How to Solve polynomials

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An algebraic expression of the form a₀+a₁x+a₂x²+….+a_nxⁿ where a₀, a₁, a₂,….a_n are real numbers, n is a positive integer is called a polynomial in x.

Polynomial consists of:-

Constants: Like -8, 3/5, 20

Variables: Like a and b or x and y

Exponents: Like the 2 in a²

Purpose of Polynomials:

Listed below is the purpose of polynomials.

• Polynomials show in a wide mixture of area of mathematics and science.
• They are used to form polynomial equations, which instruct a wide range of troubles, from simple word tribulations to complex harms in the sciences.
• The purpose of polynomials is to describe polynomial functions, which emerge in setting range from fundamental chemistry and physics to economics and social science.
• Polynomials are used in calculus and arithmetic examination to estimate extra functions.
• In advanced mathematics, polynomials are used to build polynomial rings, a mid concept in intangible algebra and reckoning geometry.

Introduction of distributive property problems

An operation is distributive if the result of applying it to a sum of terms equals the sum of the results of applying it to the terms individually.
a ( b + c ) = ( a x b ) + ( a x c )
Here, ‘a’ is multiplied with the sum of two terms ‘b and c’ in the left hand side which, gives the same answer when ‘a’ is multiplied individually with ‘b’ and ‘c’ and then added.

Distributive Property Problems Example Part - 1:

1) Solve the problem using distributive property 9(9 + x).

Solution:

=9(9 + x)

=(9 * 9 + x * 9)

=(81 + 9x)


2) Solve the problem using distributive property 2(4 + 9x)

Solution:

2(4 + 9x)

(4 * 2 + 9x * 2)

(8 + 18x)


3) Solve the problem using distributive property 7(-1 + x)

Solution:

=7(-1 + x)

=(-1 * 7 + x * 7)

=(-7 + 7x)


4) Solve the problem using distributive property 12(a + b + c)

Solution:

=12(a + b + c)

=(a * 12 + b * 12 + c * 12)

=(12a + 12b + 12c)


5) Solve the problem using distributive property 7(a + c + b)

Solution:

=7(a + b + c)

=(a * 7 + b * 7 + c * 7)

=(7a + 7b + 7c)

Distributive Property Example Problems Part - 2:


6) Solve the problem using distributive property -10(3 + 2 + 7x)

Solution:

-10(3 + 2 + 7x)
Combine like terms: 3 + 2 = 5
-10(5 + 7x)
(5 * -10 + 7x * -10)
(-50 + -70x)


7) Solve the problem using distributive property -1(3w + 3x + -2z)

Solution:

=-1(3w + 3x + -2z)
=(3w * -1 + 3x * -1 + -2z * -1)
=(-3w + -3x + 2z)


8) Solve the problem using distributive property 1(-2 + 2x2y3 + 3y2)

Solution:

=1(-2 + 2x2y3 + 3y2)

=(-2 * 1 + 2x2y3 * 1 + 3y2 * 1)

=(-2 + 2x2 y3 + 3y2)


9) Solve the problem using distributive property 5(5 + 5x)

Solution:

=5(5 + 5x)

=(5 * 5 + 5x * 5)

=(25 + 25x)


 10) Solve the problem using distributive property y(1 + x)

Solution:

=(y + yx)

=(y + xy)

11) Solve the problem using distributive property 5(x + 10).

Solution:

=5( x + 10)
=(5* x + 5 * 10)
=(5x + 50)

Calculus problems solver

All the algebra, functions, relations etc that we study in math can be termed as pre-calculus. The next huge topic in math is calculus. Calculus is distinct from pre-calculus. In calculus there is less of static properties and more of dynamic properties. We talk of one quantity tending to another in calculus.
Calculus is a wide topic that encompasses various subtopics. The most common of which are: Limits, continuity, differential calculus and integral calculus.

The origin of calculus goes back at least 2500 years to the ancient Greeks. They use to find help with calculus problems of how to find area of any polygon.  The method they used was like this: the polygon was divided into triangles and the area of the triangles was added to find the area of the entire polygon, which is very similar to what we do in integration.

In solving calculus problems like finding the area of closed region bounded by curves, it is not possible to split the region to triangles as the boundaries are curved. The Greek method was to inscribe polygons in the figure the circumscribe polygons about the figure and then let the number of sides of the polygon increase. The limiting position of these areas was the area under consideration. Eudoxus used this method to solve calculus problems pertaining to area of circle and to prove the famous formula of A = pi r^2.

The problem of trying to find the slope of the tangent line to a curve gave rise to the branch of calculus called differential calculus. In problems of finding rate of increase of quantities, velocity, acceleration etc, the idea of limits is useful.

Sample calculus problems:

1. The price of a box of pencils is as follows: 5x/(x-1). Where x is the number of boxes one buys. So that means the more then number boxes that a person buys, the lower is the per box cost. What would the price tend to if a person buys lesser and lesser number of boxes?
Solution: The above problem is same as solving the limit: lim(x->1)(5x/(x-1)). We can see that x tends to 1, the term tends to infinity. So if someone wants to buy just one box, the cost would be pretty high.

2. Radius of a circle increases at a constant rate of 4 cm/s. Find the rate of change of area of circle when radius is 6cm.
Solution: This is application of derivatives.
A = pir^2
dA/dt = 2pir.dr/dt
dr/dt=4 and r=6
so, dA/dt = 2*pi*6*4 = 48pi (cm)^2/s

Improper Fractions

Fractions form an integral part of our lives. Fraction is part of a whole. Hence, in a fraction we have two numbers. One on the top called numerator and the other on the bottom called denominator.
We classify fractions as proper fractions and improper fractions based on the values of numerator.

Improper fractions: 
Fractions with numerator greater than the denominator are called improper fractions.
Example: 13/4, 12/5
How to simplify improper fractions:
This is nothing but reducing improper fractions.
As the numerator is greater than the denominator in improper fractions, when we do the long division of this fraction, we will be getting a quotient greater than 1 and a remainder.
We can generalize the simplification by an algorithm:
Step 1: Perform the long division
Step 2:
Example: Simplify the improper fraction 18/7

Step 2:  Simplified form of the mixed fraction is


Improper fractions:

Improper fractions have two parts on simplification. The part which does not have denominator is called whole part and the part with denominator is called fractional part. The other name for the simplified improper fraction is mixed fraction.
Mixed fraction has both whole number part and a fractional part.

Adding improper fractions:
We adopt four steps to add improper fractions.
Step 1: Find the Least common multiple of the denominators
Step 2: Find the equivalent fractions of individual fractions such that the denominators of both the fractions are same.
Step 3: Add both the numerators and write the common denominator
Step 4: Simplify the improper fraction if need be
Example: 11/3 + 5/4
(11 x 4)/(3 x 4) + (5 x 3)/ (4 x 3)
= (44/12) + (15/12)
= 59/12
=4 11/12

Multiplying Improper Fractions
In this, the method adopted is as same as in multiplication of proper fractions. We follow three steps to evaluate the product of two improper fractions.
Step 1: Multiply both the numerators and write the result in numerator
Step 2: Multiply both the denominators and write the result in denominator
Step 3: Simplify if need be
Multiply 15/7 and 4/3
Step 1: Numerator = 15 x 4 = 60
Step 2: Denominator = 7 x 3 = 21
Step 3: Required product = 60/21 = 3 x 20/ 3 x7 = 20/7 = 2 6/7

Dividing Improper Fractions
In this as well, we follow the general rules of division of fractions.
Step 1: Write the first fraction
Step 2: Find the multiplicative inverse of the second fraction
Step 3: Multiply the results of steps 1 and 2 following the algorithm of multiplication of fractions.
Example:
Divide 15/ 7 by 4/3
Step 1: 15/7
Step 2: Multiplicative inverse of 4/3 is ¾
Step 3: 15/7 x 3/4 = (15 x 3)/ (4 x 7) = 45/28.
What is an improper fraction?
Improper fractions are those whose numerators are greater than denominators. The value of improper fractions is always greater than one.

Unit Conversion

Unit conversion is converting from one unit to another of the same quantity. Unit conversion is used when we have to compare or convert any quantity in any particular form. Unit conversion can be done for length, volume, time, temperature, area, energy etc. It is done using conversion factors.
For example: -
I kilogram = 1000 grams
1 foot = 12 inches
1 meter = 100 centimeters
1 minute = 60 seconds
1 day = 24 hours
What is the Metric Unit for Length?
The metric unit of length is meter.
Unit conversion length
1 meter = 1000 millimeter
1 meter = 100 centimeter
1 meter = 10 decimeter
1 meter = 0.001 kilometers
How to do unit conversions?
Unit conversions can be done by using the following steps: -
Write the given value.
For example: - Convert 582 cm to m.
Find conversion factor for the given and the desired units
In the above example the conversion factor is 100 cm = 1m.
Write it as a fraction with the given units as a denominator or in the opposite direction.
582 cm (1 m)/(100 cm)
Cancel the ‘like’ units that is cm in the numerator can be cancelled by cm in the denominator.
582 cm (1 m)/(100 cm )=582  (1 m)/100

Multiply odd units, we will be left with 582 times 1 m in numerator which gives 582 m and 100 in denominator which gives 582 m/100 = 5.82 m
5.82 m
For example: -
If we have to convert 12 millimeter to kilometer, then the steps would be: -
Convert 12 mm to km
Conversion factors are: -
1000 mm = 1 m
1000 m = 1 km
1.2 mm(1m/1000mm)((1 km)/(1000 m))
=0.0000012 km
This is how unit conversion is done when we have any quantity in a unit and we need to convert it in different unit.

Introduction to Trinomials

What is a trinomial?
To understand the concept of trinomial, we should know what a monomial is. A monomial is an algebraic expression that has only one term.

For example: - 2a, 4x, 6z are all monomials. A trinomial is a polynomial which consists of three monomial terms. For example: -
• 2a+3b-4c
• 4x-84-6z
They both are the examples of trinomials.
Factoring trinomials
Rules to factor trinomials: (how to factor trinomials)
A trinomial expression is an algebraic expression which has exactly three terms. Trinomials can either be quadratic equation or a higher order equation. In case it is a quadratic equation, it can be simplified either by factoring or using quadratic formula.

For example: - 2x^2-7x+6 can be written as 2x^2-4x-3x+6 which can be factored as (2x-3) (x-2) so this is how we factor trinomials if they are in quadratic form.
A higher order trinomial can be turned into a quadratic equation by factoring common terms and then can be factored again.

For example: - 6x^2y + 14xy + 4y
In this expression, we have 2y common in all terms, so we can take that common, so that becomes
2y (3x^2+7x+2)
And that can further be factored as 2x (3x+1) (x+2).
Following instructions should be followed for solving trinomials: -
• First, factor the common factors from all the terms. For example if we have 3x^2+24x+45 then we can take 3 common from all the terms and write the expression like this, 3(x^2+8x+15).
• Check the trinomial equation you are left with.
• If the highest power of the trinomial equation is 2, then it can be factored like a quadratic equation
• If the highest power is the higher degree, then we should look for a pattern that allows you to solve it like a quadratic equation.
• Solve the quadratic part of the equation and make the factors.

Applied Statistics

Applied Statistics is nothing but applying statistics in real life. There are different levels of measurement that are used in applied statistics. Below are few example:

Nominal
Interval
Ordinal
Ratio



For more help connect with an online statistics tutor and get your help. Not just statistics but you can get help from geometry tutors and so on as well.

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