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rebecca ailiff Posted on Oct 01, 2014
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What does polynomial function mean? - Office Equipment & Supplies

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k24674

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  • Office Equip... Master 8,093 Answers
  • Posted on Oct 01, 2014
k24674
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Let us backtrack so as to better jump.
An algebraic expression may contain one or several algebraic terms, separated by a plus sign or a or a minus and a sign.
Each algebraic term is the product of a constant coefficient and a power of some variable, or the powers of several variables.
Example of an algebraic term 3(x^2)(y^6)....
If the exponents of the various powers are positive integers, the term is called a monomial. In short no square roots, or fractionary powers of the variables appear in monomials. Thus 2/x, 3SQRT(x), or 1/x^5 are not monomials.
Finally, a polynomial is an algebraic expression made up of one or more monomials.
Example P(X)=(1/3)X^7-(SQRT(5)*X^4+ 16X-25 is a polynomial of degree 7 in the indeterminate/variable X
Q(X,Y)= 3(X^3)*Y^2 + 4X-5Y+10 is a polynomial of degree 5 in the variables X and Y.

  • k24674 Oct 02, 2014

    Yes, I said no Square roots of variables. Square or other roots of a constant coefficient (not a variable) are just numbers and they are allowed.

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Related Questions:

0helpful
1answer

Factoring program

What is the question. Your calculator FX-9750 GII does not have a Computer Algebra System or CAS, so you cannot factor a polynomial.
If you want you can try to find the zeros of the polynomial function (the values of x when the function crosses the horizontal axis) either by solving P(x)=0 or by graphing y=P(x).
Once you have the approximate roots x_1, x_2, ...,x_n, you can factor out the coefficient of the leading term (the term with the highest power) and write P(x) =a_n *G(x). here a_n is the coefficient of the leading term.
G(x) can then be written as G(x)=(x-x_1)(x-x_2)*(x=x_3)...(x-x_n)
and the original polynomial will be
P(x)=a_n(x-x_1)(x-x_2)*...*(x-x_n)
Note: Here is an example of P(x) and the corresponding G(x)
P(x)=5x^3+7x^2-13x^+29
P(x)=5(x^3+ (7/3)*x^2-(13/5)x+29/5)
G(x)=x^3+ (7/3)*x^2-(13/5)x+29/5
You should keep in mind that the roots are in general complex. Not all polynomials are factorisable in the set of Real numbers.
0helpful
1answer

Need instruction how to solve ploynomials/trinomials

You can do it two ways.
Graphically:
Open the graph utility, enter a function y1= expression(x) where expression(x) is the polynomial. Graph the function y1(x). Play with the window dimensions to display the interesting part of the graph. You can then see if there are any zeros (roots).
While the graph is displayed, press SHIFT F5 (G-Solv). Press F1:Root to look for a root. Read the value of the root at the bottom of the screen (to the left).

Solve utility
Open the QEquation solver, select F2:polynomial, or F3: solver.
The polynomial solve can solve polynomial equations up to degree 6.
The solve solve polynomial and any other type of non algebraic equations
0helpful
1answer

How to solve polynomials

That depends on the order of the polynomial. If the polynomial is of second order, the quadratic formula will give you its roots. There are similar formulae for third, fourth, and fifth order polynomials, and it has been proven that no such formulae exist for higher-order polynomials.
Some TI calculators have a numerical solver which can produce numerical solutions to arbitrary polynomials. Since you declined to specify the model of your calculator, I'm afraid I can't tell you whether yours has one or not.
If you need further assistance, please reply to this post and be sure to specify the model of your calculator.
0helpful
2answers

How do you find the derivative without values for x

I am not sure I understand what you want, but if I am guessing right, you need to calculate formulas for derivatives, where the name of some variable appears eg= d(x^2)/dx = 2 x.
Sorry, the calculator does not have a CAS, you cannot perform symbolic calculations regardless of the subject (derivatives, integrals, sums of series, factoring polynomials, expanding polynomial products, polynomial division, etc.)
0helpful
1answer

Give the polynomial function whose roots are -2 1 and 3

There are an infinite number of polynomials with those roots. Assuming you want one with the lowest degree, here are two:
x^3 - 2x^2 - 5x + 6
2x^3 - 4x^2 - 10x + 12

Since the roots of the polynomials are -2, 1, and 3, the values (x+2), (x-1), and (x-3) must be zero.
To get these polynomials, simply multiply
k * (x+2) * (x-1) * (x-3)
where k is any nonzero value.
0helpful
1answer

Give the polynomial function whose roots are -2 1 and 3

We can write this polynomial as:
  • (x-(-2))*(x-1)*(x-3)=
  • (x+2)(x-1)(x-3)=
  • (x+2)[x*(x-3)-1*(x-3)]=
  • (x+2)*(x^2-3x-x+3)=
  • (x+2)(x^2-4x+3)=
  • x*(x^2-4x+3)+2*(x^2-4x+3)=
  • x^3-4x^2+3x+2x^2-8x+6=
  • x^3-2x^2-5x+6
x^3-2x^2-5x+6 is polynomial with roots -2, 1, 3.

You can see this polynomial in following picture:

elessaelle_2.png

Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).
0helpful
1answer

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

0helpful
1answer

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.
0helpful
1answer

How do I find the roots of a third degree function using my calculator?

Here is an example for a third degree polynomial. (Screen captures are from an FX9860 G SD, so there may be some differences in screen appearance, but the procedure is the same.)
Press the [MENU] key and select [EQN]
k24674_19.jpg

Select the type of equation to solve. Press [F2:Poly] for a polynomial.
k24674_20.jpg
Choose the degree of the polynomial. For a 3rd-degree press F2

k24674_21.jpg

Enter the coefficients in the template

k24674_22.jpg

and press [F1:SOLV]

k24674_23.jpg

Here is the solution for the case I took.

k24674_24.jpg

Press [F1:REPT] if you want to enter the coefficients of another polynomial.
0helpful
1answer

Polynomial function

Hi there,

Say you have an equation x = 2y.

On a graph, you plot the points representing the values that fit into the equation.

X Y
2 1
4 2
6 3
etc.

And to throw a curve in there:
X Y
-2 -1
-4 -2
-6 -3

If have further questions repost, I'll be around for a while.

Hope it helps
Mike
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