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