Geometrical Meaning of the Zeroes of a Polynomial

We know that  a real number k is a zero of the polynomial p(x) if p(k) = 0 in a geometrical way.

The x-coordinate of the point, where the graph of a polynomial intersect the x-axis is called the zero of the polynomial.

An nth - degree polynomial intersects the x-axis of n points and therefore, has a maximum of n zeros in geometrical graph

In a quadratic polynomial ax2 + bx +c,

If a>0, then the graph is a parabola that open upwards.

If a<0, then the graph is a parabola that open downwards.

Special case to the zeroes of a polynomial in geometrical meaning


In a geometrical way the evaluation of zeroes with polynomial evaluation of the equation in geometrical meaning is follows

Case (i) :

Here, the graph cuts x-axis at two distinct points A and A′.
The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case is shown in the following figure



Case (ii) :

The graph given here cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A′ of Case (i) coincide here to become one point A is shown in the following figure.



The x-coordinates  of the  A is the only zero for the quadratic polynomial ax2 + bx + c in this case.

No zero case

The graph given here is either completely above the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point

So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also refers that the polynomial of degree 2 has at-most two zeroes is shown in the following figure.

Concept of the Derivative Online Help

In online, concept of derivative is clearly explained with the help of solved example problems. In online so many websites explain the concepts with the help of math sites. The derivative concept is clearly explained in calculus whereas derivative concept helps to find the rate of change for the given function with respect to change in the input function. The following are the example problems with detailed solution helps to explain the concept of derivative in online.


Concept of derivative online help example problems:


The following example problems explain the concept of derivative in online.

Example 1:

Determine the derivative by differentiating the polynomial function.

f(t) = 3t 3 +4 t 4  + 5t

Solution:

The given function is

f(t) = 3t 3 +4 t 4  + 5t

The above function is differentiated with respect to t to find the derivative

f '(t) = 3(3t 2 )+4(4t 3  ) + 5

By solving above terms

f '(t) = 9t 2 +  8t 3 + 5

Example 2:

Determine the derivative by differentiating the polynomial function.

f(t) = 6t6 + 5 t5 + 4 t4 + t

Solution:

The given equation is

f(t) = 6t6 + 5 t5 + 4 t4 + t

The above function is differentiated with respect to t to find the derivative

f '(t) =  6(6t 5)  +5 (5 t4 ) +4(4 t3) + 1

By solving above terms

f '(t) =  36t 5  + 25 t4  + 16 t3 + 1

Example 3:

Determine the derivative by differentiating the polynomial function.

f(t) = 2t 2 +4 t 4  + 15

Solution:

The given function is

f(t) = 2t 2 +4t 4  + 15

The above function is differentiated with respect to t to find the derivative

f '(t) = 2(2t  )+4(4 t 3 ) + 0

By solving above terms

f '(t) = 4t +16t3

Example 4:

Determine the derivative by differentiating the polynomial function.

f(t) = 5t5 +4t 4 +3t 3  + 2

Solution:

The given function is

f(t) = 5t5 +4t 4 +3t 3  + 2

The above function is differentiated with respect to t to find the derivative

f '(t) = 5(5t 4 )+4(4t 3 ) +3( 3t 2) +0

By solving above terms

f '(t) = 25t 4 +16t 3  +9 t 2


Concept of derivative online help practice problems:


1) Determine the derivative by differentiating the polynomial function.

f(t) = 2t 3 +3 t 4  + 4 t 5

Answer: f '(t) = 6t 2 +12 t3 + 20 t 4

2) Determine the derivative by differentiating the polynomial function.

f(t) = t 3+t5 + 4 t 6

Answer: f '(t) = 3t2 + 5t4 + 24 t 5

Solve Online Scale Factor

Scale is a thing which is used to measure the dimensions of the object. The terms scale factor is used to mention the dimensions of the geometric shapes. In geometry, the term scale factor is usually used to change the dimensions of the image. Using the scale factor, any shapes dimension can be increased or decreased. In online, students can learn about various topics. In online, students can learn about scale factor clearly. Online learning is very interesting and interactive. Moreover online learning is different from class room learning in different ways. In online, students have one to one learning.


More about scale factor:


A scale factor is a figure used as a figure by which another figure is multiplied in scaling.

A scale factor is new to scale shape in 1 to 3 proportions.
Scale factor can be established by the following scenario
Size Transformation:     In size change, the scale factor is the ratio of express the amount of reduction.

Scale Drawing:    In scale diagram a diagram that is like but either superior or lesser than the real object, for example, a road map or a building blueprint...

Comparing Two Similar Geometric Images:      The scale factor when compare two parallel geometric imagery is the ratio of lengths of the equivalent sides.






Solve Scale factor Problem:


Q 1:   The area of a rectangle is 5 and it is enlarged a scale factor by 5. Solve for the area of an enlarged rectangle?

Sol :  The Area of a rectangle(R) is 5 and its Enlarged by a scale factor of 5.

that means the length L and the width W are equally multiply by 5.
As well note that the area of Rectangle is the length L times the width W, so Rectangle = LW.

Because the area of a rectangle is given by Rectangle(R) = LW, and L turn out to be 4L and W turn out to be 4W,

Solve for new triangle,

the new area of a enlarged rectangle  = (5L)(5W) = 25LW = 25(R).
Now we know that Rectangle(R) = 5, so 25(R) = 125.

Q 2:  The side of the square is 6 cm and it is enlarged a scale factor by 4. Solve for the area of an enlarged square?

Sol :  The Area of a square = a2

=  6 * 6

= 36 cm2.

Scale factor is 4 cm.

New side of square  = 4 * 6

= 24 cm.

Area of enlarged square = 242.

= 576 cm2.

Answer is 576 cm2.

Study Online Interest

To study online interest is useful to calculate the interest easily. We have two types of interest Simple interest and compound

interest. Formula for simple interest is,    Interest = Principle `xx` Rate `xx` Time

Where  I = Total amount of interest paid

P = Amount lent

R = Percentage of principle charged as interest each year.

Example: rate = 2% means then 2/100 =0.02. Use this value in formula.

Formula for compound interest is,       A = p`(1+r)^(n)`


Examples of study online interest:


Ex:1

To find simple interest if principle= 3000, rate= 3%, time= 5

Sol:

To study online interest we have to write the formula first

That is Simple interest = Principle `xx` Rate `xx` Time

Apply the given values in the formula,

Interest = 3000 `xx` 3%`xx` 5               R = 3% = 3/100 = 0.03

= 3000`xx` .03`xx` 5

= 450

The answer is 450

Ex:2

Shan invested Rs.6000 for 5 years. He received the simple interest of Rs.1,500.Find the rate of interest.

Sol:

To study online interest we have to write the interest formula first

That is Simple interest = Principle `xx` Rate `xx` Time

Here principal, P= Rs.6000

Number of years, n= 5

Simple interest, I = Rs.1500

Rate of interest, r =?

We know that rate of interest, r =`(100 xx 1500)/(6000 xx 5)`

=5

Therefore Rate of interest = 5%

Ex:3

If the principal amount is $7000, the periodic rate of interest is 3% and the number of compounding periods is 3, what is the value at maturity?

Sol:

To study online interest we have to write the formula first

A = p`(1+r)^n`

Here principal, P= $7000

Rate of interest, r= 3%

= 3/100 = .03

n = 3

A = P (1 + r) n

A = 7000 (1 + .03) 3

A = 7649.089

Ex:4

If the principal amount is $2000, the periodic rate of interest is 2% and the number of compounding periods is 5, what is the value at maturity?

Sol:

To study online interest we have to write the formula first

A = p`(1+r)^n`

Here principal, P= $2000

Rate of interest, r= 3%

= 2/100 = .02

n = 5

A = P (1 + r) n

A= 2000 (1 + .02) 5

A= 2208.1616

These are the some examples to study online interest.

Learning Examples of Histograms

Introduction on learning examples of histograms:

In statistics, a histogram is a graphical display of tabular frequencies, shown as adjacent rectangles. Each rectangle is erected over an interval, with an area equal to the frequency of the interval. The height of a rectangle is also equal to the frequency density of the interval, i.e. the frequency divided by the width of the interval. The total area of the histogram is equal to the number of data. A histogram may also be based on the relative frequencies instead. It then shows what proportion of cases fall into each of several categories (a form of data binning), and the total area then equals 1.

Procedure and learning example for drawing the histogram:

To draw the histogram for a given frequency table
                  Marks    : 10-20   20-30    30-40   40-50   50-60   60-70
Number of students:    4           6            8         10          7         5  
Sol:
Histograms are drawing the first quadrant .because the first quadrant frequencies and class intervals are positive. Class intervals are marked along x axis similarly the frequency are marked along the y axis. Identify the maximum frequency in the table and then chose a suitable scale along the y- axis.
                     Draw first rectangle on the intervals 10 -20 as the base width and height of 4 units. Draw all rectangles taking the consecutive class intervals as base of the rectangles and their heights corresponding to the frequencies. The closed figure obtained is called histogram

Histogram:

This is the histogram for given frequency table.
Example: 2
To draw the histogram for a given frequency table,
 
Sol:

• Here in the frequency table, data range and frequency are given
• We take a data range in x axis and frequency in y axis

Histogram:

This is the histogram for given frequency table

More examples for learning histograms

To draw the histogram for following frequency table,

Sol:

• Here the given data deals with score(group) and frequency(count)
• We are taking score in x-axis
• Similarly the frequency is in y axis

Histogram:

This is the histogram for given frequency table.

Trigonometric Unit Circle Learning

A circle, whose radius is equal to one unit, is called as unit circle. The concept of unit circle is frequently used in trigonometry. In trigonometry, a circle with center (0, 0) and a radius of one unit is a unit circle. The equation of a circle is (x-h)2 + (y-k)2 = r2

For a unit circle, the center is (0, 0) and radius is 1, so the equation of a unit circle is x2 + y2 = 1


Learning - Properties of unit circle


Consider a point (x, y) in a unit circle.



The right triangle in the unit circle in the above diagram, Pythagoras theorem satisfies the equation of unit circle.

x2 + y2 = 1

Learning - Forms of unit circles points

Exponential form:            eit

Trigonometric form:        z = cos(t) + i sin(t)

Learning - Trigonometric functions

In a unit circle, consider a point (x,y) on the circle. If the angle formed between line joining the center (0,0) and the point (x,y) and the horizontal axis is `theta`,



Then the trigonometric functions for the angle `theta` is given by,

sin `theta` = opposite side/hypotenuse

cos `theta`= adjacent side/hypotenuse

tan `theta`= opposite side/adjacent side

csc `theta`= 1/sin`theta` = hypotenuse/opposite side

sec `theta`= 1/cos `theta` = hypotenuse/adjacent side

cot `theta`= 1/tan `theta`= adjacent side/opposite side


Example for trigonometric unit circle learning


Find the value of each of the 6 trigonometric functions for an angle theta that has a terminal side containing the point (3, 4).



By Pythagoras theorem, x2 = 32 + 42

x2 = 9 + 16

x2 = 25

x = 5

So, hypotenuse = 5, opposite side = 4 and adjacent side = 3

Then the trigonometric identities are given by,

sin `theta` = opposite side/hypotenuse = 4/5

sin `theta` = 4/5

cos  `theta` = adjacent side/hypotenuse = 3/5

cos `theta` = 3/5

tan `theta` = opposite side/adjacent side = 4/3

tan `theta` = 4/3

csc `theta` = 1/sin `theta`   = hypotenuse/opposite side = 5/4

csc `theta` = 5/4

sec `theta` = 1/cos `theta` = hypotenuse/adjacent side = 5/3

sec `theta` = 5/3

cot `theta` = 1/tan `theta` = adjacent side/opposite side = 3/4

cot `theta` = 3/4

A Factorization is the method for finding the variable factor value of the given expression. A Factor number is multiply with other number. If the factor value is a prime number then it is called prime factor value. The Factor is nothing but a value it is gives the answers to the given problem. If a polynomial factor might be written as the product of two or more expressions, then each expression is called the factor of the given polynomial. If a polynomial factor value might be written as the product of two or more expressions, then each expression is called the factor of the given polynomial. In this article we shall discuss about factoring solver.

Sample problem for factoring solver:

The following problem will help you understand the factoring solver.

Example 1:
Find the factoring value of the given function. 2x²- 12 x +16 = 0

Solution:
In the first step we find pattern factor value of the given numerical values of the x coefficients.

                                                                       32   (product)
                                                                     /    \    
                                                               - 4        - 8 
                                                                     \    / 
                                                                     -12    (sum)

2x²- 12 x + 16 = 2x² - 4x - 8x + 16
                      = 2x (x-2) - 8(x-2)
                     = (2x - 8) (x – 2)
                      = (x -4) ((x – 2)

So the factor of the given function is 2 and 4.

Example 2:
Find the factoring value of the following trinomial. 3x²- 9 x +6

Solution:
In the first step we find pattern factor value of the given numerical values of the x coefficients.

                                                                      18   (product)
                                                                     /    \    
                                                                 - 3      -6
                                                                     \    / 
                                                                      -9     (sum)
3x²- 9 x + 6 = 3x² - 6x - 3x + 6
                        = 3x (x-) - 3(x-2)
                        = (3x – 3) (x – 2)
                        = (x-1) (x-2)

So the factor of the given trinomial function is 2 and 1.

Practice problem for factoring solver:

• Find the factors of the given function. 6x² - 18 x + 12

            Answer: x = 1, 2.

• Find the pattern factors of the given function. 4x²- 24 x +32 = 0

            Answer: x = 2, 4.