Factor Polynomials Completely

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. Factorization (also factorisation in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. (Source: Wikipedia)

Example problems for factor polynomials completely

Factor polynomials completely example problem 1:

Factorize the given polynomial equation completely 4x^2 - 45x + 11

Solution:

Given polynomial equation is 4x^2 - 45x + 11

First factorize the given equation, we get

(4x^2 - 45x + 11) = (4x^2 - 44x - x + 11)

Grouping the first two terms and second two terms, we get

= (4x^2 - 44x) - (x - 11)

= 4x (x - 11) - 1 (x - 11)

= (x - 11) (4x - 1)


The factors of the given polynomial equation is (x - 11) and (4x - 1)

Answer:

The final answer is (x - 11) and (4x - 1)

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Factor polynomials completely example problem 2:

Factorize the given polynomial expression 7x^2 + 33x + 36

Solution:

Given polynomial expression is 7x^2 + 33x + 36

First factorize the given expression, we get

(7x^2 + 33x + 36) = (7x^2 + 21x + 12x + 36)

Grouping the first two terms and second two terms, we get

= (7x^2 + 21x) + (12x + 36)

= 7x (x + 3) + 12 (x + 3)

= (x + 3) (7x + 12)


The factors of the given polynomial expression is (x + 3) and (7x + 12)

Answer:

The final answer is (x + 3) and (7x + 12)

Factor polynomials completely example problem 3:

Factorize the given polynomial expression x^2 + 48x - 49

Solution:

Given polynomial expression is x^2 + 48x - 49

First factorize the given expression, we get

(x^2 + 48x - 49) = (x^2 + 49x - x - 49)

Grouping the first two terms and second two terms, we get

= (x^2 + 49x) + (- x - 49)

= x (x + 49) - 1 (x + 49)

= (x + 49) (x - 1)


The factors of the given polynomial expression is (x + 49) and (x - 1)

Answer:

The final answer is (x + 49) and (x - 1)

Practice problems for factor polynomials completely

Factor polynomials completely practice problem 1:

Factorize the given polynomial expression 3x^2 - 18x + 24

Answer:

The final answer is (x - 4) and (3x - 6)
Factor polynomials completely practice problem 2:

Factorize the given polynomial expression 10x^2 - 25x + 10

Answer:

The final answer is (x - 2) and (10x - 5)

meters to centimeters

The meter is the basic unit in mathematics. The meter is used for calculating the distance. The symbol of meter is ‘m’.

The kilo meter is a unit of distance end to end in a metric system. It is the usually used measurement unit for expressing distances between geographical places in most of the world

One kilometer = 1000 meter

One meter = 1 * 100 centimeters

One kilometer = 1* 100000 centimeters

Examples for meters to centimeters:

Example 1:

The Scale A 25 millimeter in one hour. The second scale B 8 millimeters in one ground. What is the difference between the Scale A and B?

Solution:

Here scale A and scale B having the speed in unusual units. Here we need to convert the speed in the equal unit and find the difference between Scale A and Scale B.

Scale A = one millimeter = 0.1 centimeter

So 25 millimeter = 25 * 0.1 = 2.5 centimeter

Scale B = 8 millimeter

Difference = 2.5 - 0.8 = 1.7 centimeter

Example 2:

Convert the 12 kilometers in centimeter?

Solution:

Step 1:

From the formula

1 kilometer = 100000 centimeter

Step 2:

Therefore 12 kilometers = 12 * 100000 = 1200000

Answer:

So, 12 kilometers = 1200000 centimeters

Example 3:

Convert the 250 meters in centimeter?

Solution:

Step 1:

From the formula

1meter = 100 centimeter

Step 2:

Therefore 250 meters = 250 * 100 = 25,000

Answer:

So, 250 meters = 25000 centimeters

Example 4:

Convert the 2000 mm in centimeter?

Solution:

Step 1:

From the formula

1mm = 0.1 centimeter

Step 2:

Therefore 2000 mm = 2000 * 0.1 = 200

Answer:

So, 2000 mm = 200 centimeters

Example 5:

Find the centimeters for 20 kilometers?

Solution:

Step 1:

From the formula

1 kilometer = 100000 centimeter

Step 2:

Therefore 20 kilometers = 20 * 1 00 000 = 2 000 000

Answer:

So, 20 kilometers = 2 000 000 centimeters

Example 6:

Convert the 500 meters in centimeter?

Solution:

Step 1:

From the formula

1meter = 100 centimeter

Step 2:

Therefore 500 meters = 500 * 100 = 50,000

Answer:

So, 500 meters = 50,000 centimeters

Example 7:

Convert the 10,000 mm in centimeter?

Solution:

Step 1:

From the formula

1mm = 0.1 centimeter

Step 2:

Therefore 10,000 mm = 10,000 * 0.1 = 1000

Answer:

So, 10,000 mm = 1000 centimeters

Example 8:

Find the centimeter for 216 kilometers

Solution:

Step 1:

From the formula

1 kilometer = 100000 centimeter

Step 2:

Therefore 216 kilometers = 216 * 100000 = 21 600 000

Answer:

So, 216 kilometers = 21 600 000 centimeters

Example 9:

Convert the 75 meters in centimeter?

Solution:

Step 1:

From the formula

1meter = 100 centimeter

Step 2:

Therefore 75 meters = 75 * 100 = 7500

Answer:

So, 75 meters  = 7500 centimeters

Example 10:

Convert the 15000 mm in centimeter?

Solution:

Step 1:

From the formula

1mm = 0.1 centimeter

Step 2:

Therefore 15000 mm = 15000 * 0.1 = 1500

Answer:

So, 15000 mm = 1500 centimeters

Converting Mixed Numbers

Mixed numbers is one of the basis concept of arithmetic. If we divide one number by another, we get a fraction. For example if 4 is divided by 5 we get `(4)/(5). ` The number above the bar is called the numerator and the number below the bar is called the denominator. If the numerator is less than the denominator it is called a proper fraction and if the number is greater than the denominator, it is called an improper fraction. An improper fraction can also be expressed a combination of an integer and a proper fraction. If we represent in this fashion it is called a mixed fraction or mixed number. For example `(6)/(5)`is an improper fraction and also can be represented as 1`(1)/(5)` which is a mixed fraction.

An improper fraction can be converted to a mixed fraction and we will see in detail how this can be done.

Having problem with What is an Improper Fraction keep reading my upcoming posts, i will try to help you.

Explanation for the mixed number math

The explanation for the mixed number math are given below,

There are many operations can be performed by a mixed number. They are given as,

• Addition operation mixed number
• Subtraction operation mixed number
• Multiplication operation mixed number
• Division operation mixed number

Example problem for mixed number math

Problem 1: Converting Improper Fractions to Mixed Numbers , `(13)/(2)` .

Solution:

Step 1: In the first step, we are going to divide the fraction number,

2 ) 13 ( 6      `->` Quotient Number

     12

  _____

      1        `->`  Remainder Number

  _____

where,

• 2 is called as the divisor.
• 13 is called as the dividend.
• 6 is called as the Quotient .
• 1 is called as the Remainder.

The mixed number for the given fraction is 6 `(1)/(2)` .

Problem 2: Convert the given fraction number into the mixed number , `(24)/(5)` .

Solution:

Step 1: In the first step, we are going to divide the fraction number,

5 ) 24 ( 4      `->` Quotient Number

     20

   _____

       4        `->`  Remainder Number

   _____

where,

• 5 is called as the divisor.
• 24 is called as the dividend.
• 4 is called as the Quotient .
• 4 is called as the Remainder.

The mixed number for the given fraction is 5 `(4)/(4)`.

Problem 3: Convert the given fraction number into the mixed number , `(37)/(6)` .

Solution:

Step 1: In the first step, we are going to divide the fraction number,

6 ) 37  ( 6      `->` Quotient Number

     36

   _____

       1        `->`  Remainder Number

   _____

where,

• 6 is called as the divisor.
• 37 is called as the dividend.
• 6 is called as the Quotient .
• 1 is called as the Remainder.

The mixed number for the given fraction is 6 `(1)/(6)`.

Practice problem for mixed number math

Problem 1: Convert the given fraction number into the mixed number , `(50)/(7)` .

Answer:  7 `(1)/(7)`

Problem 2: Convert the given fraction number into the mixed number , `(66)/(8)` .

Answer: 8 `(2)/(8)`

Algebra Absolute Value Inequalities

Algebra absolute value inequalities are nothing but the absolute value inequalities using the algebraic expression. Here we are going to see how to solve absolute value inequalities in algebra. We will see some example problems foe algebra absolute value inequalities. It is better to understand the inequalities. Normally absolute value mean without considering the sign of the value. For example |x| = `+-x`

Example problems for algebra absolute value inequalities:

Example 1 for algebra absolute value inequalities:

Solve the following |x + 3| `gt=` 5

Solution:

Given equation is |x + 3| `gt=` 5

We can divide these into two parts.

(x + 3) `gt=` 5 and - (x +3) `gt=` 5

First part:

If we take the first part (x + 3) `gt=` 5

Add -3 on both sides

We get x + 3 – 3 `gt=` 5 – 3

x `gt=` 2

Second part:

- (x +3) `gt=` 5

-x – 3 `gt=` 5

Add +3 on both sides

-x – 3 + 3 `gt=` 5 + 3

-x `gt=` 8

So x `lt=` -8

So the solution is -8 `gt=` x `gt=` 2

We will see some more examples for Solving Absolute Value Inequalities. It is better to understand the concept.

Example 2 for algebra absolute value inequalities:

Solve the following |x - 9| `lt= ` 2

Solution:

Given equation is |x - 9| <= 2

We can divide these into two parts.

(x - 9) `lt=` 2 and - (x - 9) `lt= ` 2

First part:

If we take the first part (x - 9) `lt=` 2

Add + 9 on both sides

We get (x – 9 + 9) `lt=` 2 + 9

x `lt= ` 11

Second part:

- (x - 9) `lt=` 2

-x + 9 `lt=` 2

Add -9 on both sides

-x + 9 – 9 `lt=` 2 - 9

-x`lt=` -7

So x `gt=` 7

So the solution is 7`lt=` x `lt=` 11

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These are some of the example for algebra absolute value inequalities. From this we can understand how to solve the algebra absolute value inequalities.

Rational Numbers Help

• Let us study about rational numbers help. There are several numbers categories in the mathematics subject.

• The clear knowledge about those types of numbers will help us to solve the different methods of math problems.

• The rational numbers are termed to be as the numbers that are represented in the simplest form of ratios with two integers as `a/b` , where the value of ‘b’ will be a nonzero value. Examples are below.

Rational numbers help:

Rational numbers help – example 1:

• Compose the addition process with the following rational numbers `4/2, 5/1, 24/6, 7/2, sqrt(4)` and 1.1

Solution:

• The given rational numbers series for composing addition process are `4/2, 5/1, 24/6, 7/2, sqrt(4)` and 1.1

• To compose the adding process with the given rational numbers follow the steps given below:

• `4/2, 5/1, 24/6, 7/2, sqrt(4)` and 1.1

• 2 + 5 + 4 + 3.5 + 2 + 1.1

• 17.6

• Thus we seemed to have the total added value of the given rational number series is found to be as ’17.6’

Rational numbers help – example 2:

• Compose the multiplication process with the following rational numbers `4/2, 5/1, 24/6, 7/2, sqrt(4)` and 1.1

Solution:

• The given rational numbers series for composing multiplication process are `4/2, 5/1, 24/6, 7/2, sqrt(4)` and 1.1

• To compose the multiplication process with the given rational numbers follow the steps given below:

• `4/2, 5/1, 24/6, 7/2, sqrt(4)` and 1.1

• 2 * 5 * 4 * 3.5 * 2 * 1.1

• 308

• Thus we seemed to have the total multiplied value of the given rational number series is found to be as ’308’

My previous blog post was on Rational Expression please express your views on the post by commenting.

 Rational numbers help – exercises:

• Compose the addition process with the following rational numbers 1`2/3, 4/1, 18/6, 9/2, sqrt(36)` and 0.1 (Answer: 21.6)

• Compose the multiplication process with the following rational numbers `12/3, 4/1, 18/6, 9/2, sqrt(36)` and 0.1 (Answer: 129.6)

study online absolute value help

Study online absolute value help is nothing but we are going to study about the absolute values and how to find the absolute values of the numbers and functions. Normally we know online mean if we are giving some keyword we will get the help on the keywords. Here we are going to get the help through online. And the keyword we are going to take here is absolute values. we can get the help on the absolute values using the examples.

Examples for Study online absolute value help:

Normally the absolute value notation is ||. If we want to find the absolute value of any number we just write the non-negative number of the particular number. This is the absolute value of the number.

Example:

For example absolute value of the number 16 is |16| = 16.

Absolute value of the number -19 is |-19| = +19

Likewise we have to write the absolute values of the number.

I am planning to write more post on Absolute Value Problems. Keep checking my blog.

From the above we can study about the absolute value of the number. We will see how to fins the absolute value of the function and the complex number.

More examples for Study online absolute value help:

First we will see an example for a function.

Study online absolute value help – function:

Find the absolute value of the following function |x – 1| = 5

Solution:

Here we have to find the absolute value of the given function.

So we have to wrote the function like

x – 1 = 5 …………. (1) And – (x – 1) = 5 …………….. (2)

Equation 1:

x – 1 = 5

Add +1 on both sides we get

x – 1 + 1 = 5 + 1

x = 6

Equation 2:

– (x – 1) = 5

–x + 1 = 5

Add -1 on both sides

We get –x + 1 – 1 = 5 – 1

So x = +4

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Study online absolute value help – complex number:

Find the absolute value of |x+ iy| = 1

Solution:

Here we have to find the absolute value of x.

We can write |x + iy| like

Absolute value = `+- sqrt(1^2 + 1^2)`

= `+- sqrt(1 + 1)`

= `+- sqrt(2)`

From the above we can study how to find the absolute value of the function and the complex number.

binomial distribution practice problems

In mathematics, binomial distribution is one of the interesting topics in probability theory and statistics. Binomial distribution is one of the main type’s theoretical frequency distributions. Binomial distribution is also called as a Bernoulli experiment.  Binomial distribution is the process of the number of success in a sequence for the n independent trails. Each trails gains the success of probability p. The following are the example and practice problems in binomial distribution.

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Binomial distribution practice problems – Mean Variance and standard deviation:

Mean:

μ = E[x] = np

Standard deviation:

σ = `sqrt(npq)`

Variance:

E[x^2] = σ^2 = npq

Binomial distribution practice problems – Example problems:

Here we solve some example problems based on the binomial distribution

Example 1:

A coin is tossed twelve times. Calculate the expected number of heads, variance and the standard deviation by using the binomial distribution.

Solution:

Given

Let coin tossed for ten times, so n = 12

If we toss a coin means, probability of getting head is p = `1/ 2`

Probability of getting tails is denoted by q

q = 1 – p                        [p + q = 1]

q = 1 – `1/ 2`

q = `1/2`

Mean:

μ = E[x] = np

= 12 (`1/2` )

= (`12/2` )

μ = E[x] = 6

Variance:

E[x^2] = σ^2 = npq

= 12(`1/2` )(`1/2` )

= 12(`1/4` )

= (`12/4` )

= 3

E[x^2] = σ^2 = 3

Standard deviation:

σ = sqrt(3)

= 1.732

σ = 1.732

Answer:

Mean = μ = E[x] = 6

Variance = σ^2 = E[x^2] = 3

Standard deviation = σ = 1.732

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Example 2:

A die is rolled for 50 times. Calculate the expected number , variance and the standard deviation by using the binomial distribution.

Solution:

Given

Let die is rolled for three times, so n = 50

If we roll a die means, probability of getting is p = `1/ 6`

Probability of not getting is denoted by q

q = 1 – p                        [p + q = 1]

q = 1 – `1/ 6`

q =` 5/6`

Mean:

μ = E[x] = np

= 50 (`1/6` )

= `50/6`

= `25/ 3`

μ = E[x] = 8.33 (or) `25/ 3`

Variance:

E[x^2] = σ^2 = npq

= 50(`1/6` )(`5/6` )

= 50(`5/36` )

= `125/18`

E[x^2] = σ^2 = `125/18` = 6.94

Standard deviation:

σ = `sqrt(6.94)`

= 2.64

σ = 2.64

Answer:

Mean = μ = E[x] = 8.33 (or) =`25/3`

Variance = σ^2 = E[x^2] = 6.94 (or) = `125/18`

Standard deviation = σ = 2.64

Binomial distribution practice problems – practice problems:

Problem 1:

A coin is tossed for twenty times. Calculate the expected number of heads, variance and the standard deviation by using the binomial distribution.

Solution:

Mean = μ = E[x] = 10

Variance = σ^2 = E[x^2] = 5

Standard deviation = σ = 2.23

Problem 2:

A coin is tossed for fifteen times. Calculate the expected number of heads, variance and the standard deviation by using the binomial distribution.

Solution:

Mean = μ = E[x] = 7.5

Variance = σ^2 = E[x^2] = 3.75

Standard deviation = σ = 1.94