Question about Texas Instruments TI-83 Plus Calculator

2 Answers

Multiplying Tried to do a polynomial problem and when trying to do the powers I.E (-3^4) it gives me a negative number.

Posted by on

  • Anonymous Mar 25, 2014

    (2x2)(3x2+5x-4

×

Ad

2 Answers

What you meant to type was (-3)^4. The calculator is doing 3^4, then applying the negative sign when you type it the way you did.

Posted on Sep 09, 2007

Ad

The calculator is mathematically correct. -3 x -3 x -3 x -3 = 81

Posted on Sep 05, 2007

Ad

1 Suggested Answer

6ya6ya
  • 2 Answers

SOURCE: I have freestanding Series 8 dishwasher. Lately during the filling cycle water hammer is occurring. How can this be resolved

Hi,
A 6ya expert can help you resolve that issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
The service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
click here to download the app (for users in the US for now) and get all the help you need.
Good luck!

Posted on Jan 02, 2017

Add Your Answer

Uploading: 0%

my-video-file.mp4

Complete. Click "Add" to insert your video. Add

×

Loading...
Loading...

Related Questions:

1 Answer

What does polynomial function mean?


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.

Oct 01, 2014 | Office Equipment & Supplies

1 Answer

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.

Oct 06, 2013 | Casio FX9750GII Graphic Calculator

1 Answer

Sslc maths


What follows is true for CONVEX polygons.
Let n be the number of sides of a convex polygon, and let n be greater than or equal to 4, then
The number of diagonals is given by n*(n-3)/2
Using this rule, write n*(n-3)/2=20. Clear the fraction, open brackets. You end up with n^2-3n-40=0.
Factor the polynomial or use the formulas for the quadratic equation to find the roots as n=-5 or n=8. Discard the negative root because n must be positive.

Oct 05, 2013 | Computers & Internet

1 Answer

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.

Oct 03, 2011 | Texas Instruments TI-84 Plus Calculator

1 Answer

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).

Oct 03, 2011 | Office Equipment & Supplies

1 Answer

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

1 Answer

How to find cyclic redundancy check ?


Cyclic Redundancy Check
The cyclic redundancy check, or CRC, is a technique for detecting errors in digital
data, but not for making corrections when errors are detected. It is used primarily
in data transmission. In the CRC method, a certain number of check bits, often
called a checksum, are appended to the message being transmitted. The receiver
can determine whether or not the check bits agree with the data, to ascertain with
a certain degree of probability whether or not an error occurred in transmission. If
an error occurred, the receiver sends a "negative acknowledgement" (NAK) back
to the sender, requesting that the message be retransmitted.
The technique is also sometimes applied to data storage devices, such as a
disk drive. In this situation each block on the disk would have check bits, and the
hardware might automatically initiate a reread of the block when an error is
detected, or it might report the error to software.
The material that follows speaks in terms of a "sender" and a "receiver" of a
"message," but it should be understood that it applies to storage writing and reading
as well.
The CRC is based on polynomial arithmetic, in particular, on computing the
remainder of dividing one polynomial in GF(2) (Galois field with two elements)
by another. It is a little like treating the message as a very large binary number, and
computing the remainder on dividing it by a fairly large prime such as
Intuitively, one would expect this to give a reliable checksum.
A polynomial in GF(2) is a polynomial in a single variable x whose coefficients
are 0 or 1. Addition and subtraction are done modulo 2-that is, they are
both the same as the exclusive or operator. For example, the sum of the polynomials

x3 + x + 1 and
x4 + x3 + x2 + x

Apr 15, 2011 | Computers & Internet

2 Answers

How to find the polynomial roots in TI-83 Plus


One way is to use the Polynomial Root Finder and Simultaneous Equation Solver app.

Press the APPS key, then select PlySmlt2 and press ENTER. Press ENTER to get past the opening screen, then select "POLY ROOT FINDER" and press ENTER. Select the order of the polynomial and other settings as desired. Press F5 (the GRAPH key) to go to the next screen. Enter the polynomial coefficients, then press F5 to solve. The next screen will show you the roots (unless you selected real roots and the polynomial doesn't have any real roots).

If you don't have the app installed, you can download it from
http://education.ti.com/educationportal/sites/US/productDetail/us_poly_83_84.html

Jan 20, 2011 | Texas Instruments TI-83 Plus Calculator

1 Answer

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

1 Answer

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

Not finding what you are looking for?
Texas Instruments TI-83 Plus Calculator Logo

Related Topics:

106 people viewed this question

Ask a Question

Usually answered in minutes!

Top Texas Instruments Office Equipment & Supplies Experts

k24674

Level 3 Expert

8064 Answers

Stevan Milanovic

Level 2 Expert

265 Answers

Donald DCruz
Donald DCruz

Level 3 Expert

17130 Answers

Are you a Texas Instruments Office Equipment and Supply Expert? Answer questions, earn points and help others

Answer questions

Manuals & User Guides

Loading...