Factoring Parametric Polynomials

[Wild_B]

Hello all,

With an exam coming out my teacher ask to see my calculator to test its capacity at resolving a command.

I would like to factor parametric polynomial expressions. Ex : x^2-k in relation to x
would anyone know how to do this? or even factor expressions with multiple variables?
For information a TI inspire does this perfectly. on TI inspire this is the command: factor(x^2-k,x)

Thank you for any help that you could provide.

P.S. I dont seem to be able to do it with microsoft math either

[Han]

(05-03-2019 02:09 PM)Wild_B Wrote:  Hello all,

With an exam coming out my teacher ask to see my calculator to test its capacity at resolving a command.

I would like to factor parametric polynomial expressions. Ex : x^2-k in relation to x
would anyone know how to do this? or even factor expressions with multiple variables?
For information a TI inspire does this perfectly. on TI inspire this is the command: factor(x^2-k,x)

Thank you for any help that you could provide.

P.S. I dont seem to be able to do it with microsoft math either

For quadratics, all factorizations can be obtained by using the quadratic formula. That said, you need to clarify if you are factoring over the real numbers or complex numbers, since certain values of k would lead to irreducible polynomials (e.g. \( x^2+1 \) cannot be factored since it has no real roots; it can, however, be factored if you are considering complex roots).

Anyway, given \( a\cdot x^2 + b\cdot x + c \), the factors are:

\[ a \left( x - \frac{-b+\sqrt{b^2-4ac}}{2a}\right) \left(x- \frac{-b-\sqrt{b^2-4ac}}{2a}\right) \]

All your TI is doing is substituting in the values for \( a\), \(b\), and \(c\), and then simplifying. You could easily put together a small program that does the same thing.

[parisse]

factor does not introduce sqrt of symbols, factorization is done over Q[x,k] here and there x^2-k is irreducible. But you can easily factor x^2-k yourself by hand, or for more complex expressions by running solve(x^2-k=0,x), see that sqrt(k) is required to factor it, then factor(x^2-k,sqrt(k)) will factor the polynomial over Q[sqrt(k)][x]
Note also that canonical_form can help you understand factorization of 2nd order polynomials.

[Wild_B]

@Han thanks for the insight. I have no experience yet in programming though, so I don't have the slightest idea on how to write that program.

@Parisse thanks for the tip, that definitely gets me to where I need to be (check the image I posted). Now to understand what's happening behind all that.