Methods of Factoring Trinomials

Introduction :

Trinomials:

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.

Trinomial expressions:

1. 3x + 5y + 8z with x, y, z variables

2. 3t + 9s2 + 3y3 with t, s, y variables

3. 3ts + 9t + 5s with t, s variables

4. Axaybzc + Bt + Cs with x, y, z, t, s variables, a, b, c nonnegative integers and A, B, C any constants.

5. Pxa + Qxb + Rxc where x is variable and constants a,b,c are nonnegative integers and P, Q, R any constants.

6. x2 + 8x + 15 where x is variable ( source : wikipedia)

Methods of Factoring Trinomials:

By using the following two methods we can factoring the trinomials.

Factoring trinomials Method 1:

The co-efficient of the first term (that is x2) is one. That is a=1.

x2+bx+c=(x-r1)(x-r2), Here r1 and  r2 are the roots of the trinomials.

We can say, (x - r1), (x - r2) are the factors of the trinomial.

Example Problems on methods of factoring trinomials :

By go through the following problem you can learn the method 1 of factoring the trinomials.

Example 1:

Factor the trinomial x2- 10 x +16

Solution:

Given, x2- 10 x +16

We need to factoring the given trinomial,

16   (product)

/    \   

- 8      - 2

\    /

-10     (sum)

So we can write the given equation into,

x2- 10 x +16 = x2 - 2x - 8x +16

= x ( x-2 ) - 8( x-2 )

= ( x - 8 ) ( x - 2 )

Answer: (x-8) and (x-2) are the factors of the given trinomial.

Verification:

(x-8) (x-2)  = x (x - 8) -2 (x - 8)

= x2-8x -2x+16

(x-8) (x-2)  = x2 -10x + 16

Example 2

Find the factors of   x2+ 3x -18.

Solution:

Given, x2+ 3x -18

We need to factoring the given trinomial,

-18 ( Product)

/     \

6       - 3

\     /

3 ( sum)

So we can rewrite the given equation into,

x2+3x - 18 = x2-3 x+6x -18

= ( x2-3x ) + ( 6x-18 )

= x ( x-3 ) + 6 ( x -3 )

= (x - 3) (x + 6)

Answer: factors (x-3) and (x+6)

Algebra is widely used in day to day activities watch out for my forthcoming posts on how to write an algebraic expression and example of algebraic expression. I am sure they will be helpful.

Methods of Factoring Trinomials:

Factoring trinomials Method 2:

The co-efficient of the first element( x2) trinomial is greater than one. That is a > 1.

The following example will help you to understand this method.

Example:

Factoring the trinomial 6x2- 3x - 3.

Solution:

Given, 6x2- 3x - 3 

To factor the trinomial, multiply the coefficient of first term with the constant i.e. 6 * (-3)=-18

- 18    (product)

/    \

- 6     3

\     /

- 3    (sum)

6x2- 3x - 3  =  6x2- 6x + 3x -3

=  6x (x - 1) + 3(x-1)

= (6x+3) (x-1)

= 3(2x+1)(x-1)

Answer: Factors of the given trinomial 3, (2x+1), (x-1)

Easy way to solve mean

Mean is nothing but average value of set of numbers. Mean almost always refers to arithmetic mean.

Mean = Total of all the elements in a data set / Total number of elements in a data set

The following steps are given below:

Step 1: Find how many elements are there in a set.

Step 2: Find the sum of all the elements in a set.

Step 3: Divide the sum by total number of elements in a set.

Step 4: Final answer is the required mean.

Easy Way to Solve Mean - Examples

Example 1: Solve and find the arithmetic mean value for following data {32, 45, 59, 62, 19, 72, 21, 42, 87}

Solution:

Step 1: Find how many elements are there in a set.

There are 9 elements, so n = 9.

Step 2: Find the sum of all the elements in a set.

32 + 45 + 59 + 62 + 19 + 72 + 21 + 42 + 87 = 439

Step 3: Divide the sum by total number of elements in a set

= Sum of all the element / Total number of elements.

= 439 / 9

= 48.77

Therefore arithmetic mean of given date is 48.77

Example 2: Solve and find the arithmetic mean for following data {21, 42, 63, 74, 56, 46, 57, 68, 79}.

Solution:

Step1: Find how many elements are there in a set.

There are 9 elements, so n = 9.

Step 2: Find the sum of all the elements in a set.

21 + 42 + 63 + 74 + 56 + 46 + 57 + 68 + 79 = 506

Step 3: Divide the sum by total number of elements in a set

= Sum of all the element / Total number of elements.

= 506 / 9

= 56.22

Therefore arithmetic mean of given date is 56.22

Example 3: Calculate the arithmetic mean for following data {14, 29, 37, 43, 26, 52, 62, 71, 24}.

Solution:

Step1: Find how many elements are there in a set.

There are 9 elements, so n = 9.

Step 2: Find the sum of all the elements in a set.

14 + 29 + 37 + 43 + 26 + 52 + 62 + 71 + 24 = 358

Step 3: Divide the sum by total number of elements in a set.

= Sum of all the element / Total number of elements.

= 358 / 9

= 39.77

Therefore arithmetic mean of given date is 39.77

Example 4: Calculate the arithmetic mean for following data {31, 43, 68, 87, 85, 69, 157, 277}.

Step1: Find how many elements are there in a set.

There are 8 elements, so n = 8.

Step 2: Find the sum of all the elements in a set.

31 + 43 + 68 + 87 + 85 + 69 + 157 + 277 = 817

Step 3: Divide the sum by total number of elements in a data set.

= Sum of all the element / Total number of elements.

= 817 / 8

= 102.125

Therefore arithmetic mean of given date is 102.125

Easy Way to Solve Mean - Practice

Solve these practice problems using easy way. These practice problems are very easy to solve

Problem 1: Calculate the arithmetic mean for following data {66, 55, 97, 88, 35, 36, 74, 82}.

Answer: 66.62

Problem 2: Calculate the arithmetic mean for following data {232, 346, 445, 556, 662, 771, 774, 321}.

Answer: 513.375

An Introduction to Regression Analysis

Let’s have an example of regression .A new plant to manufacture widgets was set up, the plant’s personnel manager advertises the employment opportunity in the plant and the very next morning it has 2000 people waiting for applying to the 100 available jobs. It will be important to select the best of 100 people who will cut above the rest employees because the training of these people will involve both time and money and firing the employee is going to be difficult as well as bad for community relations. In order to overcome this situation and to help make correct decision an analysis is adopted by the Personnel manager, this analysis is the Regression analysis.  A Regression analysis is used to predict the purposes and to understand relationship between two variables x and y. It is used when two or more variables are systematically connected by a linear relationship.  A Multivariate Regression Analysis is a technique that estimates a single regression model with more than one outcome variable.

In a Regression equation, y is the dependent variable and x is a independent variable, the Regression Formula can be given as y’= a +bx; where y’ is the estimated y which is on the y-axis from the point on the regression line for the predictor x value, x is an arbitrary chosen value of the predictor variable for which the corresponding value of the criterion variable is desired, a is the intercept point of the regression line and the y-axis and b is the slope of the regression line. a and b are calculated as follows,
a = [sigma(y)][sigma(x2)] – [sigma(x)][sigma(x)(y)]divided by {n [sigma(x2)]- [sigma(x)]2} and

b= n[sigma(xy)] – [sigma(x)][sigma(y)] divided by {n [sigma(x2)]- [sigma(x)]2}

Regression Analysis Example or Regression Example: Regression analysis is an analysis to assess the values of the given parameters for a particular function that cause the function to the best utilize a set of observed datas that are given. For example, if we take into account the value of an automobile, it decreases constantly with a certain amount each year after the purchase, and each for mile it has driven, the related linear function would predicts its value as a function of the two independent variables like ‘age’ and ‘miles’:  value = price + dep age x age + dep miles x miles; where value is the dependent variable which is the value of the car, age is the age of the car and the miles is the number of miles that the car has covered. The regression analysis done determines the best of values of the three parameters, price, the estimated value when the age is zero, dep-age, which is the depreciation that takes place each year, dep-miles is the depreciation for each mile driven.  The values of dep-age and dep-miles will be -ve as the car would be losing value as the car age and the miles covered increase.

Sum and Difference Formulas in Trigonometry

Sum and Difference Formulas for Sine and Cosine
The sum Formulas for Sine and Cosine are:
Sin(A+ B) = SinA.CosB + CosA.SinB
Cos(A+B) = CosA.CosB – SinA.SinB

The difference Formulas for Sine and Cosine are:
Sin(A – B) = SinA.CosB  - CosA.SinB
Cos(A – B) = CosA.CosB + SinA.SinB

Trigonometric Sum and Difference Formulas
Sin(x+y) = sin(x).cos(y) + cos(x).sin(y)
Cos(x+y) = cos(x).cos(y) – sin(x).sin(y)
Tan(x+y) = [tan(x) + tan(y)]/[1- tan(x).tan(y)]

Sin(x-y) = sin(x).cos(y) – cos(x).sin(y)
Cos(x-y) = cos(x).cos(y) + sin(x).sin(y)
Tan(x-y) = [tan(x) – tan(y)]/[1+ tan(x).tan(y)]

Let us solve some of trigonometric problems using trig sum and difference formulas
Solve, cos(30 degrees)cos(15 degrees) – sin(30 degrees)sin(15 degrees) without actually solving. The given trigonometric expression is in the form cos(x).cos(y) – sin(x).sin(y) which is equal to cos(x+y) a trig sum formula. Comparing the terms we get, x = 30 degrees and y = 15 degrees and hence x+ y = 30 + 15 = 45 degrees. Finally we get, cos(x+y) = cos(30+15) = cos(45) = sqrt(2)/2

Sum and Difference Formulas Trig functions sine, cosine and tangent are given as follows:
Sin(alpha+ beta) = sin(alpha).cos(beta) + cos(alpha).sin(beta)
solve sin(75 degrees)
75 degrees is not special angle, but we can split 75 to give 45 + 30, we know both 45 and 30 degrees are special angles. So, we can re-write sin(75 degrees) = sin(45+30) applying the sum formula of sine, we get
Sin(45).cos(30) + cos(45).sin(30) = (1/2)(1/sqrt2) + (sqrt3/2) (1/sqrt2) = sqrt(2)[sqrt(3) +1]/4

Cos(alpha+beta) = cos(alpha).cos(beta) – sin(alpha).cos(beta)
Solve cos(5 pi/12) = cos(pi/4 + pi/6)= cos(pi/4).cos(pi/6) – sin(pi/4).sin(pi/6) = [sqrt(2)/2 ].[sqrt(3)/2] – [sqrt(2)/2. ½]= [sqrt(6) – sqrt(2)]/4

Tan(alpha+beta)
= sin(alpha+beta)/cos(alpha+beta)
= [sin(alpha)cos(beta) + cos(alpha)sin(beta)]/[cos(alpha)cos(beta) – sin(alpha).sin(beta)]
= {[sin(alpha)cos(beta)/cos(alpha)cos(beta)] +[ cos(alpha)sin(beta)/cos(alpha)cos(beta)]}
 Divided by [cos(alpha)cos(beta)/ cos(alpha)cos(beta)] – [sin(alpha).sin(beta)/ cos(alpha)cos(beta)]
= [tan(alpha) + tan(beta)]/[1- tan(alpha)tan(beta)]
tan(alpha+beta) =  [tan(alpha) + tan(beta)]/[1- tan(alpha)tan(beta)]

Sin(alpha- beta) = sin(alpha).cos(beta) – cos(alpha).sin(beta)
Solve sin(15) = sin(45- 30) = sin(45).cos(30) – cos(45).sin(30)
                  = [sqrt(2)/2].[sqrt(3)/2] – [sqrt(2)/2].[1/2] = [sqrt(6) – sqrt(2)]/4
Cos(alpha – beta) = cos(alpha).cos(beta) + sin(alpha).sin(beta)
Verify cos(alpha – pi) = - cos(alpha). Using the above difference formula for cosine we get,
Cos(alpha – pi) = cos(alpha).cos(pi) + sin(alpha).sin(pi) we know that cos(pi) = -1 and sin(pi) = 0
Substituting the values, we get, cos(alpha – pi) = - cos(alpha) + 0 = - cos(alpha) [verified]

tan(alpha – beta)
 = sin(alpha – beta)/cos(alpha – beta)
= [sin(alpha)cos(beta) - cos(alpha)sin(beta)]/[cos(alpha)cos(beta) + sin(alpha).sin(beta)]
= {[sin(alpha)cos(beta)/cos(alpha)cos(beta)] -[ cos(alpha)sin(beta)/cos(alpha)cos(beta)]}
 Divided by [cos(alpha)cos(beta)/ cos(alpha)cos(beta)] + [sin(alpha).sin(beta)/ cos(alpha)cos(beta)]
= [tan(alpha) - tan(beta)]/[1 +tan(alpha)tan(beta)]
tan(alpha – beta)= = [tan(alpha) - tan(beta)]/[1 +tan(alpha)tan(beta)]

Derivative table

Derivative means differentiation of any function. In other words calculating a slope of a function. Here we have to discuss about derivative table. Derivative table is a collection of all differentiation formulas in summarized form in a single table. It can be different table for different function. In derivative table first type is power of X, under this type we differentiate a constant term (d/dx C=0), function X (d/dx X=1) and function x^n (d/dx x^n=n *x^n-1).

Second type in derivative table is exponential and logarithmic functions. In this type we differentiate exponential function d/dx  e^x=  e^x, logarithmic function d/dx logx= 1/x and function like b^x which is equal to bxln(b). third type in derivative table is trigonometric function, its is also know as trig derivative table, in this type we differentiate sin function d/dx sinx=cosx, cosine function, tangent function, cosec function , sec function, cotangent function. By these differential formulas we can make trig derivatives table, we can also proof trig function by algebraic method, in this method we first proof the sin function then with the help of sin proof, we can proof cosine function and in last with the help of both sin and cosine function proof we can proof tangent function.

Forth type in derivative table is inverse trigonometric function, calculus part mostly used in this type. In this type we differentiate inverse sin function d/dx sin^-1x= 1/(√1-x^2), inverse cos function, d/dx cos^-1x = -1/(√1-x^2), inverse tan function d/dxtan^-1x = 1/(1+x^2), inverse sec function d/dx sec^-1x = 1/(x*√x^2 -1), inverse cot function d/dx cot ^-1x =-1/(1+x^2). Fifth and the last type in derivative table is hyperbolic function. In this type we differentiate hyperbolic sin d/dx sinhx= coshx, hyperbolic cos function d/dx coshx= sinhx, hyberbolic sec function d/dx sechx= -tanhx*sechx, hyperbolic cosec function d/dxcosechx = -cothx* cosechx, hyperbolic tan function d/dxtanhx= 1-tanh2x, hyperbolic cot function d/dxcothx= 1-coth2x.

Double integrals in polar coordinates, in double integration we have region (D) in every trigonometry problem, region D is much easier to describe in Cartesian form. But some region such as disk, ring or any portion of disk or ring, if we use Cartesian coordinate it will be very difficult to differentiate these functions. So for these functions we use polar coordinate. Suppose in any problem disk is given with radius 2 then in Cartesian it is difficult to differentiate with limit, but when we change limits in polar coordinate in θ and r  form  then it become simple and we can integrate  it easily. Double integrals over rectangles means integration of a area made by curve, which cut x axis and y axis at equal points.

About the Newton Raphson method

Consider the following equation:
Y = x^2 + 2x – 3
If you are asked to find the roots of that equation, it is a pretty simple process of factoring the equation and then using the zero product rule to solve for possible values of x.

Illustrated as follows:
Y = x^2 + 2x – 3
 = (x+3)(x-1)
For finding zeros, y = 0
So, (x+3)(x-1) = 0
Using the zero product rule,
(x+3) = 0 ; (x-1) = 0
Thus, x = -3 or x = 1. Therefore these are the two zeros of the given equation.

Now consider the following equation:
Y = 2x^2 + x+ 5
Since in this case the left hand side polynomial is prime, we can simply use the quadratic formula to find the real roots of this equation (if any).

But now suppose if the function is like this:
Y = 3x^5 + x^3 – 2x^2  - x + 7
There are no simple methods or formulae for finding the roots (or zeros) of such a function. This is where we use some method of approximation. One such method is the Newton – Raphson method named after the mathematicians  Issac Newton and Joseph Raphson who invented the method. The idea here is to start with some approximate zero of the function and then through iterative process come to closer and closer approximations of the zeros of the polynomial function.  For a function of one variable the Newton -

Raphson algorithm can be stated as follows:
1. For a function f, we guess the zero = x0 to begin our iterations.
2. For the next best approximation of the zero of f we use the following formula:
X1 = x0 – f(x0)/f’(x0). Thus f(x1) is closer to 0 than f(x0).
3. The next best approximation would be x2 given by the formula:
X2 = x1 – f(x1)/f’(x1). Thus f(x2) is closer to 0 than f(x1)
4. The above iterative process can be continued for as many number of times as we like. Each time the zero we get would be closer and closer to the actual zero of the polynomial.
5. The general formula for the (n+1)th approximation is like this:
X(x+1) = x(n) – f(x(n))/f’(x(n))

For any newton – raphson example, we begin by guessing a zero. For functions with smaller coefficients, we can conveniently assume 0 as an approximate zero. For other functions, we may graph the function using few test values to come to some approximate first guess.

Graphing and solving logarithmic functions

A logarithmic function is the inverse of an exponential function. The general form of an exponential function would be y = b^x, where b is the base of the exponent and x is the exponent or the index. b belongs to positive real numbers and x is any real number. Therefore we see that the domain of an exponential function is all real numbers; whereas the range of this exponential function would be all positive real numbers.

The general form of a logarithmic function would be y = log_b?x, where b is the base of the logarithm. It is the same base that we used above in the exponential function. The x and y have switched places. Therefore, if we were to write the exponential function y = b^x in logarithmic form it would be log_b?y = x.

Graphs of logarithmic functions: To graph logarithmic functions online could be easy, but to graph them manually is not difficult either. Since we know that the logarithmic function is the inverse of an exponential function, we can make a table of values and plot the points, and then join the points with a curve. Let us try to see that with an example.

Example 1: Graph the function y = log_2?x.

Tabulating the results we have:
X (1/4) (1/2) 1 2 4 8
y -2 -1 0 1 2 3

Now we plot those ordered pairs on a graph sheet and run a curve through it. See picture below:

The blue curve above is the graph of the logarithmic function y = log_2?x. The red curve is the graph of the inverse exponential function y = 2^x.
Solving logarithmic functions:  There are various methods to solve logarithmic functions. One way is to convert the logarithmic function to the corresponding exponential function and then solve.
Example 2: Solve log_5?125 = y.
Solution: Converting to exponential form we have, 5^y = 125
5^y = 5^3
Since in the above equation the bases are equal, the exponents would also be equal.
Therefore, y = 3 is the answer.