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

- 224

The product of two negative numbers is positive.

FOIL. First, outer, inner, last.

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.

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.

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:



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

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

I never know if these questions are serious, but I'll play along.

Remember the definition of a power: it's a number multiplied by itself that many times. For example, 4 to the 3rd power (4^3) is 4 X 4 X 4, or 64. So the problem you list can be written

9^2 + (-7^3) + (-6^2) + 5, or

9 X 9 + (-7 X -7 X -7) + (-6 X -6) + 5

Don't forget that negative X negative = positive when multiplying.

81 + (-343) + 36 + 5 = -221

Having gone over a month with no response, I assume this is no longer a problem.

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.