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Is it possible to factor on a Ti-83 Calcultor?

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Like if I wanted to factor x^2 + 24x +80 {EXAMPLE}
    Is this possible?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 29, 2008 #2
    This is definitely possible. What are some factors of 80 that will add to 24?

    Oh whoops I just saw that you were looking for factoring on Ti-83 calculator. But why would you want to do that when you can factor by hand? I don't think you can do it anyway.
    Last edited: Jan 29, 2008
  4. Jan 29, 2008 #3
    because midterms are tomorrow and I am weak on factoring :)
  5. Jan 29, 2008 #4
    Ok calculators suck. If you can't find factors to factor your function, then just use the quadratic formula.
  6. Jan 29, 2008 #5
    You mean to say that you want to cheat, yourself and your teacher? Obviously the idea is to test your ability to factor, not your calculator's ability. Nothing cures the problem like a healthy dose of practice.
  7. Jan 30, 2008 #6
    well practicing will make you more comfortable with them. Read up on the factor theorem also.

    if my casio991MS can factor, the Ti-83 will probably be able to also.
  8. Jan 30, 2008 #7


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    You can just solve x^2 + 24x + 80 = 0 and once you have the solutions, use those to factor :)

    E.g. quadratic formula gives
    [tex]x_\pm = \frac{ - 24 \pm \sqrt{24^2 - 4 \cdot 80}}{2} = - 12 \pm 8 = -20, -4[/tex]
    So it factors as
    (x + 20)(x + 4)

    If you insist on doing it with the GC, you could use the equation solver...
    [MATH] [0 (Solver)]

    Enter the equation
    (X^2 + 24X + 80)
    and solve it (press Enter, Alpha + Enter (Solve))
    You will get a solution, e.g. X = -4
    Then divide out this solution from the original equation, so go back and enter
    (X^2 + 24X + 80)/(X + 4)
    (note the sign) -- now solve again and it will give you the other one.
    Then you have again X = -4, X = -20, so it factors as
    (X + 4)(X + 20)
  9. Jan 31, 2008 #8
    start a new program and put these lines as your program
    :Disp "Ax^2+BX+C"
    :prompt A,B,C
    :Disp "roots=",E/A,F/A
    that will do the quadratic formula
  10. Feb 1, 2008 #9


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    What is this notation [itex]x_\pm[/itex], what does it mean?
  11. Feb 1, 2008 #10


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    It means there are two solutions, which we denote by [itex]x_+[/itex] and [itex]x_-[/itex] (might as well have called them [itex]x_1[/itex] and [itex]x_2[/itex], or x and y.
    The first one is
    [tex]x_1 = x_+ = \frac{ - b + \sqrt{D}}{2} [/tex] (*)
    and the other is
    [tex]x_2 = x_- = \frac{ - b - \sqrt{D}}{2} [/tex] (**)
    where D is the usual discriminant
    [tex]D = b^2 - 4 a c[/tex].

    It is common to write the formulas (*) and (**) together in one line as
    [tex]x_{1,2} = x_\pm = \frac{ - b \pm \sqrt{D}}{2} [/tex]
    where we mean: pick either the plus or the minus sign throughout.

    The same notation also occurs in e.g.
    [tex]\sqrt{4} = \pm 2[/tex]
    which means that +2 and -2 are both solutions, so shorthand for
    [tex]\sqrt{4} = 2 \text{ or } -2[/tex].

    (Aside, we also have an opposite symbol, which can be used in for example
    "If [itex]x = \pm 1[/itex] then [itex]-2x = \mp 2[/itex]"
    which means both "if x then -2x = -2" and "if x = -2 then -2x = +2" simultaneously).
  12. Feb 1, 2008 #11
    TI-83....probably not. But you could graph the function and find the roots, which comes off pretty easy.
  13. Feb 1, 2008 #12
    Of course there's a way to do it. Like someone said earlier, you can just take the zeroes of the function and place the opposite of it in the "factored form". Just use a program to do the quadratic and multiply the answer by -1, and display "(X",D,")", and the same for the second value.

    :Prompt A
    :Prompt B
    :Prompt C
    sqrt. (B^2-4AC)->C
    :Disp "(X",D,")(X",E,")"

    I just came up with this off the top of my head, so I haven't tested it. It might take a little tinkering, most likely in the display command (I can never seem to get those to work right when I'm using variables). It should work though. If it remains in decimal form, that means it can't be factored rationally. Otherwise, it's pretty self-explanitory.
  14. Feb 4, 2008 #13

    if you use this program;
    :Prompt A
    :Prompt B
    :Prompt C
    sqrt. (B^2-4AC)->C
    :Disp "(X",D,")(X",E,")"
    you will get a incorrect answer use this
    Set Mode To A+Bi
    :Prompt A,B,C
    :sqrt. (B^2-4AC)->C
    :Disp "Roots",D>Frac,E>Frac
  15. Feb 4, 2008 #14
    why not just graph the function and find the x-intercepts...?
  16. Feb 4, 2008 #15
    by graph i meant sketch (on calculator) sorry for any misunderstandings caused. Also i agree with the other guys... Just do it by hand!
  17. Feb 4, 2008 #16
    Because for some reason, most people who aren't in a major involving math are too lazy. If you just have to put the numbers into the calculator, you don't even have to bother putting the graph function in and finding the zeroes.
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