Question about Texas Instruments TI-84 Plus Silver Edition Graphic Calculator

There are several factoring programs available at

http://www.ticalc.org/pub/83plus/basic/math/algebra/

Posted on Aug 19, 2010

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Posted on Jan 02, 2017

-54k^2+3k
I assume that you want to factor this completely.

I do it one component at a time.

What is the biggest common factor of -54 and 3. I use a trick of adding the digits of the number. If 3 and/or 9 go into that number evenly, than 3 and/or 9, respectively go into the original number. In this case, 5+4 = 9 and since 3 goes into 9 evenly, 3 goes into 54 evenly. Since the factors of 3 are 3 and 1, this must be the greatest common factor of this component.

Onto the next component, what is the greatest common factor of k^2 and k. The answer is k, because k^2 doesn't go evenly into k.

Combining the two, the greatest common factor is 3k.

Next we have to divide 3k into each term.

3k(-18k+1) or -3k(18k-1)

Good luck,

Paul

I do it one component at a time.

What is the biggest common factor of -54 and 3. I use a trick of adding the digits of the number. If 3 and/or 9 go into that number evenly, than 3 and/or 9, respectively go into the original number. In this case, 5+4 = 9 and since 3 goes into 9 evenly, 3 goes into 54 evenly. Since the factors of 3 are 3 and 1, this must be the greatest common factor of this component.

Onto the next component, what is the greatest common factor of k^2 and k. The answer is k, because k^2 doesn't go evenly into k.

Combining the two, the greatest common factor is 3k.

Next we have to divide 3k into each term.

3k(-18k+1) or -3k(18k-1)

Good luck,

Paul

Mar 29, 2017 | Homework

There is a program on page 253 of the manual (http://support.casio.com/manualfile.php?rgn=5&cid=004002013) to get the prime factors of any number. You could modify the program to get all the factors.

Another way would be to do it manually. I start with the number 1 and go up to the square root of the number. The square root of 120 is 10.95, so let's go up to 11. Using 120 as an example:

120 /1 = 120 thus factor is (1, 120)

120/2 = 60 thus factor is (2, 60)

120/3 = 40 thus factor is (3, 40)

120/4 = 30 thus factor is (4, 30)

120/5 = 24 thus factor is (5, 24)

120/6 = 20 thus factor is (6, 20)

120/7 = 1.7 thus 7 not a factor

120/8 = 15 thus factor is (8,15)

120/9 = 13.3 thus 9 is not a factor

120/10 = 12 thus factor is (10,12)

120/11 = 10.9 thus 11 is not a factor.

Good luck.

Paul

Another way would be to do it manually. I start with the number 1 and go up to the square root of the number. The square root of 120 is 10.95, so let's go up to 11. Using 120 as an example:

120 /1 = 120 thus factor is (1, 120)

120/2 = 60 thus factor is (2, 60)

120/3 = 40 thus factor is (3, 40)

120/4 = 30 thus factor is (4, 30)

120/5 = 24 thus factor is (5, 24)

120/6 = 20 thus factor is (6, 20)

120/7 = 1.7 thus 7 not a factor

120/8 = 15 thus factor is (8,15)

120/9 = 13.3 thus 9 is not a factor

120/10 = 12 thus factor is (10,12)

120/11 = 10.9 thus 11 is not a factor.

Good luck.

Paul

Jul 05, 2015 | Casio FX-9750GII Graphing Calculator

To convert a fraction to lowest terms, we must divide the numerator (the top number) and the denominator (the bottom number) by a common number until we can't reduce it any more. A method for doing this is to factor the numerator and denominator by their common factors.

So let's factor 711 and 22.

1 x 711

9 x 79

3 x 3 x 79

I don't think I can factor 79. We should check the numbers from 2 up to the square root of 79, or approximately 9.

Factoring 22,

1 x 22

2 x 11

Unfortunately, none of the factors of 711 are common with the factors of 22, so it is already in its lowest terms, and cannot be reduced.

Good luck,

Paul

So let's factor 711 and 22.

1 x 711

9 x 79

3 x 3 x 79

I don't think I can factor 79. We should check the numbers from 2 up to the square root of 79, or approximately 9.

Factoring 22,

1 x 22

2 x 11

Unfortunately, none of the factors of 711 are common with the factors of 22, so it is already in its lowest terms, and cannot be reduced.

Good luck,

Paul

Apr 09, 2015 | Office Equipment & Supplies

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.

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

G(x)=

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

All the terms in the expression have a common variable factor (y^3). factor it out to get (0.2+0.3+0.1)y^3=0.6y^3 or (6/10)y^3 or **(3/5)y^3.**

Apr 19, 2013 | Office Equipment & Supplies

This calculator is unable to factor a polynomial expression.

In general there are a few factoring methods

In general there are a few factoring methods

**Factor by grouping terms****Factor by completing the square**(quadratic polynomial)**Factor by finding two integers such their sum is equal to the coefficient of the middle term, and their product is equal to the third (constant term)**. This is valid for a quadratic polynomial where the leading coefficient (of the x^2 term) is equal to 1.**X^2+SX+P**

Jul 31, 2012 | Casio FX-115ES Scientific Calculator

The short story is that this calculator does have a computer algebra system or CAS and thus cannot factor polynomials with arbitrary (unknown) coefficients or known coefficients.

However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).

Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.

The procedure ( for a 3rd degree polynomial) is as follows:

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.

However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).

Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.

The procedure ( for a 3rd degree polynomial) is as follows:

If you want to factor a cubic polynomial P3(X) = aX^3 bX^2 cX d , you write the corresponding cubic equation as

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots

You can handle the quadratic polynomial the same way.

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

Sep 11, 2010 | Casio FX-9750GPlus Calculator

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.

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

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots

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

Unless I am failing to understand your problem, just combine like terms to get:

9x^2+2x+4 and then enter those coefficients into the quadratic solver.

That will get you the two answers -- both of them will be complex.

If you wanted to turn those two answers into factors (complex) of the original equation simply

change the signs on each and put an x in front.

Of course, the quadratic formula would also work.

CG

9x^2+2x+4 and then enter those coefficients into the quadratic solver.

That will get you the two answers -- both of them will be complex.

If you wanted to turn those two answers into factors (complex) of the original equation simply

change the signs on each and put an x in front.

Of course, the quadratic formula would also work.

CG

Apr 17, 2009 | Casio FX-115ES Scientific Calculator

oh yeah, you must've archived x, that is, you put a number onto the x variable so that the calculator calculates the answer for you, and THEN AUTOMATICALLY calculates the answer when x is in terms of that number. I am guessing you accidentally did 10[sto] x at one point. You can simply unarchived the x number and everything will be dandy!

Dec 08, 2007 | Texas Instruments TI-89 Calculator

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