Multiplying Trinomials

An Algebraic expression in the form axn is called a monomial in x, where a is a known number, x is a variable and n is a non-negative integer. The coefficient of yn and n has a number, the degree of the monomial. For example, 7x3 is a monomial in x of degree 3 and 7 is the coefficient of x3.When we have  2 monomial in the polynomial function it is known as a binomial , similarly sum of three monomial is called a trinomial. For example, 2x5 – 3x2 + 3 is a trinomial. we can perform various operations like addition, subtraction, multiplication and division. Now we are going to see multiplying trinomials.

Methords of multiplying trinomials:

We can multiply the trinomials by using the distributive method, in that there are two methods followed one is Horizontal method, and the other one is vertical method.

Horizontal method:

This horizontal method is one of the distributive methods. This is used for multiplying the trinomials in horizontal form and simplify them.

Vertical method:

This vertical method is one of the distributive methods. This is used for multiplying the trinomials in vertical form and simplify them.

These are illustrated below with examples

Example for multiplying trinomials:

Example 1:

Using the vertical method and multiply the trinomials

(a + 5) (a2 + 7a + 8)

Solution:

Given, (a + 5) (a2 + 7a + 8)

Using vertical method, we multiply the given trinomial

a2 + 7a + 8

a + 5

a3 + 7a2 + 8a                multiply the trinomial a2 +7a + 8 from the second binomial term a.

5a2  + 35a + 40    multiply the trinomial a2 +7a + 8 from the second binomial term + 8

a3 + 12a2 + 43a + 40       (Combine like terms then we get result)

Solution to the given trinomials is a3 + 12a2 + 43a + 40.

Example 2:

Using the horizontal method and multiply the trinomials

(a + 6) ( a2 + 12a + 23)

Solution:

Given

(a + 6) ( a2 + 12a + 23)

Using horizontal method, we multiply the given trinomial

(a + 6) ( a2 + 12a + 23)

Multiply the trinomial a2 + 12a + 23 from the second binomial term a.

(a2 + 12a +23) × (a)

a3 + 12a2 + 23a

Multiply the trinomial a2 + 12a + 23 from the second binomial term 6.

(a2 + 12a + 23) × 6

6a2 + 72a + 138

Group like terms

a3 + 12a2 + 23a + 6a2 + 72a + 138

a3 + 12a2 + 6a2 + 23a + 72a + 138



Combine like terms

a3 + 18a2 + 95a + 138

Solution to the given trinomials is a3 + 18a2 + 95a + 138

Antiderivative of Trigonometric Functions

In calculus, a function F is said to be antiderivative of the function f if the derivative F' = f. The process of finding antiderivatives is called as antidifferentiation.

F(x) = int f(x) dx

Learning antiderivative of trigonometric functions from the tutor is interactive and fun than a book. Students feel more convenience to study antiderivative of trigonometric functions from the tutor.. Tutors conduct the regular tests to improve the student's knowledge about antiderivatives. Following is the list of antiderivatives formulas and example problems to show how the tutor helps to learn antiderivative of trigonometric functions.

Study antiderivative formulas for trigonometric functions from tutor:

int sin x dx = - cos x + C

int cos x dx = sin x + C

int sec2 x dx = tan x + C

int cosec2 x dx = - cot x + C

int sec x tan x dx = sec x + C

int cosec x cot x dx = - cosec x + C

int tan x dx = ln |sec x| + C

int cot x dx = - ln |cosec x| + C

int sec x dx = ln |sec x + tan x| + C

int cosec x dx = ln |cosec x - cot x| + C

Study antiderivative of trigonometric functions with example problems from tutor:

Example problem 1:

Find antiderivative of a function,  f(x) = 7cos (5x)

Solution:

Step 1: Given function

f(x) =7cos (5x)

int f(x) dx = int 7cos (5x) dx

Step 2: Integrate the given function with respect to ' x',

int7cos (5x) dx = 7(sin (5x)) 1/5

= (7sin (5x))/5

Example problem 2:

Find antiderivative of a function y = 5sec2 (9x)

Solution:

Step 1: Given function

y = 5sec2 9x

int y dx = int 5sec2 9x dx

Step 2: Integrate the given function y = 5sec2 9x with respect to ' x',

int5sec2 9x dx = 5tan (9x) 1/9

= (5tan (9x))/9

Example problem 3:

Find antiderivative of a function, f(x) = 5sin (4x) + sec (6x)

Solution:

Step 1: Given function

f(x) = 5sin (4x) + sec (6x)

int f(x) dx = int 5sin (4x) + sec (6x)dx

Step 2: Separate the integral function

int 5sin (4x) + sec (6x) dx = int  5sin (4x) dx + int sec (6x) dx

Step 2: Integrate the above function with respect to ' x ',

int5sin (4x) + sec (6x) dx = (-5cos (4x))/4 + (log (sec (6x) + tan (6x)))/6 + C

Algebra Rational Exponents

Algebra is one of the most basic element of mathematics in which, we switch from basic arithmetic to variables.  Algebra covers a large number of subdivisions like polynomials, rational, exponents, logarithms, expressions etc under it. Exponents are in the form of 'ab ' where a is the base and b is the power (exponent). Exponents in rational form are called as rational exponents. For example: a 1/2 , 'a' has a rational exponent of 1/2. The rules , representation and examples on algebra ational exponents is given in the following sections.

Rational Exponents:

As said earlier rational exponents are in the form  ab/c, where b/c is the rational exponent. Algebra rational exponents can be represented in the following ways.

root(n)(x)  = x^(1/n)

root(3)(x)  = x^(1/3)

root(3)(x^2)  = x^(2/3)

an = b  ===>  a = b^(1/n)

Examples on Rational exponents:

ALgebra Example 1:

Simplify the expression(root(3)(2^3))^4

Solution:

The given expression is (root(3)(2^3))^4

It can be represented as  ((2^3)^(1/3) )4

Therefore, (2^(3/3) )4

= 24

= 2*2*2*2 = 16

Therefore, The simplified answer for the expression is 16.

Algebra Example 2:

Simplify the expression(root(2)(2^3))^4

Solution:

The given expression is (root(2)(2^3))^4

It can be represented as  ((2^3)^(1/2) )4

Therefore, (2^(3/2) )4

= 2^(12/2)

= 26 = 2*2*2*2*2*2 = 64

Therefore, The simplified answer for the expression is 64.

Algebra Example 3:

Simplify the expression(root(2)(2^3))^2

Solution:

The given expression is (root(2)(2^3))^2

It can be represented as  ((2^3)^(1/2) )2

Therefore, (2^(3/2) )2

= 2^(6/2)

= 23 = 2*2*2 = 8

Therefore, The simplified answer for the expression is 8.

Practice problems on rational exponents:

Here are few practice problems given to make sure that the students have learned  the  above mentioned rational exponents concept,

1. Simplify the expression root(5)(x) = 2 , and find the value of 'x'

2. SImplify the expression root(3)(27)

Solution:

1. x =32

2. 3

Solving Quadratic Equations

An equation of the form ax2+bx+c=y where ‘a’ is not equal to zero; a, b are the coefficients and c is a constant is called a Quadratic Equation.

There are different methods used in solving these equations. They are, factorization method, grouping factors method, completing square method, using the quadratic formula if the equation does not factor and graphing method.

Solving Quadratic Equations Examples are as follows: Solve 2x2+5x-3=0
Let us solve this using the factorization method. Here first find the product of the coefficient of x2 term and the constant term, (2)(-3)=-6. Now find the factors of -6 such that the sum of the factors is equal to the coefficient of the middle term(x). The list of factors of -6 can be given as, -2x3, -3x2, -1x6, 6x-1.


Now select the appropriate set of factors whose sum would be equal to 5.
That would be 6, -1 as 6-1=5. Let us now re-write the given equation by splitting the middle term into the factors, which gives
2x2+6x –x +3=0. Now re-grouping the terms we get, 2x(x+3)-1(x+3). Taking (x+3) results in, (x+3)(2x-1). So the factors of the given equation are (2x-1) and (x+3). Each of these factors is equated to zero and solved to arrive to the solution set of the equation. 2x-1=0; x=1/2 and x+3=0; x=-3. So, the solution set is x=1/2, x=-3.

The above method can be used only when the given equation can be factored. There are some cases where the equation cannot be factored. In such cases the Quadratic Formula Help solving Quadratic Equations. The standard Quadratic Equation being ax2+bx+c=0, the quadratic formula is given by,
x= -bsqrt[b2-4ac] divided by 2a.

There would be two solutions, one with positive square root and other with negative square root. Let us consider an example problem, solve x2+3x-5=0. Here first the given equation is compared with the standard form to get the values of a,b and c respectively.

By comparing the terms to the standard form ax2+bx+c=0 we get, a=1, b=3 and c=-5. Now these values are substituted in the quadratic formula, x= [-bsqrt(b2-4ac)]/2a which gives x= [-3sqrt(9 –(4)(1)(-5))]/2(1)
= -3sqrt[9+20]/2 = [-3]/2. The solutions for the given equation are x=[3+29]/2 and x=[3-29]/2

Thus using the quadratic formula it becomes easy to solve equations which cannot be factored. This formula can also be used to an equation that can be factored too. Some of the Quadratic Equation Practice Problems are:
Solve 2x2-4x-3=0
Solve x2-14x+45=0
Solve -3x2+x+5=0
Solve 4x-5-3x2=0
Solve x2-x-6=0

Algebra is widely used in day to day activities watch out for my forthcoming posts on Finding Volume of a Cube and cbse 11th sample paper. I am sure they will be helpful.

Composite Function Solver

In mathematics, composite function is the application of one function to the results of another. For instance, the functions f: X → Y and g: Y → Z can be composed by computing the output of g when it has an argument of f(x) instead of x. intuitively, if z is a function g of y and y is a function f of x, and then z is a function of x                                  .( Source: Wikipedia )

In the following section we are going to see properties of composite function and some problems on composite function by using composite function solver.


Composite function solver

Fig(i) Composite function

Properties of composite function solver:

The composite function solver has the following properties.

The notation for indicate the composite function is  `@` .That is (f ∘ g).Where f and g are two functions. We can also say that f `@` g as f circle g or g composed with f, g after f , g of f.

f ∘ (g ∘ h) = (f ∘ g) ∘ h

(f ∘ g)(x) =f(g(x))

If the function f and g are commute each other ,we can define (f ∘ g) = (g ∘ f)

(f ∘ f)(x) =f 2 ( x )

(f ∘ f ∘ f )(x) = f 3 (x)


Problems on composite function solver :

Problem 1:

f(x) = 2x+5 and g(x) = 4x. Find f(g(x)).

Solution:

Given, f(x) = 2x+5

g(x) = 4x

We need to find f(g(x)) .

That is in the function f(x) ,substitute x =g(x) = 4x

f(x) = 2x+5

f(g(x)) =  2(4x) + 5

= 8x + 5

f(g(x)) = 8x + 5

Answer: 8x + 5

Problem 2:

f(x) = 3x2 + 5x -6 and g(x) = x -5 .Find (f ∘ g)(x)

Solution:

Given,  f(x) = 3x2 + 5x -6

g(x) = x -5

We need to find (f ∘ g)(x),

According to composite function solver property , (f ∘ g)(x) = f(g(x))

To obtain  f(g(x)) ,Substitute x =g(x) in f(x),

f(x) = 3x2 + 5x -6

f(g(x)) = 3( x -5 )2 + 5 ( x -5 ) -6

= 3 ( x2 - 10x +25) + 5x -25 - 6

=3x2 - 30x +75 + 5x -31

= 3x2 - 25 x +44

Answer: f(g(x)) = 3x2 - 25 x +44


Problem 3:

h(x) = x+2 and g(x) = -5x .Find h(g(6))

Solution:

Given , h(x) = x+2

g(x) = -5x

g( 6 ) = -5(6)

= -30

f(g(6)) = x+2

= -30 + 2

= -28

Answer: f(g(6)) = -28

Exponential Scientific Notation

This article is showing the exponential scientific notation. Scientific notation is nothing but transform very big numbers into simplest form of a number. That is called as scientific notation. Scientific notation is nothing but exponential of the base integer 10 have expressed in the number scientific notation. Example for scientific notation is 1860000000 this can be note down as 1.86 x 109. And the one more method in decimal is 0.0000081 = 8.1 x 10-6

Steps to find exponential scientific notation:

The steps to find exponential scientific notation are given below that:

The scientific notation has one number to the left of the decimal place.

We have to set the decimal point following the initial digit.

How many digits the decimal point was moved point out during the exponential of ten (10).


Example problems based on exponential scientific notation:

The example problems based on exponential scientific notation is given below that,

Example 1:

Change 82000 in exponential scientific notation.

Solution:

Step 1:

Specified that 82000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down 8.2 x 104 = 82000 [104 = 10000]

Step 3:

After that it can be write in scientific notation for 8.2 x 104.

Example 2:

Change 456000 in exponential scientific notation.

Solution:

Step 1:

Specified that 456000.

Step 2:

Specified number is a big number, but it is simple to convert scientific notation.

First note down 4.56 this is not equivalent to specified number.

Step 3:

Identify how many numbers available following the decimal point in previous step.

Present are 5 digits subsequent the decimal point.

4.56 × (100000)

Therefore the notation will be 105.

Step 4:

The scientific notation can be note down as 4.56 x 105

Example 3:

Change 987000 in exponential scientific notation.

Solution:

Step 1:

Specified that 987000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down the number as 9.87

Step 3:

There are 5 digit following the number is there so we can write 105.

Step 4:

The scientific notation can be note down as 9.87 x 105.
Practice problems based on exponential scientific notation:

The practice problems based on exponential scientific notation is given below that,

Problem 1:

Change 8960000 in exponential scientific notation.

Answer: The concluding answer for this problem is 8.96 x 106.

Problem 2:

Change 963000 in exponential scientific notation.

Answer: The concluding answer for this problem is 9.63 x 105.

Degree of Product Differentiation

The degree of the product differentiation represents the operations in the differentiation which deals with the product of the terms. For example if we have the function like xy then the degree of the function is 2 [x^1y^1 = ^2  ]. The homogeneous equation represents the degreee of the functions in the numerator and the denominator are same. In this article we are going to discuss about the degree of product differentiation for the students in detail with the homogeneous equations in different form with clear view.


Examples for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function  dy/dx = [y^2-x^2]/[2xy]

Solution:

The function given for differentiation  is   dy/dx = [y^2-x^2]/[2xy]

Here it is present in the homogeneous form.

Since the numerator degree[ [y^2-x^2]] is two and the denominator [ [2x^1y^1]] degree is also two

Put y = vx and  dy/dx = v + xdv/dx in the equation given we get,

=>  v + x[dv]/dx   =    [(vx)^2-x^2]/[2x(vx)]

=>  v + x[dv]/dx   =    [v^2x^2-x^2]/[2vx^2]

=>  v + x[dv]/dx   =    x^2[v^2-1]/[2vx^2]

=>  v + x[dv]/dx   =    [v^2-1]/[2v]

=>   x[dv]/dx   =    [v^2-1]/[2v] - v

=>   x[dv]/dx   =    [v^2-1-2v^2]/[2v]

=>   x[dv]/dx   =    [-1-v^2]/[2v]

=>  [2v]/[-[1+v^2]] dv  =  dx/x

=>  [2v]/[1+v^2] dv  =  -1/x dx

Apply integration on both sides, we get

=>  int   [2v]/[1+v^2] dv  = int  -1/x dx

=>  int   [2v]/[1+v^2] dv  = -int  1/x dx

=>   log|1 + v^2|  =  - log |x| + log c

=>   log|1 + v^2|   + log |x| =  log c

=>   log|x[1 + v^2]|  =  log c

=>   x[1 + v^2]  =  +-   c

Put y = vx and v = y/x

=>   x[1 + [y/x]^2]  =  C_1

=>   x[1 + y^2/x^2]  =  C_1

=>   x[x^2 + y^2]/x^2  =  C_1

=>   [x^2 + y^2]/x  =  C_1

=>   x^2 + y^2  = xC_1           is the answer for the differentiation for the degree of product.

Problems for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function dy/dx = [x^2y]/[x^3 + y^3 ]

Solution:

The answer is      [-x^3]/[3y^3] + log|y| =  C