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Nov 24th, 2004, 03:43 PM
#1
Factoring Trinomials???
Is there a mathematical way to factor trinomials?
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Nov 28th, 2004, 01:55 AM
#2
Fanatic Member
An extremely complicated one, yes.
Assuming by mathematical you mean "pure equation", like the quadratic formula.
Don't pay attention to this signature, it's contradictory.
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Nov 28th, 2004, 03:37 PM
#3
Boy that is easy.
Let's factor this for example
y² - 5y + 6
1.) Set it up as a product of two to where each will hold two terms.
2.) Find the factors that go in the first positions:
Code:
(y ) (y )
Remember y*y = y²
3.) Find the factors that go in the last positions:
Code:
(y - 2) (y - 3)
*-2 and -3 are two numbers whose prod. is 6
and sum is -5
4.) FOIL to double check if you are correct.
First
Outside
Inside
Last
Code:
F + O + I + L
First:
(y - 2) (y - 3)
(y * y) + O + I + L
Outside:
(y - 2) (y - 3)
(y * y) + (-3 * y) + I + L
Inside:
(y - 2) (y - 3)
(y * y) + (-3 * y) + (-2 * y) + L
Last:
(y - 2) (y - 3)
(y * y) + (-3 * y) + (-2 * y) + (-3 * -2)
Next we combine like terms:
(y * y) + (-3 * y) + (-2 * y) + (-3 * -2)
y² + -3y + -2y + 6
y² - 5y + 6
It's that easy. Hope this helps.
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Nov 29th, 2004, 11:25 AM
#4
Eh... maybe I should have reiterated a little bit...
I'm a high schooler taking college math courses.. I know how to factor trinomials by hand.
I want to write a calculator program for it, and the thing right now that's proving difficult is factoring each coefficient. If the first term was always 1, it would be a piece of cake. But I'm looking at something like ax2+bx+c where a, b, and c are inequal to 1 or 0. That kinda makes it a bit harder...
I can do it with recursion, but that takes forever. I need an effective and quick factoring system that I can use to set up the rest of the program.
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Nov 29th, 2004, 02:34 PM
#5
Fanatic Member
VB 6.0, Access, Sql server, Asp
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