยฒ + 1/xยฒ) + 3(x + 1/x) + 5 = 0
Use
xยฒ + 1/xยฒ = (x + 1/x)ยฒ – 2
Put
y = x + 1/x
Then
yยฒ + 3y + 3 = 0
So a reciprocal quartic is reduced to a quadratic equation.
Key substitution:
For even-degree reciprocal equations use

y = x + 1/x

when the polynomial is symmetric.

3. Descartes’ Rule of Signs

Descartes’ Rule of Signs gives the possible number of positive and negative real roots of a polynomial.

Positive Roots

Arrange the polynomial in descending powers of x and count the changes of sign in the coefficients.

For
f(x) = xโด – 3xยณ + 2xยฒ + 5x – 7
Signs:
+ , โˆ’ , + , + , โˆ’
Sign changes:
+ to โˆ’ = 1
โˆ’ to + = 1
+ to + = 0
+ to โˆ’ = 1
Total = 3.Therefore the number of positive real roots is
3 or 1
because the number differs from 3 by an even integer.

Negative Roots

To investigate negative roots, calculate f(-x) and count sign changes.

If
f(x)=xยณ-6xยฒ+11x-6
then
f(-x)=-xยณ-6xยฒ-11x-6
There are no sign changes.Therefore there are
0 negative real roots.
Important: Descartes’ Rule gives the possible number of real roots, not their exact values.

4. Elementary Theorems on Roots

Consider the polynomial

aโ‚€xโฟ + aโ‚xโฟโปยน + aโ‚‚xโฟโปยฒ + … + aโ‚™ = 0
with roots
ฮฑโ‚, ฮฑโ‚‚, …, ฮฑโ‚™.
### Sum of Roots
ฮฑโ‚ + ฮฑโ‚‚ + … + ฮฑโ‚™ = โˆ’aโ‚/aโ‚€
### Sum of Pairwise Products
ฮฃฮฑแตขฮฑโฑผ = aโ‚‚/aโ‚€
### Product of Roots
ฮฑโ‚ฮฑโ‚‚…ฮฑโ‚™ = (โˆ’1)โฟ aโ‚™/aโ‚€

Cubic Equation

For
axยณ + bxยฒ + cx + d = 0
if roots are ฮฑ, ฮฒ, ฮณ:
ฮฑ + ฮฒ + ฮณ = โˆ’b/a

ฮฑฮฒ + ฮฒฮณ + ฮณฮฑ = c/a

ฮฑฮฒฮณ = โˆ’d/a

Quadratic Example

2xยฒ – 7x + 3 = 0
If roots are ฮฑ and ฮฒ:
ฮฑ + ฮฒ = 7/2
ฮฑฮฒ = 3/2
These relations are extremely useful when a question asks for expressions such as ฮฑยฒ + ฮฒยฒ, 1/ฮฑ + 1/ฮฒ, or ฮฑยณ + ฮฒยณ.

5. Cardano’s Method for Cubic Equations

Cardano’s method provides a classical formula for solving a general cubic equation.

Step 1: Start with the cubic

axยณ + bxยฒ + cx + d = 0
Divide by a:
xยณ + Axยฒ + Bx + C = 0

Step 2: Remove the xยฒ term

Put
x = y โˆ’ A/3
The equation becomes
yยณ + py + q = 0
where
p = B โˆ’ Aยฒ/3

q = 2Aยณ/27 โˆ’ AB/3 + C
This is called the depressed cubic.

Step 3: Cardano substitution

Put
y = u + v
Then
(u+v)ยณ + p(u+v) + q = 0
Using
uยณ + vยณ + (3uv+p)(u+v)+q=0
choose
3uv + p = 0
Therefore
uv = โˆ’p/3
and
uยณ + vยณ = โˆ’q
Hence uยณ and vยณ are roots of
tยฒ + qt โˆ’ pยณ/27 = 0
Thus
uยณ = โˆ’q/2 + โˆš(qยฒ/4 + pยณ/27)
and
vยณ = โˆ’q/2 โˆ’ โˆš(qยฒ/4 + pยณ/27)
Therefore one root is
x = โˆ›[โˆ’q/2 + โˆš(qยฒ/4+pยณ/27)] + โˆ›[โˆ’q/2 โˆ’ โˆš(qยฒ/4+pยณ/27)] โˆ’ A/3
Example:Solve
xยณ โˆ’ 6x โˆ’ 20 = 0
Here
p = โˆ’6
q = โˆ’20
Therefore
qยฒ/4 + pยณ/27 = 100 โˆ’ 8 = 92
So Cardano gives the real root through
x = โˆ›(10 + โˆš92) + โˆ›(10 โˆ’ โˆš92)
The expression can then be simplified or numerically evaluated.
Important: Cardano’s formula may pass through complex numbers even when all three roots are real. This is historically known as the casus irreducibilis.

6. Ferrari’s Method for Quartic Equations

Ferrari’s method is a classical procedure for solving a general fourth-degree equation.

General quartic

axโด + bxยณ + cxยฒ + dx + e = 0
First divide by a:
xโด + Axยณ + Bxยฒ + Cx + D = 0

Remove the cubic term

Use
x = y โˆ’ A/4
The quartic becomes
yโด + pyยฒ + qy + r = 0
This is called a depressed quartic.

Ferrari’s Main Idea

Rewrite
yโด + pyยฒ + qy + r = 0
so that one side can become a perfect square.The central strategy is to introduce an auxiliary quantity m and construct an identity of the form
(yยฒ + m)ยฒ = (linear expression in y)ยฒ
The resulting condition for m produces a cubic resolvent equation.That cubic can be solved using Cardano’s method.

Conceptual Flow

1
Normalize the quartic.
2
Eliminate the cubic term.
3
Convert the equation into a depressed quartic.
4
Construct a perfect-square expression.
5
Obtain the resolvent cubic.
6
Solve the resolvent cubic using Cardano’s method.
7
Factor the quartic into two quadratic equations.
8
Solve the two quadratics.
Memory Trick:
Quartic โ†’ Depress โ†’ Perfect Square โ†’ Resolvent Cubic โ†’ Cardano โ†’ Two Quadratics.

7. Descartes’ Method of Solving Quartics

Classical treatments of quartic equations also include a method associated with Descartes. The general idea is to transform the quartic into a form where it can be decomposed into quadratic factors.

Consider a depressed quartic:

xโด + pxยฒ + qx + r = 0
A useful factorization target is
(xยฒ + ax + b)(xยฒ โˆ’ ax + c)=0
Expanding:
xโด + (b+c-aยฒ)xยฒ + a(c-b)x + bc = 0
Therefore,
b+c-aยฒ = p
a(c-b)=q
bc=r
The problem is reduced to finding suitable values of a, b and c.This ultimately leads to an auxiliary cubic equation.
Core idea: A quartic can often be attacked by finding a suitable factorization into two quadratic expressions.

โšก Quick Revision Formula Sheet

TopicKey Formula / Idea
Quadratic rootsฮฑ+ฮฒ = โˆ’b/a, ฮฑฮฒ = c/a
Cubic rootsฮฑ+ฮฒ+ฮณ = โˆ’b/a
Cubic productฮฑฮฒฮณ = โˆ’d/a
Reciprocal equationUse y = x + 1/x for symmetric quartics
Descartes RuleSign changes determine possible positive/negative roots
Depressed cubicx = y โˆ’ A/3
Depressed cubicyยณ + py + q = 0
Cardanoy=u+v, uv=โˆ’p/3
Depressed quarticx = y โˆ’ A/4
FerrariQuartic โ†’ resolvent cubic โ†’ quadratic factors
Online Test

Foundations & Classical Algebra MCQ Test

Attempt all questions and check your score instantly.

1. For axยฒ + bx + c = 0, the sum of roots is:

2. The product of roots of axยฒ+bx+c=0 is:

3. In Descartes’ Rule of Signs, the number of positive roots is determined by:

4. To find negative roots using Descartes’ Rule, we examine:

5. A reciprocal quartic commonly uses which substitution?

6. Cardano’s method is primarily associated with:

7. A depressed cubic has the form:

8. In Cardano’s method, for y=u+v, we choose:

9. Ferrari’s method is used for solving:

10. A general quartic can first be normalized to:

11. To remove the cubic term from a normalized quartic, use:

12. The depressed quartic has no:

13. If a polynomial has 4 sign changes, possible positive roots are:

14. For a cubic axยณ+bxยฒ+cx+d=0, product of roots is:

15. Ferrari’s method ultimately reduces the quartic problem to:

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