Degree of Polynomial Calculator

INTRODUCTION ABOUT DEGREE OF POLYNOMIAL CALCULATOR:       

In arithmetic, a polynomial is an expression of finite length constructed from variables and constants, using only the addition, subtraction, multiplication operations, and non-negative, whole-number exponents. The maximum power of the variable in a polynomial is called the degree of the polynomial. A term with no variables is called a constant term and its degree is zero.I like to share this Polynomial Solver with you all through my article.

Find the Degree of Polynomial:

CONDITION FOR POLYNOMIAL:

In polynomial the variable should not involve in division operation.

Ex: `x^2 -4x+7, x^3 +5x^2-x+6, ` it is a polynomial.

Ex: `x^2 -4/x+3/2,` not a polynomial.

General form for degree of Polynomial Calculator:

A polynomial function is a function that can be defined as,

`f(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_2x^2+a_1x+a_0`

Where n is a non-negative integer and `a_0, a_1,a_2,` ...,` a_n ` is constant coefficients.

First Degree of polynomial calculator:  

The first degree polynomial is the polynomial that has only one variable. The variable may be x, y or z and it is easy to find the value for the variable. The combinations of two first degree polynomials are called as second degree polynomial.

Second degree polynomial calculator:

In this equation 6x2+5x+2  the power of x is 2. so it is a second-degree polynomial or a polynomial of degree 2.

POLYNOMIAL CALCULATOR:

To solve the polynomial in calculator, first we have to enter the two polynomials then click anyone of the button like add, subtract, multiply, division and factor and the equivalent operation will be performed and answer will be displayed. Thus the polynomial calculator is shown below.



Example and Practice Problem for first and Second Degree Polynomial Calculator:

First degree polynomial calculator:

Example 1:

Determine the value for the first degree polynomial 2x+10 = 12.

Solution:

The given first degree polynomial is 2x+10 = 12.

For the above polynomial, subtract 10 on both sides of the equation.

2x +10-10 =12-10

2x = 2

Divide by 2 on both sides of the above equation.

`(2x)/2 = 2/2`

x = 1

The value for the first degree polynomial 2x+10 = 12 is 1.

Second degree polynomial calculator:

Example 1

Find whether the given expression is a second degree polynomial or not.

(x - 1) (x - 2) = 0

Solution:

Multiply two expressions in the left hand side,

(x - 1) (x - 2) = 0

x(x-2) -1(x-2) =0

x2 -2x -x + 1(2) = 0

x2 -3x +2 = 0

So, the highest exponent of variable x is 2.

So, the given polynomial equation is second degree polynomial equation.

Practice Problem for first Degree Polynomial:

1.Find the value for the first degree polynomial of 2x+ 5 = 6.

Answer: x= 1/2.

2.Compute the value for the first degree polynomial 4x+12 = 24.

Answer: x=3.


Practice Problem for Second Degree Polynomial:

1.    Find whether the given expression is a second degree polynomial or not.

(x - 10) (x - 5) = 0

Answer: The polynomial equation is second degree polynomial equation.

Lower Quartile

Introduction to lower quartile:

In descriptive statistics, a quartile is any of the three values which divide the sorted data set into four equal parts, so that each part represents one fourth of the sampled population. Lower Quartile is one type of quantile.In that quartile, the 25 % of data which lie in the lower half of the data set is called as lower quartile. We can also define it as the middle value of the lower half.

For example,

21, 5, 22, 33, 17, 36, 14, 42, 15, 35, 27


Arrange the data in ascending order,

5, 14, 15, 17, 21, 22, 27, 33, 35, 36, 42

In the above Data set,

15 is lower quartile.

Calculation of Lower Quartile:

Lower Quartile for Odd data set:

Let us see how to calculate the lower quartile of the data set that contains odd numbers.

Step1:  First find the median of the data set. The median divide the data set into upper and lower part.

Step2:  Find the median for lower part of the data set .That median is called as lower quartile.                    

Ex: Data Set: 8, 49, 53, 19, 45, 44, 11, 41, 47, 51, 39

Ordered Data set: 8, 11, 19, 39, 41, 44, 45, 47,49,51,53

Median (Mid value) =44

Lower part = 8, 11, 19, 39, 41

Upper part = 45, 47,49,51,53

Lower quartile (Median of lower part) = 19

Lower Quartile for Even Data set:

Let us see how to calculate the lower quartile of the data set that contains even numbers.

Step 1: Find the Median of the data set by calculating the average of the data set.

Step 2: Next in the lower part, find the median by calculating the average. That value is known as lower quartile.

Ex: Data set: 13, 6, 24, 33, 9, 38, 16, 46, 19, 53, 26, 65

Ordered data set: 6,9,13,16,19,24,26,33,38,46,53,65

Median: (24+26)/2= 25

Lower quartile:  (24+33)/2= 28.5



Practice Problems on Lower Quartile

Find the lower Quartile for the following;

1.   17,11,47,27,69,20,6,34,12,93,33

2.   7,4,18,23,30,20,14,36,41,11,48,64

Answer Key:

1.12

2.12.5

Compound Dependent Events

This page is based on compound dependent events which is a study under dependent events in statistics. In probability, an event is a one or more possible outcomes from an experiment. An event consisting of one or more simple events is called compound event. An event is called dependent event, if an event does affect the other event.

Compound Events Dependent - Examples

Ex 1: A bag contains 20 bulbs, in which 5 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Sol:

Let S be the sample space, n(S) = 20

Number of good bulbs = 20 - 5 = 15

A be the event of getting a good bulb, n(A) = 15

B be the event of getting a defective bulb, n(B) = 5

P(A) = `(n(A))/(n(S))` = `15/20`= `3/4`

Now 19 balls will be remaining in the box, so

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) = `3/4` · `5/19` = `15/76`.

P(Good and Defective bulb) = `15/76` .

Ex 2: A jar contains 7 yellow candies, 5 green candies, and 8 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Sol:

Let S be the sample space, n(S) = 7 + 5 + 8 = 20

A be the event of getting a yellow candy, n(A) = 7

B be the event of getting a green candy, n(B) = 5

P(A) = `(n(A))/(n(S))` = `7/20`

Now 19 candies will be remaining in a jar.

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) =P(A) · P(B) = `7/20` · `5/19` = `7/76`

P(Yellow and Green) = `7/76`

Practice Problems

Pr 1: A bag contains 15 bulbs, in which 7 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Pr 2: A jar contains 10 yellow candies, 6 green candies, and 4 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Ans: 1) ` 4/15` 2) `3/19`

Poisson Normal Distribution

Poisson normal distribution an Introduction:

Normal distributon:

In probability theory and statistics, the normal distribution or Gaussian distribution is an absolutely continuous probability distribution with zero cumulants of all orders higher than two.

Poisson distributon:

In probability theory and statistics, the Poisson distribution  (or Poisson law of large numbers) is a discrete probability distribution that expresses the probability of a number of events occurring in a fixed period of time if these events occur with a known average rate and independently of the time since the last event.


Poisson Distributon in Poisson and Normal Distribution:

The three basic assumptions on poisson distribution are as follows

The probability of one photon arriving in ?t is proportional to ?t when ?t is very small.

P(1;?t) = a?t for small ?t

where a is a constant whose value is not yet determined.

The probability that more than one photon arrives in ?t is negligible when ?t is very small.

P(0;?t)+P(1;?t) = 1 for small ?t

The number of photons that arrive in one interval is independent of the number of photons that arrive in any other non-overlapping interval.

Variance of the Poisson Distribution:

var(k) = E[k2] - E2[k] = at

Note that for a poisson distribution mean and the variance are equal .

Algorithm for poisson distribution:

Algorithm :Poisson random number:

init:

Let L ? e-?, k ? 0 and p ? 1.

do:

Generate uniform random number u in [0,1] and let p ? p × u.

while p > L.

return k - 1.



Normal Distribution in Poisson and Normal Distribution:

Description of normal distribution:

For any of situation in which the absolute value of a continuous variable is changed randomly from trial to trial.

The random improbability or random error.

Some properties of normal distribution Bell curve:

The curve will be Symmetric,

It will be unimodal,

The value will be extends to +/- infinity,

And the area under the curve is always one (it will be the fixed value).

The normal distribution bell curve is fully based on the mean and standard deviation functions.

These are some of the important points of  poisson and normal distribution.

Numerator of Fractions

Introduction to fractions

A fraction is a number that can represent part of a whole. A fraction consists two parts, a numerator and a denominator, the numerator represents the number of equal parts and the denominator represents number of equal parts make up a whole.

For example, `2/3` is a fraction. Here, the number above the fraction bar is called as the numerator (2) and number below the fraction bar is called as the denominator (3) of the fraction.

Here we are going to study about fractions

Sample Problems on Fractions

Here we are going to study about adding and subtracting fractions with same denominators and different numerators.

Example 1

`2/11` + `5/11`

Solution

Here the given problem is to add the fractions `2/11` and `5/11`

Notice that the denominators of both fractions are same, so we can just add the numerators.

So,

`2/11 + 5/11` = `(2 + 5)/11`

= `7/11`

So, the sum of `2/11` and `5/11` is `7/11`

Example 2

`6/14 - 3/14`

Solution

Here the given problem is to subtract `3/14` from `6/14`

Here the denominators are same, so we can just subtract the numerators.

`6/14 - 3/14` = `(6 - 3)/14`

= `3/14`

Few more Sample Problems on Fractions

Here we are going to study about adding and subtracting fractions with different numerators and denominators.

Example 1

`2/5 + 1/4`

Solution

Here the given problem is to add `2/5` and `1/4`

Notice that the denominators are different in the given fractions and we cannot add the fractions directly. The given fractions can be added with the help of least common denominator.

The least common denominator of `2/5` and `1/4` is 20. Because, 20 is the least common multiple of the denominators 5 and 4.

So, the fractions can be re-written as,

`2/5` = `2/5` * `4/4` = `8/20`

`1/4` = `1/4` * `5/5` = `5/20`

Now the problem becomes,

`8/20 + 5/20` = `(8 + 5)/20`

= `13/20`

Example 2

`3/7 - 1/3`

Solution

Here the given problem is to subtract `1/3` from `3/7`

Notice that the denominators are different in the given fractions and we cannot subtract the fractions directly. The given fractions can be subtracted with the help of least common denominator.

The least common denominator of `1/3` and `3/7` is 21. Because, 21 is the least common multiple of the denominators 3 and 7.

So, the fractions can be re-written as,

`3/7` = `3/7` * `3/3` = `9/21`

`1/3` = `1/3` * `7/7` = `7/21`

Now the problem becomes,

`9/21 - 7/21` = `(9 - 7)/21`

= `2/21`

Antiderivative of Log X

Introduction to anti-derivative of log x:

The natural logarithm is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718. The natural logarithm is generally written as ln(x), loge(x) or sometimes, if the base of e is implicit, as simply log(x). Formally, ln(a) may be defined as the area under the graph of `1/x ` from 1 to a, that is as the anti-derivatives or integral,

ln a  = `int_1^a(1/x)dx`

This defines a logarithm because it satisfies the fundamental property of a logarithm:

ln (ab) = ln a + ln b

Source Wikipedia.

Anti-derivative Logarithmic Formulas:

1. `int` `(1/x)` dx = log x + c

2.  `int` e x dx = e x + c

3. `int (dx) / (a^2 - x^2) ` = `(1/(2a)) log [(a + x) / (a - x)] + c`

4. `int (dx) / (x^2 - a^2)` = `(1/(2a)) log [(x - a) / (x + a)] + c`

5. `int (dx) / sqrt(a^2 - x^2)` = `sin^-1(x / a) + c`

6. `int (dx) / sqrt(x^2 - a^2) ` = `log [(x + sqrt(x^2 - a^2)] + c`

Anti-derivative Logarithmic Problems:

Anti-derivative logarithmic problem 1:

Find the anti-derivative of given logarithmic function, log x  with respect to x

Solution:

Given logarithmic function, ` int ` log x. dx

Let,    u = log x                            dv = dx.

`(du)/(dx) ` = `1/x`                                    v = x

du = `1/x` dx

We know anti-derivative parts formula,  `int ` u dv = uv - `int ` v du

`int ` log x. dx  = log x . x - `int`` x ((dx) /x) `

= x. log x - `int` dx

= x. log x - x + c

= x( log x - 1) + c

Answer: Anti-derivative of log x is    x( log x - 1) + c

Anti-derivative logarithmic problem 2:

Find the anti-derivative of given logarithmic function, `(1 + 25x)/x^2`   with respect to x

Solution:

Given function, ` int` `(1 + 25x)/x^2` . dx

`int``(1 + 25x)/x^2`. dx  =` int` `dx/x^2` + `int` ` (25x)/x^2 ` dx            

= `int` `x^(-2) ` dx + ` int `` 25x^(-1)` dx               

= `x^(-1)` +  25 log x + c

Answer:  Anti-derivative of  `(1 + 25x)/x^2`  is x-1 + 25 log x + c

Anti-derivative logarithmic problem 3:

Find the anti-derivative of given logarithmic function, `e^(3x)/(1- e^(3x))` with respect to x

Solution:

Let u = 1- e3x           du = - `3` `e^(3x)` dx      

So, substitute the u and du

`int `  `e^(3x)/(1- e^(3x))` dx = `int`` (-1/3)(du)/u`                                 

= `(-1/3)` ` int` `1/u` du

= `(-1)/3` ln u + c                                                      we know u = 1- e3x

=` (- ln (1-e^(2x)))/3` +c

= `((-1)/3)` ln(e3x -1) + c

Answer:  Anti-derivative of `e^(3x)/(1- e^(3x))` is  `((-1)/3)` ln(e3x -1) + c

Subtracting Mixed Fractions

Introduction subtracting mixed numbers:

Natural number and a fraction is called mixed number. If numerator is less than denominator then it is said to be proper fraction. Improper fraction is formed when numerator is greater than denominator. For subtracting mixed numbers we have to first convert it in to proper fractions and then subtract the numbers so obtained.

Problem Subtracting Mixed Numbers with Different Denominators:

Pro 1 : 2 `3/5` - 3` 5/7`

Sol : Here the both are mixed fraction,

Step 1:  convert into the proper fraction and subtracting mixed numbers

2 `3/5` first we have to change into the proper numbers `(5xx2+3)/5` , we get

`13/5` is the proper numbers

3 `5/7=` `[(7xx3)+5]/7` = `26/7 ` is the proper number

Here we round mixed numbers and adding both functions

`13/5` -` 26/7`

In this step subtracting mixed numbers taking the Least Common Multiplier of 5 and 7 is 35. `[(13 xx 7) - (26 xx 5)]/(35)`  = `(91-130)/35 = (-39)/(35) `= -1.1142

Problem Subtracting Mixed Numbers with same Denominator:

Ex 1:      2 `3/4` - 3 `5/4`

Sol:  Here the fraction 4 is the equivalent fraction mixed number, subtracting both mixed numbers

2 `3/4` first we have to change into the proper numbers `(4xx2+3)/4` we get

`11/4 ` is the proper numbers

3 `5/4 = [(4xx3)+5]/(4) = (17)/(4)`

Here we round mixed numbers and adding both functions

`11/4 ` - `17/4 = (11-17)/(4) = (-6)/(4) = (-3/2) = -1.5`