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

A short note on derivative of Cot function

Cot is the short for the trigonometric function cotangent. It is the complementary function of the tan also called the tangent function. Cot is defined as the ratio of the adjacent side to the opposite side of the acute angle in a right triangle. It is also the reciprocal of the tan function.

Therefore symbolically it can be written like this:
Cot(x) = adjacent side/opposite = 1/tan(x)
We can graph the cot function using a table of values as follows:
X -pi -3pi/4 -pi/2 -pi/4 0 Pi/4 Pi/2 3pi/4 pi
Cot(x) Inf 1 0 -1 Inf 1 0 -1 inf

The graph would look as follows:

 From the graph we note that the cot function is not defined at the points –pi, 0 and pi. That is because at these points, the tan function has the values of 0,0 and 0 respectively. Since the cot function is the reciprocal of tan function, the cot is not defined at these points.

Derivative Cot X  function:

The above graph is of the function y = Cot(x). Now the derivative of Cot X would be the slope of tangent to the above curve at any point x. Therefore for example if we were to find the derivative of Cotx at the point x = pi/4, then the graph of the tangent line to the curve at that point would like this:

The blue line in the above picture is the tangent to the cot function at the point x= pi/4. The slope of this tangent is the derivative of cot x. In other words we can also say that the derivative of the cot function is the rate of change of cot function at a given point. This is also called the instantaneous rate of change of cot function at a point x=a. (In this case it is x = pi/4). Another way of stating the same thing is like this: the gradient of the curve of the function y = cot x at a point x=a, is called the derivative of the function at that point.

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The derivative of the function y = cot x at the point x = 0 can be found by making a tangent to the curve at the point x=0. Again from the above graph we see that, at x=0, the tangent would be vertical. So slope of tangent = not defined. Therefore derivative at this point is not defined.

Homogeneous and non homogeneous differential equation

Non Homogeneous Differential Equation: A differential equation is a linear differential equation if it is expressible in the form P0 d^n y/dx^n + P1 d^n-1 y/dx^n-1 + P2 d^n-2 y/dx^n-2 + … + Pn-1 dy/dx + Pn y = Q, Where P0, P1, P2, …., Pn-1, Pn and Q are either constants or functions of independent variable x.Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient  of the various terms are either constants or functions of the independent variable, then it is said to be linear differential equation. Otherwise, it is a non-differential equation.

Homogeneous differential equation: A function f(x, y) is called a homogeneous function of degree n if f(lambda x, lambda y) = lambda^n f(x, y).

For example, f(x, y) = x^2 – y^2 + 3xy is a homogeneous function degree 2, because f(lambda x, lambda y) = lambda^2 x^2 – lambda^2 y^2 + 3. Lambda x. lambda y = lambda^2 f(x, y).Let us understand Homogeneous Differential Equation Examples. Suppose we have to Solve x^2 y dx – (x^3 + y^3) dy = 0 . The given differential equation is x^2 y dx – (x^3 + y^3) dy = 0, dy/dx = (x^2 y)/(x^3 + y^3)……(i), Since each of the functions x^2 y and x^3 + y^3 is a homogeneous function of degree 3, so the given differential equation is homogeneous. Putting y = vx and dy/dx = v + x dv/dx in (i), we get  v + x dv/dx = vx^3 /(x^3 + v^3 x^3), v + x dv/dx = v/1 + v^3, x dv/dx = [v/(1 + v^3)] – v, x dv/dx = (v – v – v^4)/(1 + v^3), x dv/dx = -v^4/(1 + v^3), x (1 + v^3) dv = -v^4 dx, (1 + v^3)/v^4 dv = -dx/x, (1/v^4 + 1/v) dv = -dx/x, Integrating both sides we get v^-3/-3 + log v = -log x + C=> -1/3v^3 + log v + log x = C=>-1/3 x^3/y^3 + log(y/x . x) = C=-1/3 . x^3/y^3 + log y = C, which is the required Homogeneous Solution Differential Equation.

Linear Homogeneous Differential Equation:  A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree. The general form of a linear differential equation of first order is  dy/dx + Py = Q, where P and Q are functions of x (or constants).For example, dy/dx + xy = x^3, x dy/dx + 2y = x^3, dy/dx + 2y = sin x etc. are linear differential equations. Let us Find the General Solution of the Homogeneous Differential Equation  (1 + y^2) dx + (1 + x^2) dy = 0. Now   (1 + y^2) dx + (1 + x^2) dy = 0 => dx/(1 + x^2) + dy/(1 + y^2) = 0.On integration, we get  tan^-1 x + tan^-1 y = tan^-1 C, (x + y)/(1 – xy) = C, X + y = C (1 – xy).

P values

P values
P values are obtained from either the tables or programmable calculators. We can use standard excel sheet as well to calculate the P values. Rather it is tedious to evaluate manually and requires a great mathematical skills.


From Z score
Step 1: Under Normal Distribution, we try to identify the z score for the respective mean and standard deviation value of the null hypothesis using the formula
z= (Test value – mean)/ standard deviation
Step 2: Refer to the Z table and find the P value directly.
To read the P values from the normal distribution table take the first two digits of the z score and locate it on the left most column, move along that row to locate the intersection of third digit of the z score along the column.


Table of P values:
We have two z tables to find the p values. As the normal distribution is symmetric about its mean, and assumes a bell shaped curve, the area right of the mean is equal to the area left of the mean. We have table for the z values varying between 0 and positive infinity, and for the values varying between negative infinity to positive infinity.

P Values Table

P values Significance
The importance of P values in statistical analysis is to test the authenticity of the hypothesis. We need to identify the hypothesis related to the study. In case of null hypothesis, the P value signifies whether the null hypothesis is true or not subject the allowed deviation.
It is imperative to note that the P values calculated should not be less than the allowed significance level to accept the null hypothesis else the alternate hypothesis is accepted.


Finding P values
We adopt the following algorithm to evaluate the P value
Step 1: Propose a null hypothesis
Step 2: Indentify the means and standard deviation
Step 3: Find the Z score
Step 4: Find the P value from the table.

P values in Statistics:
We need a parameter to take an impartial decision. In statistical significance testing, P value gives us the probability of obtaining maximum value of the test statistic as the one that is actually either desired or observed with an assumption that the null hypothesis made is true. We reject the hypothesis if the p- value obtained is less than the significance level, which is usually, 0.05 or 0.01.When the null hypothesis is rejected, and the result proposed is statistically significant.

What is absolute value?

What is absolute value?
Absolute value of number is defined asthe difference between it and zero and is denoted by symbol of |x| or sometime we can use abs(x). Absolute valueismost commonly used by mathematicians and scientists as a tool which is used to separate magnitude and direction when only magnitude matters. Absolute value is positive. The absolute value of a real number is the difference between that real number and zero

If x is the real number then the absolute value for real number |x| = x, when it is positive. |x| = -X, when x is not positive. The absolute value is the distance of x from 0.
The equation for absolute value is,
K = |x-b| ------------- (1)
If x-b is positive
Then k = x - b
X = b + k   ------------- (2)
If x-b is negative
-k = x - b
X = b – k ---------------- (3)
Finding the absolute values.
Example: | x-5 | = 7
From equations 2 and 3 the absolute values are
| x-5 | = 7 | x-5| = -7
X = 7 + 5 x = -7 + 5
X = 12 x = -2
So the absolute values are {12, -2}

Definition for absolute value

The absolute value of a number measures its distance to the origin/zero on the real number line. From above the figure x is 5 units away from the zero. The absolute value is always positive, so it can change the sign of the negative number in to positive number. But the absolute value of zero has no sign, since it is equal to its absolute value.

Absolute value equations:
Isolate the absolute value expression to solve absolute value equations. The expression is called as k. Then write both the equations in which the expression provided within the absolute value symbol is one of the two expressions.  Write the first equation k should be equal to the expression inside the absolute value symbol, and then write the other equation should be equal to –k. Then we get the set of solution obtaining the solution for the equations.

Example: | 2x+12 | = 4x
| x – b | = k
From above equation 2 and 3
| x – b | = k | x – b | = - k
2x + 12 = 4x 2x + 12 = -4x
X = 6 x = -2
So the solution of equation is {-2, 6}

Absolute value in algebra
Absolute value is defined as; it is functions which measure the size of the element as a field or integral domine (D).
|x| = 0
|x| = 0 if and only if x =0
|xy| = |x||y|
|x+y| = |x| + |y|

Example: |1|  = 1
  |-1| = 1