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Old 02-08-2005, 07:51 PM
Biskmatar Biskmatar is offline
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factoring completely ax2+bx+c

HOW? I am confused!!
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Old 02-08-2005, 09:22 PM
Tohopekaliga Tohopekaliga is offline
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Re: factoring completely ax2+bx+c

Without specific numbers, it is much easier to show with variables going the other way, with more variables.

^2 means 'squared,' cx^2 means 'c times x squared'

(ax+b)^2 = a^2 * x^2 +2abx + b^2

(ax + b)(cx + d) = acx^2 + adx + bcx + bd

Therefore,

(x + 1)^2 is the same thing as x^2 + 2x + 1

I can explain more, step-by-step if really needed.
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Old 02-09-2005, 07:36 PM
Biskmatar Biskmatar is offline
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Re: factoring completely ax2+bx+c

thanks tahts helpful
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Old 02-18-2005, 10:37 PM
The Dark Side The Dark Side is a male United_States The Dark Side is offline
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Re: factoring completely ax2+bx+c

Factoring messes me up too, thanks Kokiri Kid
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Old 03-03-2005, 01:26 AM
Xizor United States Xizor is offline
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Re: factoring completely ax2+bx+c

That would be the quadratic formula, right? If it is, (I hope it is, otherwise I'm rambling), then you don't need to factor it, and it comes down to this:

x = -b + the square root of b squared - 4(a)(c)
____________________________________________
2a

so I hope that helps, otherwise I just wasted my time.
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Old 03-03-2005, 07:30 AM
Tohopekaliga Tohopekaliga is offline
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Re: factoring completely ax2+bx+c

Factoring is different from the quadratic formula. They serve different purposes, and often the instructions for problems is "factor completely," meaning you need to know how to do that. Besides, it's easier to work with if you don't have a calculator than that formula.

And you should write that formula as x = (-b±root[b^2 - 4ac])/(2a)
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Old 03-03-2005, 12:06 PM
Nox Nox is a male United States Nox is offline
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Re: factoring completely ax2+bx+c

I see in your example there that you were using the FOIL method, correct KK? Thanks for helping, even if I don't have that yet in my math class, because I've always wanted to learn it and wasn't that good at it really.
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Old 03-03-2005, 04:52 PM
Gracie United_States Gracie is offline
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Re: factoring completely ax2+bx+c

Hey, I could have helped. Yeah, anyways I think that was FOIL. We just recently went over polynomials and quadratic forumula... Very confusing, but I got a 99 on one test and 100 on the other. ^_^
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Old 03-05-2005, 06:31 AM
Matt Matt is a male United States Matt is offline
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Re: factoring completely ax2+bx+c

That factoring was confusing, I was taught a different way, and to me it is a lot easier.

Lets say we have

2xsquared + 3x - 20.

What we do is multiply the 2 and the -20, and get -40 we keep the ten the same.

Next we need two numbers that multiply to - 40 and add to POSITIVE 3, now I just made this problem up, so I don't know if it will work, lemme see here we write out the factors for 40, which are...

1, 40
2, 20
4, 10
5, 8>>> this works right. -8 x 5 = -40, and -8 + 5= 3

So we go back to rewrite the problem, as

2xsquared - 8x + 5x - 20>>>>>>>>>> from here we do GCF to group

2x(x-4) + 5(x-4)>>>>>>>>>>>> Now just group

(x-4)(2x+5)

If you use the foil method on this binomial, you will end up with the problem from the beginning, meaning it is correct, so I'l check my work to make sure it is right.

(x-4)(2x+5)>>>>>>>>. to make things easier for me here, I change x-4 to

x + -4

x times 2x = 2xsquared + 5x + -8x + -20>>>>>>>>>> combine like terms

5x plus -8x = 3x>>>>>>>>>. lookie here we are left with

2xsquared + 3x - 20 (I changed the plus negative back to minus positive)

Look at that, that is the exact same problem we started with, meaning that we have factored correctly. Its alot, but it is hell of a lot easier than what I saw up there, that just confused me. And one last bit, you all should know, if there is just a letter, and you need the number from it, the number is a 1.

Oh that's the bell, class practice this method, to see how easy it is. Remember the hardest part is getting the times a number, and add to get the dsecond number, after that, it all lays out like a hero sandwich
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Old 03-05-2005, 07:07 PM
Alonely Alonely is a female United States Alonely is offline
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Re: factoring completely ax2+bx+c

x² + 6x + 9

When factoring a trinomial problem as such, you multiple the first term and the third term.

x² ∙ 9 = 9x²

Now, take out the first term temporarily. In this case, that would be the x².

6x + 9x²

Now we need to find two numbers that MULTIPLY to get the third term, 9x², but ADD to get the middle term, 6x.

We know already that a high positive number added to a low negative number comes out to be a positive number. And we also know that two negatives and two positives become a positive number.

To tell if the numbers you will multiply and add are negative or positive, look at the signs in the problem.

x² + 6x + 9

Now we look at the sign between the middle and third term. It's a plus sign. That means that the answer is made up of two positive OR two negative terms, since a positive multiplied by a positive and a negative multiplied by a negative are both positive.

To find out if the answers are both negative answers or both positive answers, look at the sign between the first and middle term. It's also a plus. We also know in this problem that the two answers are both the same sign. We also know that two negatives cannot equal a positive, so the numbers derived from the problem are both POSITIVE.

Those two numbers are +3x and +3x.

3x + 3x = 6x
and
3x ∙ 3x = 9x²

Now, the next step is to replace the MIDDLE TERM with the two numbers you just got, and place them into the ORIGINAL problem. The 9x² would become 9 because it was 9 in the first place, not 9x².

x² + 3x + 3x + 9

Now you would factor the two.

x² + 3x becomes x(x+3)

3x + 9 becomes 3(x+3)

Now place them into the problem as normal.

x(x+3) + 3(x+3)

Now, take the term in parenthesis and place it as your first term in your answer.

(x+3)(?+?)

Eliminate the (x+3) in the problem x(x+3) + 3(x+3) because you've already used it.

Now you would take the leftover problem, which is x+3, and put it as your second term in your answer.

The answer would be (x+3)(x+3) or (x+3)².



Does that help?
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Old 03-19-2005, 12:30 AM
Halcyon Hero Halcyon Hero is a male United States Halcyon Hero is offline
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Re: factoring completely ax2+bx+c

Oh dear god why did I click here? I hate math. I would rather be dropped into a dark bottomless hole than do anything more complicated than mutiplication and division.
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Old 04-01-2005, 10:07 PM
Neo Neo is a male United States Neo is offline
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Re: factoring completely ax2+bx+c

a(b+c)(d+e) this is complete factorization. example: (2x+10)(4x+6) this is incomplete. To complete it, notice how everything is divisible by 2. Thus giving you this 2(x+5)(2x+3). Notice how you divide all the numbers by 2? HTH! (That's Hope this helps, if you dunno)
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Old 04-02-2005, 01:07 AM
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Re: factoring completely ax2+bx+c

There are many different ways you can factor an equation, which method are you on now?
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Old 04-24-2005, 07:19 AM
Evil Tomato United_States Evil Tomato is offline
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Re: factoring completely ax2+bx+c

There's a method called the "box method" that works on very difficult problems, and it's simple and easy to use.
Too bad it confuses me. ^-^;

When we were learning the quadratic formula in the eigth grade, we sang this to the tune of Pop! Goes the weasel.

X is equal to mi-nus b, plus or minus the squaaaaare root, of b squared minus 4ac, all over 2a.

Had that stuck in my head for days, I did.
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Old 04-27-2005, 08:13 PM
Rx.cl0ne Rx.cl0ne is offline
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Re: factoring completely ax2+bx+c

u mean -b plus or minus square root of b^2-4ac/all over 2a
its fun stuff give me anything and ill solve it like things how it can boggle u?
practice = perfect

solve for x
x^2+6x=-9
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