Category: Maths Notes PPT

Profit and Loss percentages are used to show the amount of profit and loss incurred in terms of percentage, which can be a percentage of profit or percentage of loss. Also, as we all know, that percentage is one of those methods which are used for comparison of two qualities. In our day-to-day life, we come across a lot of situations where we calculate or compare anything in “per-cent.” Here are some common examples: let relate the percentage of your results; everyone compares your results with your classmates or your partners.

Profit and Loss Percentage Example

Another most common example is related to buying and selling of goods. To calculate profit or loss on an item, one has to calculate it in percentage.

So, here in this article, we are going to discuss the concepts and importance of percentage in profit and loss.

To learn the concept of profit and loss percentage, we need to memorize terminologies for sales and purchase of goods.

Cost Price (CP): The price at which the item is purchased by us is known as the cost price. It is given by CP. 

Selling Price (SP): The price at which the item/good can be sold is known as the selling price. It is denoted or given by SP. 

Note: The profit or loss of a product during the sales and purchase of an item depends completely on cost price and selling price. 

Profit Percentage 

If the cost price of an item is less than the selling price, this is the only condition for profit on the item.  

               SP > CP 

Net Profit 

The net profit can be calculated as the difference between the selling price and cost price. 

             Net Profit = SP – CP

Profit Percentage Formula 

Profit % =[frac{SP-CP}{CP} times 100 = frac{Net Profit}{CP} times 100]

Loss Percentage 

If the cost price of the item is more than the selling price of the item, then the item is said to be sold at a loss. 

             SP < CP 

Net Loss 

Net Loss can be calculated as the difference between the cost price and selling price. 

             Net Loss=CP−SP

Loss Percentage Formula 

The amount of loss is sometimes expressed as a percentage after it has been computed in some cases. It is a percentage that is used to express the amount of loss that has been incurred in a business transaction. When comparing two quantities, this is extremely useful. The formula for calculating loss percentages are as follows: 

Loss % = [frac{CP-SP}{CP}]x100 = [frac{Net Loss}{CP}]x 100 

  

Note- It is strictly noted that the Profit Loss percentage is calculated on the Cost Price of the item, until and unless it is mentioned to calculate the percentage on Selling Price. 

Solved Examples

Question 1.   Find whether the following transactions are in profit or loss. Also, find the profit loss percent for each case. 

(i)   A General knowledge book was bought for Rs 250 and sold for Rs 325 

(ii)  Ranveer bought a motorcycle for Rs 12,000 and sold the same for Rs 13,500 

(iii)   A cupboard bought for Rs 2,500 and sold at Rs 3,000 

(iv)  A skirt bought for Rs 250 and sold at Rs 150 

Solution:

(i) C.P = RS 250 

     S.P = Rs 325 

     Here, S.P > C.P 

     So,  

     Profit = S.P – C.P 

                = 325 – 250 

                = Rs 75 

     Profit Percentage =[frac{{{text{Profit}}}}{{C.P.}} times 100% ]

                                     = [frac{{75}}{{250}} times 100% ]

                                     = [frac{{75}}{{25}} times 10% ]

                                     = [frac{{15}}{5} times 10% ]

                                     = 3 [ times ] 10%

                                     = 30%

 

     (ii) C.P = Rs 12,000 

           S.P = Rs 13,500 

           Here S.P > C.P 

           So, 

           Profit = S.P – C.P 

                       = Rs 13,500 – 12,000 

                       = Rs 1,500  

           Profit Percentage = [frac{{{text{Profit}}}}{{C.P.}} times 100% ]

                                           = [frac{{1,500}}{{12,000}} times 100% ]

                                           = 15/120 [ times ] 100%

                                           = 15/12 [ times ]10%

                                           = 5/4 [ times ] 10%

                                           = 5/2 [ times ] 5%

                                           = 25/2%

                                           = 12.5%

  

     (iii) C.P = Rs 2,500 

            S.P = Rs 3,000 

            Here S.P > C.P 

            So, 

            Profit = S.P – C.P 

                       = Rs 3,000 – 2,500 

                       = Rs 500 

            Profit Percentage = [frac{{{text{Profit}}}}{{C.P.}} times 100% ]

                                            = [frac{{500}}{{2,500}} times ]100%

                                            = 5/25 [ times ] 10%

                                            = 1/5 [ times ]10%

                                            = 20%

 

     (iv) C.P = Rs 250 

            S.P = Rs 150 

            Here C.P > S.P 

            So, 

            Profit = C.P – S.P 

                       = Rs 250 – 150 

                       = Rs 100 

            Loss Percentage =[frac{{{text{Loss}}}}{{C.P.}} times 100
% ]

                                          =[frac{{100}}{{250}} times ] 100%

                                          = 10/25 [ times ] 10%

                                          = 2/5 [ times ]10%

                                          = 2 [ times ] 20%

                                          = 40%

                              

Question 2: Juhi sells a washing machine for Rs 13,500. She loses 20% in the bargain. At what price Juhi bought the washing machine? 

Solution: 

S.P = Rs 13,500 

C.P =? 

Loss Percentage = 20% 

Now, 

 Loss = C.P – S.P 

C.P =S. P + Loss 

C.P = 13,500 + Loss           ….(i) 

  

Loss Percentage = [frac{Loss}{CP} times 100]%

[20 = frac{Loss}{13,500+Loss} times 100 %]

[frac{20 %}{100 %} = frac{Loss}{13,500+Loss}]

[frac{20 }{100 } = frac{Loss}{13,500+Loss}]

[frac{2}{100} = frac{Loss}{13,500+Loss}]     

[frac{1}{5} = frac{Loss}{13,500+Loss}]                                          

Cross Multiplying

                1 [ times ] (13,500 + Loss) = 5[ times ] Loss

                13,500 + Loss = 5 Loss

                5 Loss = 13500 + Loss

                5 Loss – Loss = 13500

                4 Loss = 13500

                Loss  = [frac{{13,500}}{4}]

                Loss =  Rs 3,375

From (i)

      C.P = 13500 + 3375

             = Rs 16875 

So, she bought it at Rs 16,875                             

                              

Question 3. 2. Ron purchased a table for Rs 1260, and due to some scratches on the top of the table, Ron has to sell it for Rs 1197. Find the loss percent.  

Solution:  

CP =  Rs.1260  

SP = Rs 1197.  

Since,  

      (SP) < (CP), Ron makes a loss.  

      Loss = Rs (1260 – 1197)  

               = Rs 63.  

      Loss Percentage =  [frac{Loss}{C.P}]x100%

                                    = [frac{63}{1260}] ×100  

                                    = 5% 

  

Question 4.  A man purchases a fan for Rs. 1000 and then sells it at a loss of 15% on the sale of the fan. In what range does the fan’s selling price fall?

Solution: The cost of the fan is Rs. 1000, which is the solution.

The percentage of losses is 15 percent.

As we all know, loss percentage  = (Loss/Cost Price) x 100

15 = (Loss/1000) x 100.

As a result, Loss = 150 rs.

As we are all aware,

Loss = Cost Price – Selling Price

So, Selling Price = Cost Price – Loss

= 1000 – 150

Pricing: R.850/- 

Question 5: Mukesh bought two watches at the same price and sold one at a 20% profit and the other at a 22.5 percent profit. What is the cost price of each of the watches if the difference in selling prices is Rs 150?

Solution: Let the cost of the watches be equal to 100x. 

Then the selling price of the first watch is 120x, and the selling price of the second watch is 122.5x.

The difference between selling prices is given as 150.

So, 122.5x – 120x = 150

2.5x = 150

x= 150 2.5

Therefore, x= 375

Substituting the x value in the original cost price, we get 100x = 100 (375) = Rs.37500.

Therefore the price of each watch is Rs. 37500.

Conclusion 

Profit and loss formulas are used to calculate the profit or loss generated by the sale of a specific product. A product’s cost price is the price at which it is purchased. A product’s selling price is the price at which it is sold. Profit is defined as the difference between the selling price and the cost price when the selling price is the more than the cost price. The difference between the selling price and the cost price is referred to as loss when the selling price is less than the cost price. When calculating profit and loss percentages for an article, it is important to remember that after purchasing the article, one may have to pay additional fees for transportation, repairing charges, local taxes, and so on. These additional costs are referred to as overheads. 

The number system is considered to be one of the most innovative and exciting inventions by human beings. Using number system concepts, experts have invented various tricks and puzzles involving numbers. These puzzles make a student curious and boost his/her will to study. These puzzles are not only for fun, but it also enhances the thinking process of students, thus turning maths easy for them. The objective of introducing puzzles in maths is to test the knowledge, intelligence and thinking ability of students and how much they are clear on the different concepts of mathematics. In this article, we have some tricky maths puzzles with answers that will help a student to enhance his/her knowledge.

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Number System

In order to solve number puzzles, it is necessary that a student should have some knowledge of the number system. Without any knowledge on the number system, it is impossible to solve maths number puzzles. In this article, you will also get an overview of the number system along with simple math puzzles.

A number system is considered to be a system of writing that expresses different numbers. The number system is considered as a mathematical notation used to represent numbers of a given set with the use of digits and other symbols in a consistent manner. A number system is responsible for providing a unique representation of every number and also represents the algebraic and arithmetic structure of various figures. Number system allows a student to use arithmetic operations like addition, subtraction and division.

The value of any digit can get determined by:

Types of Number System

In mathematics, there are various types of number systems, among which the four most common ones are:

Decimal Number System:

Decimal number systems have base 10 because this type of number systems uses ten digits which starts from 0 to 9. In the decimal number system, the positions that are present successive to the left of the decimal point represent units, tens and so on. Due to this, the system is always expressed in decimal numbers.

Binary Number System:

The number system which has a base 2 is known as a binary number system. In this case, only two binary digits exist, which are 0 and 1. The usual base 2 is considered as a radix of 2. Figures that are mentioned under this binary system are termed as binary numbers which are a combination of 0 and 1.

Octal Number System:

The number system which has a base 8 is known as an octal number system. In this case, the number system uses 8 numbers that are from 0 to 7. Octal numbers are mainly used in computer applications. Conversion of an octal number into decimal is similar to the decimal conversion.

Hexadecimal Number System:

In the hexadecimal number system, numbers are written and represented with 16 as a base. In this number system, the numbers are first represented in the same way as the decimal number system, and after that, they are represented as alphabets from A to F.

Rules for Solving Puzzles

• In a puzzle, each letter represents only one digit which means that one letter stands for one digit.

• In the case of puzzles, the first digit of a number can never be zero. For example, students write sixty-six as 66 and not as 066.

Maths Puzzle Questions With Answers

1. Find the value of Q from the following

4  Q  1

3  8   Q

———-

8   0   3

———-

Solution:

In column one (starting from the right), from Q + 1, we get 3 at the units place.

⇒ Q + 1 = 3

⇒ Q = 2

In the middle column,

Q + 8 gives a number in such a way that it has 0 at its units place. So Q = 2. This is verified by the fact that when Q that is 2 is added to 8, it results in 10 and hence 1 is carried forward. In the third column, the result will be 4+3+1 = 8 ( 1 is carried).

2. Find the value of A:

  A

+A

+A

——

BA

——

Solution:

A is a number whose thrice sum of itself also results in A. Therefore, A + A = 0. This case is possible only when A = 0 or A = 5.

In this case if A = 0, 

The entire sum will be 0  

Thus, B = 0.

This will lead to A = B which is not possible.

We know that different letters represent different digits. Therefore we have to consider the case in which A = 5.

Hence  5 + 5 + 5 = 15

Therefore, B = 1.

To understand Quota Sampling, you need to first understand the meaning of sampling. Used primarily in statistics, sampling refers to selecting a section or subset in a population and then analyzing it based on several factors. Sampling is generally done to be time and cost-efficient. It is because of sampling that we get to identify similarities and preferences in a population; based upon their attire, age-groups, likes, etc. Useful in a variety of scenarios, like in healthcare research, or understanding the target groups in a country’s population for brands, sampling can be done in several ways, including the quota sampling method.

Quota Sampling Definition

The meaning of Quota Sampling refers to the event when we gather samples within a group based on their specific characteristics or behaviors, depending upon their population size. The quota sampling process can be defined as a non-probability sampling method that is useful in making a sample for researchers, based on the general features in a given population. 

Here’s When to Use Quota Sampling: 

The above chart depicts the population of women from different age groups, scattered across the community. For the fashion designer to understand the behavior and traits of women from specific age groups, gathering data without segregating the age groups would prove meaningless for gaining insights. 

Different Types of Quota Sampling

Overall, quota sampling can be divided into two types, namely: Controlled Quota Sampling and Uncontrolled Quota Sampling.

Controlled Quota Sampling

Such situations refer to the cases when the researcher or survey conductor is limited to the sample choices. For example, if a school bag maker wants to survey students’ preferences for school bags, in general. Here the research would be limited to the children in their school-going age. 

Uncontrolled Quota Sampling

Any situation where the researcher or analyst doesn’t have any constraints or limitations for the sampling process is called Uncontrolled Quota Sampling. An example of an uncontrolled quota sampling can be that of a study conducted by medical workers to understand the overall public health and well-being in a nation-state at a regular period of interval. Here, the study would comprise samples from all age-groups, place of residence, gender, and more in the general population. 

Quota Sampling Process

In general, probability sampling techniques can comprise several steps to form meaningful samples. Since quota sampling is a non-probability sampling technique, various steps go into selecting samples carefully. They are:

Step I: Data Assessing and Division into Subgroups 

Also known as the stratified sampling, here the researcher goes through the population and divides it into mutually exhaustive subgroups based on the traits of: 

Step II: Calibrating the Weightage of SubGroups 

After the researcher has decided upon the subgroups, he/she needs to evaluate the proportion of the subsets compared to that of the population. This process helps in determining the number of samples to be taken from each group for further analysis. 

The General Formula for the Proportion Followed Will Be

The number of samples taken from a group = 

(Number of items present in the group) x [frac{Total;Samples ;to ;be ;Taken}{Total; Items; Present; in; the; Population}]

Step III: Appropriate Sample Selection 

Here the researcher needs to select the overall size of the samples, in alignment with the correct proportions derived from the previous step. 

Here’s an example of the sample size for a perfume manufacturer conducting a study to identify buying trends in men and women:

The above chart shows the relative proportions of various age groups in the population and the number of samples needed from respective age groups. 

Step IV: Survey Underway according to Set Quotas 

As a researcher, you need to stick to the already-set quotas for actionable and relevant results. 

Uses of Quota Sampling in Different Fields

• Quota sampling helps in achieving the best data representation model in terms of its fixed quotas or samples. 

• The quotas are the closest to the population trends and interests in terms of reality.

• Quota sampling helps researchers to track any underlying relationships between different subsets and identify various trends in specific people. 

• It is with the help of quota sampling that researchers get surveys done in a short period and reduced involved costs.

Quota Sampling Example

Let us consider a perfume maker who wants to understand the buying preferences of people for his perfume. (see Quota sampling process, step III)

As the researcher needs to go deep into the shopping choices of buyers – men and women, he needs to subset the population into respective samples. Therefore, once he completes the necessary research process, he finds that out of the total footfalls/buyers of his perfume was 10,000. 

In them, there were 4000 men (40%) and 6000 women (60%). The sample that we select needs to reflect the same percentages to stay relevant for the study. Therefore, every time the researcher considers 1000 of his buyers, he needs to take account of 400 men and 600 women, respectively. 

A rational number is formed when the common factor between the denominator and numerator is only 1. But the combination is that the denominator is always positive, including that a rational number can be standard when the numerator is associated with a positive sign. If these conditions are fulfilled, these numbers can be called rational numbers in standard form.

Based on this definition, some examples, theories, and processes demonstrate rational numbers in their standard form. Hopefully, these examples will better understand the conflict of rational numbers.

A rational number is said to be in its standard form when the common factor between the numerator and the denominator is only 1 while that denominator is always positive. In addition, the standard form of a rational number is satisfied when the numerator contains a positive sign. Such Numbers are what we call Rational Numbers in Standard Form. Below are a few theories and examples which illustrate the process of expressing rational numbers in standard form and will help you get acquainted with the concept even better.

What is the Standard Form of a Rational Number?

From the above definition, we can say that generally, x/y can be asserted as a rational number in its standard form. But there must not be any common factor except 1 between the numerator and denominator. Apart from that, the denominator, which is ‘y’ here, should always be positive.

How to Convert a Rational Number into Standard Form?

Below is a step-by-step listed guideline to express the rational number in standard form. The detailed step-by-step procedure is explained for enhanced understanding of the learners and they are along the lines.

• Step 2: The condition of the rational number in its standard form is that the denominator should be positive. Find out if your assigned denominator of the fraction is favorable or not. If it is positive, then the condition will be fulfilled, and if it is not clear, you can divide or multiply both the numerator and denominator with -1. By doing that, it will turn the negative denominator into a positive.

• Step 3: After that, you have to find out the GCD ( Greatest Common Divisor) of the absolute value of both denominator and numerator.

• Step 4: The final step is to divide the numerator and the denominator with the GCD. You have already got the GCD in the earlier stage. After the division, you will finally get the standard form of the given rational number.

Solved Examples on Rational Numbers Standard Forms

Example1: Identify whether the given Rational Numbers are in Standard Form or Not?

(A)  -8/23

(B)  -13/-39

Solution:

A. -8/23 can be said to be a rational number in standard form. Look closely, and this option fulfills the conditions of rational numbers in standard form. Here both the numerator and the denominator have none but one common factor that is 1. Apart from that, the denominator contains a positive sign. By n analysing this, we can conclude that -8/23 can be called a rational number in its standard form.

B. -13/-39 can not be said to be a rational number in its standard form. Both the denominator and numerator contain a common factor beside one, 13. Apart from that, the denominator here is -39, which means it doesn’t have any positive sign. Therefore -13/-39 can not be declared a rational number in its standard form.

Example 2: Show the Rational Number 18/45 in the Standard Form?

Solution:

Here, the rational number is 18/45.

Here we have to give the GCD ( Greatest Common Divisor) of the numerator 18 and denominator 45. As the denominator 45 is not negative. Here we do not need to do anything.

GCD (18, 45) comes out to be 9

GCD (18, 45) = 9

So if you want to convert 18/45 into its standard form, as we get from the rule, we have to divide both numerator and denominator by 9.

= 18/45

= (18÷9)/ (45÷9)

= ⅖

Hence, the rational number can express 18/45 in its standard form as ⅖.

Example 3: Identify the Standard Form of the Number 12/-18?

Solution:

In the rational number 12/-18, the denominator -18 is negative; hence, we must multiply the numerator and denominator with -1 to be positive.

12/-18 = 12 × (-1)/-18 × (-1)

= -12/18

Now we have to find out the GCD of the absolute value of the numerator and denominator. 

Hence, GCD (12, 18) comes out to be 6

GCD (12, 18) = 6

After that, both the denominator and the numerator should be multiplied and divided by the GCD that is 6.

= -12/18

= (−12)÷6)/(18÷6)(−12)÷6)/(18÷6)(-12) ÷6)/ (18÷6)

= -⅔

Hence, the standard form of Rational Number 12/-18 will be -⅔.

Example 4: How do We Reduce the Number 3/15 to its Standard Form?

Solution:

Here, we are given that the rational number is 3/15.

The denominator 15 is positive, so there is no need to change it.

We have to find out the GCD of numerator and denominator absolute values.

GCD (3,15) comes out to be 3GCD (3,15) = Therefore the numerator and denominator will be divided by the GCD.

3/15 = (3÷3)/ (15÷3)

= ⅕

So, 3/15 deduced to its standard Form is ⅕.

Reducing equations is the process of simplifying them. Since not all equations are presented in a linear form, it is vital to reduce them in simpler forms that are easy to understand. Performing it includes various mathematical approaches, and they vary as per the need of a particular equation.

The primary aim of this process is to ease the calculation process. Simplifying such complicated equations into their linear form means effortless calculation, and lesser mistakes. In this chapter of reducing method class 10 students will come across various examples of such equations, and the process to abridge them to perform a hassle-free calculation.

What is a Linear Equation?

Before moving on to the simplification process, one needs to learn more about linear equations. These equations are known as the first order. Also, these are known as equations for a straight line.

A linear equation is an algebraic expression where every term is a product of single variables and constants, or they remain constant themselves. It carries the first-order power of variables. Such equations are typically represented as Xa+Y=0, where X and Y are constants and X is not equal to 0.

Reducing method class 10 teaches students to learn converting different complicated equations into linear or simpler form.

Types of Linear Equation

Here are different calculation methods of the same:

• Some of them have one variable on the left-hand side, like 4x + 5 = 30.

• Some may have one variable, but on both sides, like 4x + 5 = 20 + 6x.

Apart from these, some linear equations may have non-linear forms. It requires reducing equations to linear form to make the calculation process simple. An example here is (2y + 5)/(4y + 2) = 1/4. Such equations are not easy to solve; it requires simplification.

What is Reducing Equation?

Reducing Equations is the process that converts non-linear ones into linear ones. Since every equation is not always available in a simple and straightforward format, it is essential to break them down to make solving easy.

Solving these equations requires usage of some mathematical applications such as cross multiplication, division, etc. on both sides. It helps to convert complicated equations to their linear forms. Following this conversion, it becomes easy to find the value of the variables.

Tactics of Simplification

Among many tactics to simplify non-linear equations, cross multiplication is one of the prominent ones. In this method, students can multiply the numerator of one fraction with another’s denominator, and vice-versa.

Cross multiplication of equation reducible to linear form example includes the following:

(a – 2)/(a + 8) = ⅔

Now, cross multiplying this equation will result in, 3(a – 2) = 2(a + 8).

Following this cross multiplication, one needs to implement another mathematical operation to move a step closer to solve this equation. It is known as opening the brackets. Additionally, another law used here is called distributive law. Under this, students need to multiply any value within the brackets with the one outside of it.

Now, on using this law on the above mentioned equation, one will get:

3a – 6 = 2a + 16

After implementing distributive law, one needs to arrange the variables on one side and constants on one side. While performing this step, students need to remember that, when they move any value from the RHS to the LHS, it will shift from its negative value to a positive one, and vice-versa. Implementing that in this equation results in:

3a – 2a = 16 + 6

a = 22

A point to note here is that, if students do addition, or subtract, or even perform multiplication with the same value on either side, they will get the value of a variable without changing the final equation.

Now, this is a relatively simple example of the concept of reducing method class 10. There are more complex examples as well, where students need to employ more mathematical applications like LCM to find the desired result.

Point to Note: Equations are a condition of a particular variable.

Reducing method class 10 is an essential chapter of mathematics, and helps students get a clear idea of solving equations. Since it is a vital chapter for the upcoming board exams as well as for higher studies, one must learn it in detail, and thoroughly.

Along with the traditional textbooks, and practice sets, online platforms like can be a big help for students. The availability of exam notes, mock question papers, study material coupled with live online classes, and doubt clearing sessions let individuals better their exam preparations.

In Mathematics, the Remainder Theorem is a way of addressing Euclidean’s division of polynomials. The other name for the Remainder Theorem is Bezout’s theorem of approaching polynomials of Euclidean’s division. The remainder theorem definition states that when a polynomial f(x) is divided by the factor (x -a) when the factor is not necessarily an element of the polynomial, then you will find a smaller polynomial along with a remainder. The resultant obtained is the value of the polynomial f(x) where x = a and this is possible only if f(a) = 0. In order to factorize polynomials easily, the remainder theorem is applied.

Example: If p(x) = x3-12x2-42 is divided by x – 3. The quotient is x2-9x-27  and the remainder is – 123.

Assuming, x – 3 = 0.

x=3

Substituting x’s value, we get: 

P (x) = -123

Therefore, this proves and satisfies the remainder theorem. 

Remainder Theorem Definition

The Remainder Theorem Definition states that when a polynomial is p ( a ) is divided by another binomial ( a – x ), then the remainder of the end result that is obtained is p ( x ).

Example: 2a2 – 5a – 1 is divided by a – 3

Solution: Here p (a) = 2a2 – 5a – 1 

and the divider is ( a – 3 )

Remainder Theorem Formula

Consider a polynomial f ( a ) where f is the polynomial and a is the variable. The polynomial f (a) is now divided by a binomial (a – x), where x is a random number, according to the theory. The polynomial is divided by (a – x) in this case, and the remainder is r. ( a ). The above definition can be expressed as:

F ( a ) / ( a – x ) = q ( x ) + r ( x )

Factor Theorem 

To find the roots of a polynomial equation, the factor theorem is applied to factorize the equation. With the help of synthetic division, you can solve problems and also you can check for a 0 remainder. When f ( a ) = 0, then y – a can be considered as the factor of the polynomial f ( a ). When the y – a is a factor of the polynomial f (  a ), then the polynomial f ( a ) = 0.

The Factor Theorem and How to Apply It

The following are the steps to use the factor theorem to identify the factors of a polynomial:

• Step 1 : If f(-c)=0, then (x+ c) is a factor of the polynomial f(x).

• Step 2 : (cx-d) is a factor of the polynomial f(x) if p(d/c)= 0.

• Step 3 : If p(-d/c)= 0, then (cx+d) is a factor of the polynomial f(x).

• Step 4 : If p(c)=0 and p(d)=0, then (x-c) and (x-d) are polynomial factors p(x).

Rather than using the polynomial long division method to find the factors, the factor theorem and synthetic division method are the best options. This theorem is generally used to eliminate known zeros from polynomials while keeping all unknown zeros unaffected, allowing the lower degree polynomial to be readily found. The factor theorem can also be defined in another way. We usually get a reminder when a polynomial is divided by a binomial. When a polynomial is split by one of its binomial components, the quotient resulting is known as a depressed polynomial. The factor theorem is demonstrated as follows if the remainder is zero:

If f(c)= 0, the polynomial f(x) has a component (x-c), where f(x) is a polynomial of degree n, and n is larger than or equal to 1 for any real number, c.

Remainder Theorem Proof

The remainder theorem is applicable only when the polynomial can be divided entirely at least one time by the binomial factor to reduce the bigger polynomial to a smaller polynomial a, and the remainder to be 0. This is one of the ways which are used to find out the value of a and root of the given polynomial f ( a ).

Proof: 

When f ( a ) is divided by  ( a – x ), then: 

F ( a ) = ( a – x ) . q ( a ) + r

Consider x = a;

Then, F ( a ) = ( a – a ) . q ( a ) + r

F ( a ) = r

Therefore, the above proves the remainder theorem. 

The Steps Involved in Dividing a Polynomial by a Non-Zero Polynomial

• Step 1: The polynomial (the dividend and the divisor) is arranged in the decreasing order of its degree. 

• Step 2: With the first term of the divisor, divide the first term of the dividend in order to find out the first term of the quotient. 

• Step 3: Now multiply the first term of the quotient with the first term of the divisor and with the obtained result, subtract the result from the divided to find out the remainder. 

• Step 4: Next, divide the remainder with the division. 

• Step 5: Repeat Step 4 until you cannot divide the remainder anymore.

Working of Remainder Theorem

Let’s look at a general scenario to see how the remainder theorem works. Let a(x) be the dividend polynomial and b(x) be the linear divisor polynomial, and q(x) and r be the quotient and constant remainder, respectively. As a result,

a(x) = b(x) q(x) + r

Let’s use k to represent the zero of the linear polynomial b(x). As a result, b(k) = 0. If we insert in x as k in the stated relation above, we have a(k) = b(k) q(k) + r

It’s worth noting that this is permissible since the starred connection remains true for all x values. It’s a polynomial identity, in reality. We’re left with a(k)=r since b(k)=0. In other words, when x equals k, the remainder equals the value of a(x). That’s exactly what we found! The remainder theorem is exactly what it sounds like: When a polynomial a(x) is divided by a linear polynomial b(x) whose zero is x equal to k, the remainder is given by r=a(k).

Remainder Theorem Examples

Question 1: Find the root of the polynomia a2 -3a -4 

Solution:  Consider the value of a to be 4. 

Substituting the value of a = 4 in the polynomial, we get:

F ( 4 ) = 42-3 (4)- 4 

F ( 4 ) = 16 – 12 – 4

Therefore, f ( 4 ) = 0.

Question 2: Find the r ( d ) of the polynomial d4– 2d3 + 4d2 -5 if it is divided by d – 2

Solution: 

D – 2  = 0

D = 2

Substituting d value in polynomial we get:

R ( 2 ) = 24 – 2 (2)3 + 4 (2) – 5 

R ( 2 ) =  16 – 16 + 8 – 5

R ( 2 ) = 3

Question 3: Find the r ( d ) of the polynomial 4d2 -d + 9 if it is divided by d – 1

Solution: 

D – 1  = 0

D = 1

Substituting d value in polynomial we get:

R ( 1 ) = 4 (12) – 1 + 9

R ( 1 ) =  4 – 1 + 9

R ( 1 ) = 12

Remainder Theorem is a way of addressing Euclidean’s division of polynomials. It states that when a polynomial is p(a) is divided by another binomial (a – x), then the remainder of the end result that is obtained is p(x).