Rational Zeros of a Function

Normally zeros of a function mean when we plug the values for the variables the functions values tends to be zero.  Let us consider if we are having a function with variable x and we have a set of solution p(x) if we plug the solution for the given variables present in the function we will get the function f(x) = 0. To find the rational zeros we have to use the rational zeros theorem.

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Rational Zeros of a Function – Examples:

The rational theorem states that if we are having the polynomial p(x) wit the integer coefficients and we are having the zeros of the polynomial `p / q` then we can say `p(p / q) = 0` . Here p is nothing but the constant term of the polynomial and the q I nothing but the leading coefficient of the polynomial p(x). We will see some examples for finding the rational zeros of a function.

Example 1 for rational zeros of a function:

Find the rational zeros of the following function.

2x2 + 12x + 10

Solution:

The given function is

2x2 + 12x + 10

First we have to find the rots of the constant term. ±1, ±2, ±5, ±10

Now the leading co – efficient of the constant term is 2. So we have to divide by 2.

±`p / q` = ±`1/ 2` , ±`2/ 2` , ±`5/ 2` , ±`10/ 2`

= ±`1/2` , ±1, ±`5/2` , ±5

Now we have to use the synthetic division method to find the rational zeros.

1 / 2 | 2           12        10

|               1         13/2

|_____________________

2           13        33/2      = not a zero

-1 / 2 | 2           12        10

|               -1         -11/2

|_____________________

2           11        9/2      = not a zero



1 | 2           12        10

|               2         24

|_____________________

2           24        34      = not a zero

-1 | 2           12        10

|               -2        -10

|_____________________

2           10        0      = is a zero

5/2 | 2           12        10

|               5        85/2

|_____________________

2           17        105/2      = not a zero

- 5/2 | 2           12        10

|               -5        -35/2

|_____________________

2           7        15/2      = not a zero

5 | 2           12        10

|              10        110

|_____________________

2           22        120      = not a zero

-5 | 2           12        10

|              -10      -10

|_____________________

2           2        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -5

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Rational Zeros of a Function – more Examples:

Example 2 for rational zeros of a function:

Find the rational zeros of the following function.

x2 + 4x + 3

Solution:

The given function is

x2 + 4x + 3

First we have to find the rots of the constant term. ±1, ±3

Now the leading co – efficient of the constant term is 1. So we have to divide by 1.

±`p / q` = ±`1/ 1` , ±`3 / 1`

= ±1, ±3

Now we have to use the synthetic division method to find the rational zeros.

1 | 1           4        3

|               1       5

|_____________________

1           5       8      = not a zero

-1| 1           4        3

|              -1      -3

|_____________________

1            3        0      = is a zero

3 | 1           4        3

|              3        21

|_____________________

1          7       24      = not a zero

-3| 1           4        3

|             -3      -3

|_____________________

1            1        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -3

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Interval of Convergence for Taylor Series

The interval of convergence for the given series is the set of all values such that the series converges if the values are within the interval and diverges if the value exceeds the interval.
The interval of convergence series must have the interval a - R < x < a + R since at this interval power series will converge.

In this article, we are going to see few example and practice problems of Taylor series to find interval of convergence which help you to learn interval of convergence.
Example Problems to Find the Interval of Convergence for Taylor Series:

Example problem 1:

Solve and determine the interval of convergence for Taylor seriessum_(n = 0)^oo(x^n)/(n!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo (x^n)/(n!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | ((x^(n+1))/((n+1)!))/((x^n)/(n!)) |

= lim_(n->oo) |   (xn!)/((n + 1)(n!))  |

= lim_(n->oo) |   x/(n + 1)  |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).

Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Example problem 2:

Solve and determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | (-1)^(n + 1)(x^(2(n + 1)+1))/((2n + 1)!) |

= lim_(n->oo) |  (-1)^(n + 1)(x^(2n + 3))/((2n + 1)!)   |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).


Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Practice Problems to Find the Interval of Convergence for Taylor Series:

1) Determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n))/(2n!)  .

2) Determine the interval of convergence for Taylor series sum_(n = 0)^ooxn.

Solutions:

1) The interval of convergence for the given Taylor series is (-∞, ∞).

2) The interval of convergence for the given Taylor series is |x| < 1.

Number Properties

Number properties and number operations is nothing but we are going to learn about the basic number properties and number operation. We are having the three basic properties and we are having the four basic operations. In this we are going to learn about the number properties and number operations using some example. It is very useful to performing the operations on the numbers easily.

Number Properties:

Basically we are having three basic number properties. We will see all the properties using some examples.

Distributive property:

Distributive property is nothing but giving the values. We can easily remember the distributive property using the following words multiplication distributes over addition.

2 X (3 + 5) = 2 X 3 + 2 X 5

2 X (8)        = 6 + 10 = 16

We are getting the equal value.

Associative property:

Associative property is nothing but grouping the numbers. We can use this for addition and multiplication.

2 + (3 + 5) = (2 + 3) + 5

2 + 8          = 5 + 5 = 10

This is the associative property of numbers.

Commutative property:

Commutative property of numbers is nothing but the value of numbers won’t change when the place of the numbers change.

2 + 3 = 3 + 2 = 5
Number Operations:

In numbers we can perform four types of operations.  Those operations are addition operation, subtraction operation, multiplication operation and division operation.



Addition operations

Addition operation is nothing but we are adding the value of any two numbers. If we take 5 and 6 if we want to perform addition operation we have to add the quantities of the numbers. The addition operation is denoted by the symbol`+`

So 5 + 6 = 11

Subtraction operation

Subtraction operation is nothing but we are going to subtract the value of numbers. The subtraction operation is denoted by `-`

Example:

9 – 6 = 3

Multiplication operation

Multiplication operation is nothing the repeated addition. The multiplication operation is denoted by `xx`

Example: 5 X 3

5 X 3 = 5 + 5 + 5 (We have to add the value of 5 three times)

= 15

Division operation

Division operation is nothing but the repeated subtraction. The division operation is denoted by `-:`

For example take 15 `-:` 3

We have to subtract the value of 3 from 15 again and again.

So 15 – 3 = 12

12 – 3 = 9

9 – 3 = 6

6 – 3 = 3

3 – 3 = 0

So totally this operation performed in 5 steps we got the remainder as 0.

So the quotient is 5 and the remainder is 0.

Decimal Numbers

The decimal number system (also called base ten or occasionally denary) has ten as its base. It is the numerical base most widely used by modern civilizations.

Decimal notation often refers to the base-10 positional notation such as the Hindu-Arabic numeral system; however it can also be used more generally to refer to non-positional systems such as Roman or Chinese numerals which are also based on powers of ten.

(Source – Wikipedia.)
Example Problems for Dividing Decimal Numbers:

Example problems 1 for dividing decimal numbers:

Divide 3.5 by 0.7

Solution:

Step 1: First we begin with 3.5 /0.7

Step 2: and Then we divide without decimal points, we attain

= 35 / 7

= 5

Step 3: and then we put decimal places

3.5 have one decimal place.

0.7 has one decimal place.

Step 4: Final answer has two decimal places 5

Answer 5.

Example problems 2 for dividing decimal numbers:

Divide 4.4 by 4

Solution:

Step 1: First we begin with 4.4 / 4

Step 2: and Then we divide without decimal points, we attain

= 44 / 4

= 11


Step 3: and then we put decimal places

4.4 have one decimal place.

4 have no decimal place.

Step 4: Final answer has one decimal places 1.1

Answer 1.1.

Example problems 3 for dividing decimal numbers:

Divide 1.2 by 0.2

Solution:

Step 1: First we begin with 1.2 / 0.2

Step 2: and Then we divide without decimal points, we attain

= 12 / 2

= 6

Step 3: and then we put decimal places

1.2 have one decimal place.

0.2 have one decimal place.

Step 4: Final answer has one decimal places 6.

Answer 2.

Example problems 4 for dividing decimal numbers:

Divide 6.4 by 8

Solution:

Step 1: First we begin with 6.4 / 8

Step 2: and Then we divide without decimal points, we attain

= 64 / 8

= 8

Step 3: and then we put decimal places

6.4 have one decimal place.

8 have no decimal place.

Step 4: Final answer has one decimal places 8

Answer 0.8.
Practice Problems for Dividing Decimal Numbers:

Practice problem 1 for dividing decimal numbers:

1) Divide 3.6 by 4

Ans: 0.9

Practice problem 2 for dividing decimal numbers:

2) Divide 7.2 by 0.8

Ans: 9

Practice problem 3 for dividing decimal numbers:

3) Divide 6.2 by 2

Ans: 3.1

Practice problem 4 for dividing decimal numbers:

4) Divide 42 by 0.7

Ans: 60

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Practice problem 5 for dividing decimal numbers:

5) Divide 72 by 0.8

Ans: 90.

Factoring Polynomials

Polynomials:

A polynomial is called an algebraic expression having the power of each variable is a only positive integral numbers.

For example,  8x2 -  16x3y2 + 24 y4 + 10, 9y2 - 16, y2 + 20x + 100.

These examples are a polynomial expression in two variables x and y and polynomial expression in one variable y.

Let us discuss about factoring polynomial .
Factoring Polynomials:

Factoring Polynomials:

Factoring a polynomial is the reverse process of multiplying polynomials.

When we factor a real number, First we are getting for prime factors for real number which multiply together to give the same real number.

For example,   28 = 7 x 2 x 2

When we are going to factor a polynomial,  First we need  to look for simpler polynomial expressions which are multiplied each other and give the same polynomial expression what we started with.

For example,  y5x2 + 25xy = 5xy(x + 5)

Let us solve sample problems on factoring polynomials.
Sample Problems: Factoring Polynomials

Steps to factoring Polynomials:

Step 1: First we factor the polynomial

Step 2: Expand the factorized polynomial using algebraic identities formula.

Step 3: Solve it if possible.

Problem 1:

Factor the polynomial expression: 25m2 – 64

Solution:

The given polynomial expression is 25m2 – 64.

Step 1: Factoring it,

= 52m2 – 82

= (5m)2 - 82

Step 2: Expand it

This is in the form of a2 – b2 = (a + b) (a – b)

So,

= (5x + 8)(5x – 8)

Therefore, the factors are (5x + 8)(5x – 8).

Step 3: Solving it,

5x + 8 = 0

5x = - 8

x = - `8/5`

5x - 8 = 0

5x = 8

x = `8/5`

Therefore, Solutions are x = `8/5` , `- 8/5` .

Problem 2:

Factor the polynomial expression: 9p2 – 81

Solution:

The given polynomial expression is 9p2 – 81.

Step 1: Factoring it,

= 32p2 – 92

= (3p)2 - 92

Step 2: Expand it

This is in the form of a2 – b2 = (a + b) (a – b)

So,

= (3p + 9)(3p – 9)

Therefore, the factors are (3p + 9)(3p – 9).

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Step 3: Solving it,

3p + 9 = 0

3p = - 9

p = -` 9/3` = - 3

3p - 9 = 0

3p = 9

p = `9/3` = 3

Therefore, Solutions are p = 3 , - 3.
Problem 3:

Factor the polynomial expression: A2 – `1/49`

Solution:

The given polynomial expression is A2 – `1/49` .

Step 1: Factoring it,

= A2 – `1/7^2`

= A2 - `(1/7)^2`

Step 2 : Expand it,

This is in the form of a2 – b2 = (a + b) (a – b)

So,

= (A + `1/7` )(A – `1/7` )

Therefore, the factors are (A + `1/7` )(A – `1/7` ).

Step 3 : Solving it,

(A + `1/7` )(A - `1/7` ) = 0

A + `1/7` = 0

A = - `1/7`

X - `1/7` = 0

X = `1/7`

Therefore, Solutions are A = `1/7` , - `1/7` .

Problem 4:

Factor the polynomial expression: y2 + 2y - 15

Solution:

The given polynomial expression is y2 + 2y  - 15.

Step 1 : Factoring it,

y2 + 2y - 15

Step 2 : Expand the polynomial expression

= y2 - 3y + 5y - 15

= y(y - 3) + 5(y - 3)

Taking out common factor,

= (y - 3)(y + 5)

Therefore, the factors are (y + 5)(y – 3).

Step 3 : Solving it,

(y + 5)(y - 3) = 0

y + 5 = 0

y = - 5

y - 3 = 0

y = 3

Therefore, Solutions are y = 3 , - 5.

Problem 5:

Solve for x in the perfect square trinomial: 9x2 + 24x + 16 = 0

Solution:

The given perfect square trinomial is 9x2 + 24x + 16 =0

Step 1: First factoring this perfect square trinomial

(3X)2 + 2·3.4x + 42 = 0

(3x + 4)2 = 0

Step 2: Expand it

This is in the form (ax + b)2 = (ax + b) (ax + b)

(3x + 4)(3x + 4) = 0

Step 3: Solving x

3X + 4 = 0

3X = - 4

Divide by 3 each side.

X = `- 4/3` .

Prime Factorization Of Composite Numbers

Converting the composite numbers in to multiplication of numbers is known as factorization, simplifying till the prime numbers is known as prime factorization.

Prime Number

    A whole number greater than 1 that cannot be divided by any other number except the number and the same number is known as the prime number.
    Some of the prime numbers are : 2, 3, 5, 7, 11, 13, and 17 ...

Composite numbers:

      It is an integer that has more than one prime factor. They can be expressed as unique set of prime numbers. The first composite number is 4. Besides 1 each other natural number is either prime or composite number.



Prime Factorization of Composite Numbers
Prime Factorization is the method of splitting the composite number into prime numbers

Factors:

The factor is the numbers that are multiplied to get another number.

  5 x 3 = 15

 (5, 3 factor)

while factoring if we get the factored numbers as the prime number, then the numbers are called as prime factors.

Example: The prime factors of 15 will be 5 and 3 ( 5 x 3 =15, and 5 and 3 are prime numbers).

Examples of Prime Factorization of Composite Numbers:

Example1.

What are the prime factorization of 225?

Solution:

Here 225 is a composite number, we are going to convert into a equivalent prime number.

It is best to start with the smallest prime number that can divide the number, which is 3, so let's check:

225 ÷ 3 = 75

since 75 is not a prime number we have to factor 75

75 ÷ 3 = 25

since 25 is not a prime number we have to factor 25.

25 ÷ 5 = 5

here, 5 is a prime number, so we can stop with this

225 = 3 x 3 x 5 x 5

The prime factorization of 225 is 3 x 3 x 5 x 5



Example 2.

What are the prime factorization of 490?

Solution:

Here 490 is a composite number, we are going to convert into a equivalent prime number.

It is best to start with the smallest prime number that can divide the number, which is 5, so let's check:

490 ÷ 5 = 98

since 98 is not a prime number we have to factor 98

98 ÷ 2 = 49

since 49 is not a prime number we have to factor 49.

49 ÷ 7 = 7

here, 5 is a prime number, so we can stop with this

490 = 5 x 2 x 7 x 7

The prime factorization of 490 is 5 x 2 x 7 x 7

Antiderivatives: An Introduction to Indefinite Integrals

The rate of change of a function at a particular value x is known as the Derivative of that function. Anti-derivatives as the name suggests is the opposite of derivatives. An Anti derivative is commonly referred to as an Indefinite Integral. We can define an indefinite integral of F as follows: Any given function G is an indefinite integral of F or an indefinite integral of a function g if the derivative of that function G’ equals g. The notation of an indefinite integral of F or the indefinite integral of a function is, G(x) = Integral g(x) dx. From this notation, we can conclude that G(x) equals integral[g(x)]dx  if and only if G’(x) = g(x)

From the above we can understand that an Antiderivative is basically an Integral of a given function, which is set into a formula which helps us to take the indefinite integral of F. So, when we say the Integral it means indefinite integral of F. Here, we have to remember to add a constant “c” as every integral has an unknown constant which is added to the equation. Let us consider an example for a better understanding, given function y = x^2 + 3x + 5. The derivative y’ would be 2x +3. Let us now find the antiderivative of y’, that gives us integral[2x +3] dx = 2. X^(1+1)/(1+1) + 3. x^(0+1)/(0+1) + c = 2x^/2 + 3x/1 + c = x^2+3x +c, this function is same as the original function except that the constant 5 is missing, this is the reason why we need to add the constant “c” to the Integral of a function.  From this we can conclude that Anti Derivative is the reverse derivative or the indefinite integral.

When we solve an Integral, we eliminate the integral sign and dx to arrive to a function G(x), this function is the antiderivative.  For instance, indefinite integral of F of the function x^3 is given by x raised to the power (3+1) whole divided by (3+1), that is, integral(x^3)dx = x^(3+1)/(3+1)= x^4/4. In general we can write the indefinite integral of F of x^n  as x^(n+1)/(n+1)
Antiderivative of Sec X
indefinite integral of F is the Integral of Sec X
Multiplying sec(x) with 1 which is [sec(x)+tan(x)]/[tan(x)+sec(x)]
Integral[sec(x)] dx = Integral[Sec x][sec(x) +tan(x)]/[tan(x)+sec(x)]      

Let u= sec(x) + tan(x)
Differentiating on both sides,
du = [sec(x)tan(x) + sec^2(x)]dx
Substituting u= [sec(x) + tan(x)]du = [sec(x)tan(x) + sec^2(x)]dx ,
Integral[sec(x)][sec(x)+tan(x)]/[tan(x)+sec(x)] dx
 = Integral[sec^2(x)+sec(x)tan(x)]dx/[sec(x)+tan(x)]
= Integral[du/u]
Solving the integral we get,
 = ln|u|+ c
Again substituting u= sec(x) + tan(x)
= ln|sec(x)+tan(x)| + c
So, the Antiderivative of Sec X is the Integral[sec x]dx = ln|sec(x)+tan(x)| + c