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# How to find a length, breath an height using polynomials method

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Posted on Jan 02, 2017

SOURCE: numerical method

Go to this site

http://oncampus.richmond.edu/~cstevens/301/Excel4.html

OK?

Posted on Jan 01, 2008

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## Related Questions:

Since you have the coordinates of the three vertices, the most straightforward method is to calculate the length of the sides using the distance formula
d(P_1,P_2)=SQRT((X_1-X_2)^2+(Y_1-Y_2)^2)
where SQRT is the square root function, X_1, Y_1) are the coordinates of point P_1, etc.
With the three lengths available, use Heron's (sometimes called Hero's) to find the area.
Here is Heron's formula.
Let's call the lengths a, b, and c
Let p be the semi-perimeter p= (a+b+c)/2
Then
Area= SQRT [ p(p-a)(p-b)(p-c) ]
Make sure that there is a matching ) parenthesis to the one in the SQRT.

Alternatively,
You can choose the base as the side opposite the vertex (0,0)
Calculate the equation of the line that supports the base.
Calculate the equation of the line issuing from (0,0) and perpendicular t the base.
Calculate the coordinates of the intersection point , call it H, of the base and its perpendicular line (coming from (0,0)).
Calculate the distance OH, that is the height relative to the chosen base.
Use the formula Area= base*height/2

Now it is up to you to choose one of the two methods and calculate the area of that triangle. The second method involves more calculations than the first, and more possibilities of errors. Good Luck

Nov 06, 2013 | Mathsoft Computers & Internet

### What is the area of the triangle with vertices (0,0),(-3,5),(2,-7)

Since you have the coordinates of the three vertices, the most straightforward method is to calculate the length of the sides using the distance formula
d(P_1,P_2)=SQRT((X_1-X_2)^2+(Y_1-Y_2)^2)
where SQRT is the square root function, X_1, Y_1) are the coordinates of point P_1, etc.
With the three lengths available, use Heron's (sometimes called Hero's) to find the area.
Here is Heron's formula.
Let's call the lengths a, b, and c

Let p be the semi-perimeter p= (a+b+c)/2
Then
Area= SQRT [ p(p-a)(p-b)(p-c) ]
Make sure that there is a matching ) parenthesis to the one in the SQRT.

Alternatively,
You can choose the base as the side opposite the vertex (0,0)
Calculate the equation of the line that supports the base.
Calculate the equation of the line issuing from (0,0) and perpendicular t the base.
Calculate the coordinates of the intersection point , call it H, of the base and its perpendicular line (coming from (0,0)).
Calculate the distance OH, that is the height relative to the chosen base.
Use the formula Area= base*height/2

Now it is up to you to choose one of the two methods and calculate the area of that triangle. The second method involves more calculations than the first, and more possibilities of errors. Good Luck

Nov 06, 2013 | The Learning Company Achieve! Math &...

### Wat is special product

Here, We deal with Some Special Products in Polynomials.

Certain products of Polynomials occur more often
in Algebra. They are to be considered specially.

These are to be remembered as Formulas in Algebra.

Remembering these formulas in Algebra is as important
as remembering multiplication tables in Arithmetic.

We give a list of these Formulas and Apply
them to solve a Number of problems.

We give Links to other Formulas in Algebra.

Here is the list of Formulas in
Polynomials which are very useful in Algebra.
Formulas in Polynomials :

Algebra Formula 1 in Polynomials:

Square of Sum of Two Terms:

(a + b)2 = a2 + 2ab + b2
Algebra Formula 2 in Polynomials:

Square of Difference of Two Terms:

(a - b)2 = a2 - 2ab + b2
Algebra Formula 3 in Polynomials:

Product of Sum and Difference of Two Terms:

(a + b)(a - b) = a2 - b2
Algebra Formula 4 in Polynomials:

Product giving Sum of Two Cubes:

(a + b)(a2 - ab + b2) = a3 + b3
Algebra Formula 5 in Polynomials:

Cube of Difference of Two Terms:

(a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3
Algebra Formula 8 in Polynomials:

Each of the letters in fact represent a TERM.

e.g. The above Formula 1 can be stated as
(First term + Second term)2
= (First term)2 + 2(First term)(Second term) + (Second term)2

Jul 02, 2011 | Computers & Internet

### Finding eigenvalue whit algebra

I do not own that calculator so I checked the manual. In the chapter Manual calculations it explains the use of matrices but not a single reference to eigenvalues or eigenvectors. I looked also in the advanced functions chapter but nothing on eigenvalues. I believe you will have to use the general algebraic methods for calculating eigenvalues: characteristic polynomial and its solutions. Set up the equation det (Mat A -lambda Mat I)=0 where A is your original matrix and I is the Identity matrix with same dimensions as A. The determinant will give you a polynomial equation in the eigenvalue lambda. the degree of the polynomial will be equal to the dimension of the matrix. You might want to feed the equation to the polynomial solver to find some eigenvalues.

Jun 30, 2011 | Casio Algebra FX 2.0 Calculator

### How to find roots of a polynomial using fx 115 ms

It depends on the degree of the polynomial.
If polynomial is od degree 2 or 3 you can use the EQN mode (the equation MODE) by pressing [MODE][5:EQN] to enter Equation mode then press [3] for quadratic polynomial or [4] for a cubic one.
You will then be prompted for the various coefficients. The canonical form of these polynomials is aX^2 plus bX plus c= 0, and aX^3 plus bX^2 plus cX plus d=0.

If polynomial is of degree higher than 3, or for a general non-linear equation you must use the Solve( feature. See example #017 on page 6 of the appendix to the manual.

Nov 28, 2010 | Casio FX-115ES Scientific Calculator

### How to factor a polynomial equation on casio fx-300es?

Hello,
The Casio FX-300ES does not handle symbolic algebra. So it cannot factor a general polynomial expression. The methods can be found in any book on Algebra.
However if you are interested in approximate factorization of quadratic and cubic polynomials, you can use the calculator to do that. It can solve aX^3 +bX^2+cX+d =0 and the quadratic equations.

If you want to factor a cubic polynomial P3(X) = aX^3+bX^2+cX+d , you write the corresponding cubic equation as aX^3+bX^2+cX=d =0 , then you divide all terms of the equation by a to obtain

X^3+(b/a)X^2+(c/a)X+(d/a)=0.

You use the calculator to solve (approximately) this equation.
Suppose you find the 3 roots X1,X2,and X3. Then the polynomial X^3+(b/a)X^2+(c/a)X+(d/a) can be cast in the factored form (X-X1)(X-X2)(X-X3) and the original polynomial P3(X) can be written as

P3(X) = a*(X-X1)(X-X2)(X-X3)

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots

Hope it helps.

Sep 27, 2009 | Casio fx-300ES Calculator

### How do you Factor a Polynomial on a Casio fx-300ES

Hello,
Sorry, but you cannot use this calculator to factorize a general polynomial.
1. It does not know symbolic algebra.
2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients (no letters) you can set [MODE] to equation and use the equation solver to find the real roots of 2nd degree or 3rd degree polynomials. Assuming that your polynomials have real roots (X1, X2) for the polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3, then it is possible to write
P2(X) =a*(X-X1)(X-X2)
P3(X)= a(X-X1)(X-X2)(X-X3)
This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)
where a is the coefficient of the highest degree monomial aX^2 +...
or aX^3 +....

But I have a hunch that this is not what you wanted to hear.

Good luck.

Mar 08, 2009 | Casio fx-300ES Calculator

### How to factor with this model

Hello,
Sorry, but you cannot use this calculator to factor a general polynomial.
1. It does not know symbolic algebra.
2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients (no letters) you can set [MODE] to Equation and use the equation solver to find the real roots of 2nd degree or 3rd degree polynomials. Assuming that your polynomials have real roots (X1, X2) for the polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3, then it is possible to write

P2(X) =a*(X-X1)(X-X2)
P3(X)= a(X-X1)(X-X2)(X-X3)

where a is the coefficient of the highest degree monomial aX^2 +...
or aX^3 +....

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

While the [MODE][5:Equation] only handles quadratic and cubic equations, the [SHIFT][SOLVE=] solver finds the roots of arbitarry expressions (not limited to polynomials). In principle you can use it to find the roots of an expression. If it is a polynomial of dgree higher that 3 you can factor it (approximately).

But I have a hunch that this is not what you wanted to hear.

Hope it helps.

Dec 09, 2008 | Casio fx-300ES Calculator

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