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BatBlaster
Member
Joined: 10 Oct 2007
Posts: 160
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 Posted: 15 Nov 2007 09:52:45 pm Post subject: |
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| I need help factoring 9y^2+6y-8. |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 15 Nov 2007 09:58:22 pm Post subject: |
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Well, the easiest direction to go is to complete the square. The first term is (3y)2 and the second term is 2(3y)(1) so we want the constant term to be 12. We can achieve this by replacing the -8 by +1-9, which is equivalent. Now we have (3y)2+2(3y)(1)+12-9, which we know from the expansion of (a+b)2 is (3y+1)2.
We're still not done, because we have (3y+1)2-9, which isn't factored. So write 9 as 32. Then we have (3y+1)2-32 which factors as a difference of squares - a2-b2=(a+b)(a-b). In this case, this is (3y+1+3)(3y+1-3) which boils down to (3y+4)(3y-2).
Of course, we could get here easily by observing that 9 is 3*3 and 8 is 4*-2, and that the middle term happens to end up 6 this way. But guess and check is not in general the best way to go about math problems, and this method is guaranteed to yield a solution if there is one.
Last edited by Guest on 15 Nov 2007 09:59:58 pm; edited 1 time in total |
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BatBlaster
Member
Joined: 10 Oct 2007
Posts: 160
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| Posted: 15 Nov 2007 10:01:52 pm Post subject: |
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| Thanks for the fast help. On my wii. Gtg |
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luby
I want to go back to Philmont!!
Calc Guru
Joined: 23 Apr 2006
Posts: 1477
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| Posted: 16 Nov 2007 08:03:31 am Post subject: |
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| There is a calc program that I made that will factor a quadratic, among other things. I can upload it to the UTI files if there is a demand. |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 16 Nov 2007 09:42:56 am Post subject: |
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Here's a trick for factoring quadratics with a non-1 coefficient for the squared term.
If you have Ax2+Bx+C, you can multiply A and C, and factor it as if the quadratic was X2+Bx+AC. But instead of having the result of the form (x+d)(x+E), you'll have (Ax+d)(Ax+e)/A
So with your original: 9y2+6y-8
9*-8=-72
12*-6=-72, 12+-6=6
(9y-6)(9y+12)/9
=(9y-6)/32
=((9y-6)/3)*((9y+12)/3)
=(3y-2)(3y+4)
This looks complicated but it's actually simple, I only wrote out every step to make it clearer.
What DarkerLine did was derive the quadratic formula, which can also make factoring easier. |
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JoostinOnline
Active Member
Joined: 22 Aug 2007
Posts: 559
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| Posted: 16 Nov 2007 06:57:05 pm Post subject: |
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Just use (-B+{1,-1}√(B²-4AC)/(2A to find the x-intercepts.
Last edited by Guest on 04 Sep 2010 08:39:55 pm; edited 1 time in total |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 16 Nov 2007 08:34:53 pm Post subject: |
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If this were a TI-Basic question, that'd be helpful. However, I'm sure this type of question would be appearing on his test at the end of the unit, where he might be expected to show his work.
BatBlaster, see if this might help you: Two Methods of Factoring Quadratics
Last edited by Guest on 16 Nov 2007 08:47:22 pm; edited 1 time in total |
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Chasney913
Member
Joined: 28 Aug 2007
Posts: 117
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| Posted: 17 Nov 2007 09:50:21 am Post subject: |
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| Yeah, I wrote a BASIC program to factor a trinomial of this form, by deriving the two roots using the quadratic formula, and working from there. It's kind of difficult, so that you can't do it on paper, but it is a cool program. I can post the code, if you'd like, but I sense a "show your work" method would be more appropriate. Pencil-and-paper and such, as opposed to calculator. |
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JoostinOnline
Active Member
Joined: 22 Aug 2007
Posts: 559
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| Posted: 17 Nov 2007 04:09:10 pm Post subject: |
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He could still use the formula. I only wrote {1,-1} because I didn't know how else to illustrate "plus or minus" in a formula other than in code.
Last edited by Guest on 04 Sep 2010 08:39:41 pm; edited 1 time in total |
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bananaman
Indestructible
Calc Guru
Joined: 12 Sep 2005
Posts: 1124
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| Posted: 17 Nov 2007 07:07:11 pm Post subject: |
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This may be interesting to some people. I wrote a program in junior high that would factor any polynomial. I looked around my computer and I still had it. I took one glance at the code and said eww. I couldn't believe how poorly optimized the code was and how many memory leaks were in it. Here it is for those of you who are interested in what method I used.
Code:
Lbl PF
Disp "DEGREE
Prompt X //Input "DEGREE:",X
1→S
X+1→dim(L₁
ClrHome
Disp ""
For(I,X,0,–1
Output(1,1,"COEF OF X^
Output(1,11,I
Input ">",A
A→L₁(X-I+1)
End
1→A
L₁(1)→N //Wow, this was stupid
If L₁(1)<0 //what was I thinking
–L₁(1)→N //abs(L₁(1→N
For(I,1,√(N
If fPart(N/I)=0
Then
A+1→dim(L₂
I→L₂(A
N/I→L₂(A+1)
If I²=N
A→dim(L₂
A+2→A
End // This is not an efficient way to get the factors of a number
End
1→A
L₁(X+1)→N
If L₁(X+1)<0
–L₁(X+1)→N //Another absolute value
For(I,1,√(N
If fPart(N/I)=0
Then
A+3→dim(L₃
I→L₃(A
–I→L₃(A+1)
N/I→L₃(A+2)
–N/I→L₃(A+3)
If I²=N
A+1→dim(L₃
A+4→A
End
End
Text(57,32,"THINKING
0→R // Why wasn't I using DELVAR
0→Q
dim(L₂)→D
dim(L₃)→E
Lbl FP
For(I,1,D
For(J,1,E
0→S
For(K,1,X+1
(L₁(K)*((L₃(J)/L₂(I))^(X+1-K)))+S→S
End
If S=0
Then
Text(7+R,Q,"(",L₂(I),"X+",–L₃(J),")
R+7→R
For(L,1,X
L₁(L)*(L₃(J)/L₂(I))+L₁(L+1)→L₁(L+1)
End
X-1→X
X+1→dim(L₁)
If L₂(I)≠1
Then
For(L,1,X+1
L₁(L)/L₂(I)→L₁(L
End
End
If R>37
33+Q→Q
Goto FP
End
End
End
Text(57,32,"
If dim(L₁)=1 and L₁(1)≠1
Text(7+R,Q,"(",L₁(1),")
If dim(L₁)=2
Text(7+R,Q,"(",L₁(1),"X+",L₁(2),")
If dim(L₁)=3
Text(7+R,Q,"(",L₁(1),"X²+",L₁(2),"X+",L₁(3),")
If dim(L₁)=4
Text(7+R,Q,"(",L₁(1),"X+",L₁(2),"X²+",L₁(3),"X+",L₁(4),")
Goto θG
If you can't tell what it does, it takes the prime factors of the first and last coefficient and then scans through the original polynomial and checks the combinations and discovers if one is a factor. It is mathematically proven that this will find the factors. My code is terrible, but I got such a kick out of it that I had to post it. |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 17 Nov 2007 07:34:43 pm Post subject: |
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This uses the Rational Root theorum, right? (the method usually referred to as "that annoying p over q thing")
It only works for integer coefficients and only finds the rational roots (try 0=x2-3x+1, which is factored into (2x-3-√5)(2x-3+√5) ).
I just had an idea, anyone think that a polynomial factoring program would be a good teaser? |
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angel14995
Member
Joined: 13 Oct 2007
Posts: 181
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| Posted: 17 Nov 2007 07:36:32 pm Post subject: |
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Well, I was working on one last year, but never got past the checking if any of the values, negative or not, would equal out.
And in there you said that you should have used DelVar. Delvar and 0/->/ are the same size and do the same things. Just using certain ones at certain points is more preferable. If you are ending a program, Delvar is a better choice to clean up, if you are going to use that variable again, 0/->/ is a better choice because storing a value to an already existing variable is faster than creating it then storing it.
Lemme see if I can make mine real fast  |
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bananaman
Indestructible
Calc Guru
Joined: 12 Sep 2005
Posts: 1124
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| Posted: 17 Nov 2007 10:32:29 pm Post subject: |
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I think it uses the rational root thereom. It has been so long that I don't know what I based it off of. I wrote the program over 6 years ago.
Delvar may be slower than 0 ->, but it takes up less bytes b/c you don't need to start a new line after it. The program doesn't take long enough to worry about optimizing for speed, so size should be the focal point. |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 18 Nov 2007 01:41:32 am Post subject: |
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I just made one that finds all rational roots of a polynomial (including multiple roots), it's 292 bytes, it's fairly fast, yet it looks really ugly. I understand why code for a game looks ugly, since the commands are designed for math, but even the code for a math program looks weird :confused: Funny, isn't it? 
@angel14995, how did you check to see if the values worked? I just used synthetic division, which simultaneously tests to see if it is a root and finds the resultant polynomial. After that it just loops back and does synthetic again. |
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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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| Posted: 19 Nov 2007 04:46:47 am Post subject: |
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As it has been previously said, Horner's Rule (a.k.a. synthetic division) is very efficient, both in dividing out monomial factors and evaluating polynomials at suspected roots. :)
thornahawk |
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