Set Theory in Grade IX

Hello friends, in today's session we are going to learn about Set Theory. Set theory is a branch of mathematics that study sets which is a collection of objects. As the set can be a collection of any type of objects but the Set theory is a collection of objects that are relevant with mathematics.
The theory begins with a fundamental relation between the object O and a set S. If our object o belongs to our set S then it is represented as O ∈ S. where the symbol ∈ says “ belongs to”.
When we take two sets into consideration then we have different results known as set inclusion.
Like if all the members of set B belongs to set A, then it is called B is a subset of A and the set representation is done as A ⊆ B.
The members of set B could be 1, 2, 3 and then the members of set A would be 1, 2, 3, 4 ,8.
So we can see that all the members of set B are in set A.
Like this we have some simple relations that we need to know.

1. Union – when we do the union then we do is simply write the members of both the sets together. For Example the members of Set A are 1, 2, 3 and the members of set B are 1, 2, 3, 4. Then the Union of these sets will have 1, 2, 3, 4 as its members. And it is represented as A ∪ B.
   
2. Intersection – for intersection we just take the common term from the given sets. For Example- set A has 1, 2, 3 and set B has 1, 2, 3, 4. so the intersection of these two will be 1, 2, 3 and it is represented as A ∩ B.
   
3. Set Difference – Like if we have two sets, A and B, denoted as A / B. if Set A = 1, 2, 3 and set B = 2, 3, 4. so A/B = 1, 2, 3/ 2, 3, 4 = 1 and B/A = 2, 3, 4/ 1, 2, 3 = 4.
   
4. Symmetric difference – if we have two sets A and B then their symmetric difference is given by (A ∪ B) (A ∩ B).
   
5. Cartesian product – it is generally denoted as A × B. is the set whose members are all possible ordered pairs. It is written as (a, b) where a is the member of set A and b is the member of set B.
   
6. Power set – If we have A as our set whose members are all possible subsets of A. Like A = 1, 2 then its power would be , 1, 2, 1, 2 .
   

Some basic sets which do come are the empty sets, set of natural numbers and the set of whole numbers.
So now you seems to have got all the basic knowledge what set is and you would be able to solve questions based on it.

Coordinate System in Grade IX

Hello friends, today we are going to learn about an interesting topic of mathematics known as Coordinate System. A coordinate system is a system in which we use one or more numbers to identify the position of a given point. There are different types of coordinate systems like Cartesian coordinate system, Homogeneous coordinate system, polar coordinate system, etc.

But what we have to learn are the Cartesian coordinate and the polar coordinate systems.

The Cartesian coordinate system is made with two planes or axis known as x and y Axis. Both the axes are perpendicular to each other. So both of them form two different planes. If any point lies in the Cartesian coordinate system then its position can be given by (x, y), where x and y are the distance of the point from x and y axis with respect to the origin.

In this system X is the horizontal axis in two dimensions. In this axis, points are called as the x coordinates.. Y is the vertical axis. In this axis the points are called as the y coordinates. Both these axis meet at one single point known as the Origin or the graph center.

In Cartesian coordinate system the distance between two points could also be found, like if the two given points are (x ₁, y ₁) and (x ₂, y ₂). then the distance between these two points could be found by using the formula

in this formula d = distance between the points, and the rest are known.

We can even find the equation of Line every easily in the Cartesian system, if we know the two points that lie on the line. It is done by using the two point form. If (x ₁, y ₁) and (x ₂, y ₂) are the two given points. Then the formula we will use will be

(y – y ₁) = (x ₂ – x ₁/ y ₂ – y ₁) ( x – x ₁)

The x, y and z axis are generally represented with i, j and k.

For Example (2, 3, 5) can be written as 2 i + 3 j + 5 _k.

We write it in this form so that it becomes easy for us to solve and there is no confusion

The next one is the polar coordinate system, in this system a point is chosen as a pole and a ray from this axis is known as the polar axis.

In this coordinate system a point is represented by an angle to the ray and the distance from the pole i.e. (r, θ).

We can also convert the polar coordinates in the Cartesian form by using simple formulas. If in polar coordinates we know the angle and the ray distance, then the value of

x = r cos θ.

y = r sin θ.

So if you are comfortable with Cartesian coordinate system then you can easily find the value of x and y and then solve the problems. The conversion of one system to another is known as transformation process.

Pythagorean Theorem in Grade IX

Hello friends, in today's session we are going to learn about Pythagorean Theorem. The theorem gives a simple relation between the three sides of a right angle triangle.

According to the theorem, it states that:

In any Right Angle Triangle, the area of the square whose Side is the Hypotenuse is equal to the sum of the areas of the square whose sides are the two legs. In algebraic form the theorem can be written as: a² + b² =c², where a and b are the small sides and c is the hypotenuse of the right triangle.

In a right triangle if we know any two sides then we can easily find the other one with the Pythagoras equation.

If we generalize this theorem then we will get the law of cosines, which makes possible the computation of the third side of the triangle.

The converse of the theorem is also true:

For any three positive numbers a,b, and c such that a ² +b ² = c ², there exists a triangle with sides a,b and c, and every such triangle has a right angle between the sides of lengths a and b.

The Pythagoras equation is also used to find whether a triangle is acute, obtuse or right angle.

If a ² + b ² = c ², then the triangle is right angled.

If a ² + b ² > c ², then the triangle is acute angled.

If a ² + b ² < c ², then the triangle is obtuse angled.

Pythagorean triplet is a set of three positive integers a, b and c. These triplets satisfy the Pythagoras equation a ² + b ² = c ².

For example - (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37) are some of the triplets below 100.

Now let's solve some examples based on Pythagoras theorem.

1. if two sides of the triangles are 5 and 12, then find the length of the hypotenuse?

By using Pythagoras theorem. a ² + b ² = c ². in here a = 5 and b = 12, so the value of c will be c ² = 25 + 144.

c ² = 169

c = 13.

2. if one side of the triangle is 9 and the hypotenuse is 15, then find the other side.

In this problem a = 9 and c= 15, then the third side will be b.

b ² = c ² - a ^2.

b ² = 225-81

b ² = 144

b = 12.

3. check whether the triangle with given sides is a right triangle or not. a = 10, b = 24 and c = 26.

We can check whether it is a right triangle or not by using the Pythagoras theorem.

26² = 10² + 24^2.

676 = 100 + 576.

676 = 676,

so these three sides satisfy the Pythagoras theorem, so the triangle formed will be right angled.

Quadratic Equation in Grade IX

A Quadratic equation is a polynomial equation of second degree. In general form a quadratic equation is written as a x ² + b x + c. In this x is our variable and a, b and c are our constants. But we should keep one condition in mind that is a ≠ 0, because if a = 0 then our equation is no more a quadratic equation.

For example -

3 x ² + 4 x + 6,

9 x ² + 4,

8 x ² + 4 x.

A very important thing that we should keep in our mind is that a quadratic equation always has two roots or in a simple manner there are two values of x. The two solutions may or may not be distinct.

To find the value of the variable x in all the above equations, we can use the quadratic formula.

The two values of x can be written as,

We can even get complex values.

In the above formula the term b ² - 4ac is known as the discriminant. It is written as

if D = 0, then we have only one real root of the equation and it is

x= -b / 2a.

The other method of solving a quadratic equation is by splitting the middle term.

In this method we split the middle term i.e. the coefficient of x in such a manner that we easily get a common term and we can find the value of x. This method is also called as the factoring method.

The general equation is ax ² + b x + c. We follow some general steps to solve the quadratic equation.

1. Multiply the constants a and c without forgetting their signs.

2. Write down b, with its sign.

3. Write all the possible factors of the product of ac and find which pairs add up to give b .

4. Now rewrite the x terms as the sum of two terms with these selected numbers as the coefficients.

So by this we can find the solution of any quadratic equation.

For Example –

1. Solve x ² + 5 x + 6.

Answer – the possible combination of the middle term is 1 x + 4 x and 2 x + 3 x.

The product of a and c is 6 x 1= 6. So the second combination is the suitable option for this quadratic equation. So now it can be written as x ² + 3 x + 2 x + 6.

Now we take x common from the first two terms and 2 from the last two terms.

Now the quadratic equation becomes x ( x + 3 ) + 2 ( x + 3 ).

This time we take (x + 3) common, so the equation now becomes (x + 2 ) ( x+ 3 ).

The value of x = -3, -2.

The solution for the quadratic equation x ² + 5 x + 6 is x = -2, -3.

So now we have all the basic knowledge we need for solving problems based on quadratic equations.

Sequences and Series in Grade IX

Hii guys, we are going to learn about Sequences and Series in our today's session.
A list of numbers when arranged in a proper fashion then the arrangement is called as a sequence. In such a case each member either comes before or after another member. If this arrangement is disturbed then it is no more a sequence.
A series is formed when we do sum of sequence of terms. In a more easy language a series is the list of numbers with the addition operation between them.
For Example -
( a₁, a₂, a₃, … ), or ( b₀, b₁, b₂, … ), or ( c₀, c₂, c₄, … ) are known as sequences and
(a₁+ a₂ + a₃ +....), or (b₁ + b₂ + b₃ + ....), or (c₁ + c₂ + c₃ + …) are known as series.

The sequences are categorized as Finite, Infinite and Multiplicative.
A more formal definition of a finite sequence with terms in a set S is a function from 1, 2, ..., n to S for some n > 0. An infinite sequence in S is a function from 1, 2, ... to S.
The sequence is multiplicative if f( x y) = f(x) f(y) for x and y such that x and y are co prime.
A series is always written with a summation sign in front of it. It is represented as ∑.
For Example -

This is an example of infinite sequence.
And


This one is for an infinite sequence.
There are many types of series, but what we have to learn about is the arithmetic, geometric and Fibonacci series.
Arithmetic series or arithmetic progression is a sequence of numbers such that difference between two consecutive terms always remains constant. It is represented in general as

where a_n is any number of a sequence.
The sum of this series is given by


where n is the number of terms and d is the common difference between them.
A geometric series is one with constant ratios between its terms.
For example

It is represented by:


here a is the first term, r is the common ratio and n is the number of terms.

and the sum of this GP is given by a formula:
s = 1/ (1- r).
where s is the sum and r is the common ratio. But this formula is only for an infinite geometric progression.
The Geometric series is the best example of an infinite series.
Fibonacci series is the series in which each successive term is the addition of the terms preceding it.
The first two terms are 0 and 1, then term next to it is addition of these two.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 and so on.
If you are not able to memorize then you can find them using the formula
F_n = F_n-1+ F_n-2. 

Tessellations in IX Grade

Hello friends, in today's session we are going to learn a very interesting topic which is Tessellations.
Tessellations means featuring different images which can fit together in repeated patterns just like a jigsaw puzzle, where no overlapping and spacing comes. To speak in an easy language a tessellation is something in which we have few geometric shapes and we attach them in such a way that they form a regular plane without any gaps and overlaps. We can see tessellations throughout our history, from ancient architecture to modern art.
The topic that we are studying today corresponds with the everyday term tiling which refers to applications of tessellations, generally made of clay. It also exist in the nature if you would have seen the Honeycomb its structure is also known as a tessellation. Where regular hexagons are joined together to form a tessellation.
So now we are going to learn about the different sub topics that come under it.
When discussing tessellation, we can even find multicolors ones, so it becomes very essential for us to specify whether the color is a part of the tiles or it is just a part of an illustration.
Let's now learn about the only theorem in our topic known as the four color theorem.
The theorem states that every tessellation in, with a set of four available colors if each tiled when colored with one different color in such a manner that no tiles of same color meet at a curve of positive length. We should always keep note that when the coloring guaranteed by the four-color theorem will not generally respect the symmetry of the tessellation. For this coloring to be done, we would need at least seven colors for filling.
When we take Quadrilateral into consideration, we can easily form a tessellation by using 2-fold rotational center at the mid point of all the four sides of the Quadrilateral. In the same way as we have done for the Quadrilateral we can construct a parallelogram which is subtended by a minimal set of translation vectors, it is going to start from the rotational center. We can easily divide it into two with the help of its diagonal, and consider one half of the triangle as our fundamental domain. A triangle formed using the similar process will have the same area as of that of the quadrilateral and can be easily constructed by this method, which is cutting and pasting.
Now we will look that how many types of tessellations exist. We generally classify them on two bases which are – the regular tessellation and irregular tessellation.
A regular tessellation is one which is highly symmetric and is generally made using the congruent geometric figures.
In our mathematics syllabus there are only three regular tessellations which exist, which are formed by Equilateral triangles, squares and regular hexagons. The more accurately formed Tessellation without any defects in it, is made by the edge-to-edge joining of the shapes. The tessellation formed by this method shares its one full side with the full side of the another polygon. No types of partial sharing is done in this method.
Like if you would have heard about the penrose tiles which are generally used to add some beauty to the walls in our houses, hotels etc. is also a tessellation which is periodic in pattern. There also exist a self dual tessellation which is not as important but the example will make you understand of what it is.
The Honeycomb structure which we have discussed above is known a self dual tessellation. It is also a natural Tessellation.
The tessellations are even used in our hardwares to store data by breaking them into parts like in the computer models. In the computer graphics systems, the technique is used to rearrange the datasets of different polygons and divide them into polygon structure so that it becomes easy for them to get stored. This is generally used for rendering process. The data is generally tessellated into geometric figures like triangles. The process by which it is done is also known as triangulation.
The Computer Aided Designing in our computers is generally represented in analytical 3D curves and surfaces, limited to the faces and the edges to constitute a continuous boundary of a 3D body. This is because the 3D bodies are generally difficult to analyze directly. So for them we can do the approximation with it, which then becomes really easy to analyze.
The mesh of a surface is usually generated per individual faces and edges so that the original limit vertices are included into mesh. For further processing and approximation, there are only basic parameters that needs to be defined for the surface mesh generator, these are as follows-

• The maximum distance that can be allowed between the planar approximation polygon and the surface. This parameter would ensure that mesh generated is similar to the original analytical surface.
  
• The maximum size that can be allowed for the approximation of polygon. This parameter ensures enough detail for further analysis.
  
• The maximum angle that is allowed between two adjacent approximation polygons. This parameter would ensure that even very small humps or hollows can have significant effect to analyze.
  

As we have learned above that the self dual tessellation like Honeycomb, there are many others which occur naturally.
When basaltic lava flows, it slowly cools down due to the contraction forces between its surfaces. When the lava cools down, it generally have cracks on its surface. These cracks are often broken into hexagonal shapes forming a tessellation like figure.
So what we have read till now is all about what a tessellation is, and we just have to mug it up all. Now we come to the mathematical part, which is the number of sides of a polygon and number of sides of vertex.
If we assume an infinite long tiling, then let A be the number of sides of a polygon and B be the average number of sides which meet at the vertex. Then (A – 2 ) ( B – 2 ) = 4. For example, we have the combinations (3, 6), (31/3, 5), (33/4, 42/7), (4, 4) and (6, 3) for the tilings.
For the types of tiling that repeats itself, one can take the averages over the repeating part. For general consideration we take the averages as the limits for a region occupying the whole plane. In some cases where we have infinite row of tiles, or tiles getting smaller and smaller outwardly, the outside is not negligible and should also be taken into consideration while taking the limit. In extreme cases the limits may not exist, or depend on how the region is expanded to infinity.

For a finite tessellation and a polyhedra we have

( A – 2 ) ( B – 2 ) = 4 (1 – X/f ) ( 1 – X/v ).
in this expression f represents the number of faces of a polygon and v represents the number of vertex and X is the euler characteristic.
The above formula tells us that the number of sides of a face, summed over all faces, gives twice the total number of sides in the entire tessellation, it can be easily expressed in terms of number of faces and number of vertices. Similarly the number of sides at a vertex, summed over all vertices, also gives twice the total number of sides.
In most cases the number of sides of a face is same as the number of vertices of a face, and the number of sides meeting at a vertex is the same as the number of faces meeting at a vertex. However, in a case like two square faces touching at a corner, the number of sides of the outer face is 8, so if the number of vertices is counted the common corner needs to be counted twice. Similarly the number of sides meeting at that corner is 4, so if the number of faces at that corner is counted the face meeting the corner twice has to be counted twice.
A tile with a hole, filled with one or more other tiles, is not permissible, because the network of all sides inside and outside is disconnected. However it is allowed with a cut so that the tile with the hole touches itself. For counting the number of sides of this tile, the cut should be counted twice.
The another topic that we are going to study is the Tessellation of other spaces. As well as tessellating the 2-dimensional Euclidean plane, it is also possible to tessellate other n-dimensional spaces by filling them with n-dimensional polytopes. Tessellations of other spaces are often referred to as honeycombs.
So now I can expect that you will be able to solve the simple problems that comes under this topic, and would be well able to explained to anyone else what a tessellation is.
So thats all what we have learn in today's session and I will come back again with some other topic for you.

Percent and Rates in IX Grade

IX grade mathematics
Friends today we all are going to understand the basic concept behind few of the most important topics of mathematics that are Intervals and percents. Interval concept need to be known by each of the student because it plays an important role in graphing, number line and other important topics. So before going further I would like to discuss about Intervals in mathematical world.
Intervals are basically a chunk of the real line. Most commonly we define for real numbers that are “a” and “b”.

1. [a , b] gives the set of all the numbers “x” satisfying a ≤ x ≤ b.
   
2. (a , b) gives the set of all numbers “x” satisfying a < x < b.
   
3. [a , b) the set of all numbers “x” satisfying a ≤ x < b.
   
4. (a , b] the set of all numbers “x” satisfying a < x ≤ b.
   

Let's take an example to understand it better. The example in which the interval (3 , 5) is basically the set of all the numbers which are greater than or equal to 3 and less than 5. The following are the numbers 3, 4 and 4.5 which are in the interval and the numbers 5, 6 and 9.8 are not in the interval.
The above mentioned notations may be a little confusing, but what student needs to remember that square brackets used in the notations mean the end point is included whereas the round parentheses used in the notations mean it is excluded. If both the end points are included in the interval then it is said to be closed and if they are both excluded then we can call it to be open. In any of the case if one of them is included and the other one is excluded then the interval is half open or we can also call it as half closed, depending on student preference.
Moving further the things are getting a little dark because the above mentioned notation is also used with “b” replaced with Ñ„ or “a” replaced with - Ñ„ and only the round parentheses are placed at the end. This notation states that the interval is unlimited or we can say that interval is infinite on the right or left respectively.
Let's take an example to understand it better. In the notation (1 , infinity (Ñ„)) is basically a complicated way of describing the set of all numbers greater than 1, and (- Ñ„, pie ) means that the set of all numbers less than or equal to . The set of all real numbers can be expressed as (-Ñ„ , Ñ„)
Now we have done with Intervals so moving towards the main topic that is percents in mathematical world. Before proceeding further let's just discuss about basic definition of Percent. In simplest of mathematical manner we can say that percent means per hundred. The basic way is to understand it is that student divide something into 100 equal parts and after dividing consider so many parts of it.
The most important word and the item that creates the greatest confusion in percentage calculations is
“something”. The something of which you calculate a certain number of percent is rarely identified, and it often changes even within the same problem.
Let's take an example to understand it better. Suppose a person earns Rs. 10 an hour and acquires a raise of 20 percent, followed by another raise of 20 percent. The first raise is basically applies on the person hourly wage of Rs. 10 and the second to the new hourly wage of the person. In the above values 20 percent of Rs. 10 is Rs. 2 and so the new hourly wage of the person is Rs. 12. And now the second raise of Rs. 12 which gives Rs. 2.40 and so the hourly wage of the person after second raise is Rs. 14.40. What student need to note is the phrase that receive "a raise of 20%". It's pretty clear that it's 20 percent of the person current salary and his or her CEO's salary, or the gross national product, but that fact is not stated explicitly. One of the most noticeable thing is that even though both raises are 20 percent and they translate into different Rupees amount because they are applied to different hourly wages.
Now let's take an another example along the same lines : If any of the person is making a certain income and he or she receive a raise of about 100 percent, followed by a pay cut of 100 percent, then the new income of the employee or person is not the same what he/she started with, but rather it is zero.
In real life it is well possible to have more than 100 percent of something. The most real example is the bosses of company which probably makes more than 100 percent of the employee income.
Before proceeding further let's discuss the basic concept behind the percent calculations. The number of which we compute the percent is called the base number. In the above example Rs. 10 and then Rs. 12 are the two base numbers. The number of percent is the rate (in above example 20 is the rate) and the rate applied to the base number is the part. These all depends on the application the part can have many different names. For example, when the problem is about money, it could be a raise, a discount, a fee, a commission, etc.
Now the basic formula of calculating percentage is denoting the whole by “P”, the rates by “R” and the base by “b” and the basic formula is:
P = R / 100 x b
Generally percentage problems become distinct by what is known and what needs to be calculated, and to recheck their difficulty usually axis from describing the base number only inherently. Another difficulty which might come is that commonly the part is not of interest in itself, but what is required that it needs to be added or subtracted to the base number to get the result of interest.
Let's take an example to understand it better:

1. Unknown Part: If a person buys a gadget at a 25 percent discount. The list price is Rs. 2000. What is the total price?
   

Solution : Here the base number “b” = 2,000
“R” = 25%
So : P = 25/100 x 2000 = 500
Finally the purchase price = 2000 – 500 = Rs. 1500.
2. Unknown Part: If any person invest money at an annual interest rate of P percent. The interest is paid monthly. Thus according to the rules every month the bank adds:

to C which is the amount of money in your account at the beginning of the month. Equivalently, every month your money is multiplied with the factor
1 + p / 12,00
Unknown base number. If any person buys a car for Rs. 4 lakh and the dealer tells the buyer that you have purchased it at a 25% discount. What is the list price of the car? As we all know that buyer received a 25 percent discount buyer purchased the car for 75 percent of its list price. Let's just denote the list price by L we get
400,000 = 75/100 x L
Now we get :
L = 400,000 x 100/75 = 533333.33

4. Unknown Rate: The population of the town is 17,000. A year after it becomes 17,678. Find the annual rate by which the population is growing ? The rate R:
   17,678 – 17,000 = 678 = R/100 x 17,000
   

We get
R = 678 x 100/17000 = 4 (approximately)
The growth rate in your town is about 4 percent.
Generally problems involving percentage involve more than one base, rate and part.
Now moving towards another concept that is never divide by zero. Generally problems involving percentage involve more than one base, rate and part. I suggest you all that a daily practice and hard work is what more commonly needed. The most important thing is that there is no alternative to hard work.
Division by zero is undefined. If anyone going to assign a value or number to the outcome of dividing by zero he/she might run into conflict and mathematical world would become useless. In next session we would go through this in detail.
This is all for today in the next session we will go through the above concept along with some complicated percentage problems. Division by zero is an interesting topic which should tell you all the basic fundamentals of mathematics and also tell you that how a zero can change the whole course of the mathematics. We all know without zero there is no mathematical terms and here with zero the mathematics becomes useless. So it would be one of the most interesting chapter.

Factorization in Grade IX

Hello friends, today we are going to learn about the Factoring Expressions.

Factoring is just the reverse of expanding but a bit complex than a (b + c) = ab + ac, but the procedure followed is the same.

Factoring is generally an expression which is written as the product of two or more expressions.

For Example - 15 = 5 x 3, where 5 and 3 are both the factors of 15. The most basic ways to factor an expression is by finding the common factors.

Now what are common factors?

If each term of the expression has several factors and has atleast one factor that is common in them, then the common term is known is its common factor.

For Example – x² y⁴ + 2 x².

In this expression we can see that x² is common in both the terms. So it can be written as

x² ( y⁴ + 2 ). So now x² is the common factor of the above expression or it can also be said as it is the Greatest Common Factor( GCF) of the above expression.

The general way of writing the common factors is GCF x ( remaining expression).

We take another example for more understanding.

Let the expression be 8Y³B² + 16Y²B.

Now to find the common factors of this we have to see that what is common, so it seems like the term 8Y²B is common in the above expression. So it can be written as 8Y²B ( YB + 2 ). This is the same way as we have written the general formula GCF x ( remaining expression).

Now as I said above factoring is just opposite of expanding. So let's now verify this.

We take two terms ( x + 5) ( x + 1).

The expansion of this is x² + 6x + 5. And the factoring of this answer gives us ( x + 5) ( x + 1). So these are the factors of the given expression. We do this by splitting the x term in two parts. There are different ways of expanding but one among them helps us to find the common factor.

The 6x can be broken as 5x+1x, 2x+4x and 3x +3x. It satisfies x + 5x, so that we get our common factors. For doing this we have to follow some general rules.

Let the general quadratic equation be ax² + bx + c. So the general rules are.

1.multiply the constants and c without forgetting their signs.

2.Write down b, with its sign.

3.Write all the possible factors of the product of ac and find which pairs add up to give b .

4.now rewrite the x terms as the sum of two terms with these selected numbers as the coefficients

So by this we can find the factors of the quadratic expression. The above method we followed is known as the decomposition method.

To make perfect squares. Any expression of the form x² + 2ax + a² can be written as a perfect square (x + a)². To see whether an expression is a perfect square or not we first see that whether the constant term in it is a whole square or not, if it is a perfect square then take square root of it and multiply it by 2. If the result is equal to the coefficient of x then the term could be easily written as a perfect square.

For Example – Factorize x² + 8x + 16.

We can see that the constant term is 16, and it is a perfect square of 4.

Now 2 x 4 is equal to 8 which is the coefficient of x. So our given expression can be written as

(x + 4)².

Now we move to differences of square.

The differences of the square term are written in a general way as (something)² – (something)².

Let's understand it with the help of a simple example. Let's take (a²x² – b²), it can be factorised as (ax + b)(ax – b). We can see that the factors are identical except the sign between them.

For Example-

Factorize 9 – r².

To factorize it we use the above expression as the general function and find the value of a, x and b.

Here a² = 1, thus a = 1

x² = 9, thus x = 3

b² = r², thus b = r,

so the above expression can be written as ( 3 + r ) ( 3 – r ). To understand it more we will take another example.

Factorize 9p² – 4.

Here also we follow the above method and find the values of a, x and b.

a² = 9, thus a = 3x² = p², thus x = pb² = 4, thus b = 2,

so the above expression after factorization gives ( 3p + 2 ) ( 3p – 2 ).

There is one more thing that every expression cannot be factorized by this method, so we use the Quadratic Formula to find the roots and then we write the factors.

Now we move to the Difference of cubes.

It is very much similar to the difference of squares, the only difference is that it has cubes in the exponents in place of squares.

We write it in a general way as a³ – b³. Which can be factorized as (a - b)(a² - ab + b²). The factorization seems to be very difficult to remember, so what we have to remember is that if we see an expression like a³ – b³, then (a – b) is a common factor and the rest we can find out using the long division method.

Sum of Cubes.

The general way of writing this is a³ + b³. And when factorized we see that the result is

(a + b)(a² - ab + b²). In this also we have to remember only that when we see an expression like

a³ + b³, then we should remember that (a + b) is its common factor and the rest we can find out using the long division method.

Now as we have performed factorization on expressions with square and cube exponents, so now we move to the expressions with higher degrees. Sometimes students get panicked seeing higher degrees, but there is no need to panic as the expressions with higher degrees are as simple as the quadratic and cubic ones. There will always be a common factor in them because of which the remaining term will be converted into a quadratic expression.

For Example take (9x⁴ - 2x³ + 10x²) as your expression.

Now we can see that x² is common from the above expression. So we can finally write it as

x²(9x² - 2x + 10). Now the expression within the brackets can be solved by using the decomposition method. If we see that there is a constant term in it, then the simplest way is by using the long division method.

Let's understand this with the help of an example.

Solve x³ - 2x² - 5x + 6 with (x + 2) as one of its factors.

We will use the long division method to solve the above expression.

x² - 4x + 3

-------------------

(x + 2)| x³ - 2x² - 5x + 6)

x³ + 2x²

---------

-4x² - 5x

-4x² - 8x

---------

3x + 6

3x + 6

------

0

so the factor other than x+2 is x² - 4x + 3.

now we have to solve x² - 4x + 3 by using the decomposition method.

So now split -4x as -x and -3x. So we get the factors as (x - 1) and (x – 3).

so at the end the answer to x³ - 2x² - 5x + 6 is (x + 2)(x - 1)(x – 3).

In this question we were given one factor of the expression, what if there is no factor given in such type of questions. Then we could still factorize by assuming that ( x – k) as its common factor. Where k is some value which satisfies the given equation. If we find the value of k by hit and trial then we wil get our one factor and the rest could be done by long division method.

So let's understand this with the help of an example.

Solve x³ + 1.

So we have no common expression given with this question.

We assume the one factor as (x – k), now by hit and trial we can see that -1 satisfies the above equation so our factor is (x + 1). Now by using long division method we can find the rest.

x² - x + 1

-----------------

(x + 1)| x³ + 0x² + 0x + 1)

x³ + x²

-------

-x² + 0x

-x² - x

-------

x + 1

x + 1

-----

0

So the another factor is x² - x + 1.

Now the solution to x³ + 1 is equal to ( x + 1) (x² - x + 1).

So now I hope that you would be able to perform the operations on Factoring expressions

Monomials and Polynomials of IX Grade

Hello friends, today we are going to learn about the Monomials and Polynomials. From its name the topic might look very difficult to a Grade IX student but it is as simple as addition and subtraction of numbers.

Let's start with the definition,

Monomial is the product of variables and the numbers(either positive or negative) with their exponents as whole numbers. The monomial terms does not contain any addition or subtraction terms in it.

In general the monomial can be written as xⁿ where n is a positive integer. If we take multiple variables like x, y and z then the monomial would be x^a y^b z^c with a, b and c as non negative integers.

Example- 3 a ² b ⁴, b d ³, –17 a b c are all monomials with no negative powers and no addition and subtraction terms.

As know we have the basic knowledge of what a monomial is so we can now perform different operations on it.

To start with the easy one we perform addition of the monomials.

Addition is done by taking out the common terms from the different monomials and writing the remaining ones in addition form.

Let's take different monomials ax ³ y ², 5b ³ x ³ y ² and c ⁵ x ³ y ². Now to perform addition on it we write these terms in addition form a x ³ y ² + 5 b ³ x ³ y ² + c ⁵x ³ y ² . Now from this addition equation we see that x ³ y ² are there in all the three terms so we take them common and write the remaining terms in addition form. So the result of this addition is ( a + 5b ³ + c ⁵ ) x ³ y ² . Now we can see that no more addition can be done, so this is our final answer of monomial addition.

ax ³ y ²+ 5 b ³ x ³ y ² + c ⁵ x ³ y ² = (a + 5b ³ + c ⁵) x ³ y ².

Subtraction is done in a similar way as the addition operation.

We take the same polynomials used above and perform the process of subtraction on them.

The polynomials are a x ³ y ², 5 b ³ x ³ y ² and c ⁵ x ³ y ². Now by writing them in subtraction form we get a x ³ y ²- 5 b ³ x ³ y ^{2 –} c ⁵ x ³ y ². We again see that the term x ³ y ² are common so we take them common and write the remaining terms in the subtraction format. So the result of this is ( a - 5b³ - c⁵ ) x ³ y ².

as no more operation could be done on it so this is the final answer of subtraction of monomial.

a x ³ y ²- 5 b ³ x ³ y ²- c ⁵ x ³ y ² = (a – 5b ³ – c ⁵) x ³ y ².

In the process of multiplication of two, three or multiple monomials, the numerical coefficients are just multiplied in the simpler manner and the exponents with common bases are just added and written .

Example – we take two monomials 5ax ³ z ⁸and –7a³ x ³ y ². Now to perform multiplication as stated above we just multiply the constant coefficients and the exponents for a,x and y are added and written.

The exponent of a becomes (1+3 =) 4, the exponent of x becomes (3+3 =) 6, the exponent of y becomes (0+2 =) 2 and the exponent of z becomes ( 8+0=) 8. So our final answer for the multiplication of these two terms is – 35a ⁴ x ⁶ y ² z ⁸.

5ax ³ z ⁸( –7 a ³ x ³ y ²) = –35 a ⁴ x ⁶ y ² z ⁸.

Just like addition and subtraction are similar to each other, in the same manner division and multiplication are also related to each other. In multiplication the exponents are added and the constant terms are multiplied, in a same way in division the constant terms are divided in simple way and the exponents gets subtracted.

Example- we take 35a⁴ x ³ z ⁹ and 7ax²z⁶ as our monomials to be divided. The first one is to be divided is called as the dividend and the one from which we are dividing is called as the divisor.

Dividend- 35 a ⁴ x ³ z ⁹.

Divisor- 7ax²z⁶.

Now as stated above we divide the constant coefficient in a simple manner and subtract the exponents with common base. The exponent of a becomes (4-1 =) 3, of x changes to (3-2 =) 1 and for z it changes to (9-6 =) 3. So our final answer is 5a³ x z ³.

35a ⁴ x ³ z ⁹: 7a x ² z ⁶ = 5 a ³ x z ³ .

Now we come over to polynomials.

Let's start with the definition,

When different monomials are added together then they form a Polynomial.

There are two types of polynomial - binomial and trinomial.

If two monomials are added then they are called as binomial. In general we write a binomial as x^a+y^b .

Examples - 3x+1, x² – 4x, 2x + y, or y – y².

If three monomials are added together then they are called as trinomial. In general a trinomial is written as x^a+y^b+z^c. Examples - x² + 2x + 1, 3x² - 4x + 10, 2x + 3y + 2.

If three or more than three monomials are added then they are represented in general as polynomials.

Examples - x² + 2x, 3x³ + x² + 5x + 6, 4x - 6y + 8.

Since now we have the basic knowledge of what a polynomial is we perform the basic operations on it also.

But first what we have to understand the concept of like and unlike terms.

Like terms are those terms which have the same variables and the exponents, there is no need for the constant terms to match. For example, 8xyz² and −5xyz² are like terms as they have same variables and powers.

Unlike terms are those which do not have the same variables or different exponents for same variables.

Example-5x³z⁸ and 7x²z⁶ are unlike terms.

To start with the simple one, we take addition.

In the process of addition we take different polynomials and the like terms get added while the unlike terms are left as they are.

Example- (y²- 3y+ 6) + (y- 3y²+ y³⁾

y²- 3y+ 6+ y- 3y²+ y³

Now we combine the like terms together,

y³ + y²- 3y²- 3y+ y+ 6,

Now we add the like terms and keep the unlike terms as they are

-y³- 2y²- 2y+ 6.

so this our final answer.

To perform Subtraction.

In subtraction also the like terms are subtracted while the unlike terms remain untouched.

For Example- (5x² + 2x +1) - ( 3x² – 4x –2 )

we remove the parenthesis,

5x² + 2x +1 - 3x² + 4x +2

and put the like terms together --

5x² - 3x² + 2x+ 4x+1 + 2,

and we get our final answer

2x²+ 6x +3.

Multiplication of polynomials is done by simply multiplying each term of one polynomial with each term of the other polynomial.

For example - ( p+ q+ r ) a = pa+ qa+ ra.

Now taking a complex one,

= ( x+ y+ z )( a+ b )

multiply each term of first polynomial with the each term of second polynomials.

= x( a+ b )+ y( a+ b ) + z( a+ b )

and now open the parenthesis and multiply to get the final answer.
= xa + xb + ya + yb + za + zb .

The process of division is also as simple as the other operations done on the polynomials.

In the division process we just separate the terms in the numerator and write the denominator in each one of them.

For Example – 2x+4 : 2

So separating the numerator we get,

2x/2 + 4/2 =

so the final answer becomes.

x+2.

Now we move forward to complex problems.

For Example - 21x ³ – 35x² : 7x.

We separate the numerator in the same manner as we have done above. So it becomes,

21x ³:7x – 35x²: 7x.

Now simply dividing this, we get

3x²- 35x.

So this is our final answer.

So now it is expected that with this basic knowledge about the monomial and the polynomial you would be able to perform the different operations on it.