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Thumbs down Factoring. MATH HELP PLEASE :| - January 6th 2009, 04:16 PM

Okay, so I have an ISU(independant study unit) due TOMORROW.
Does anybody know how to factor...

a) 5x squared-14x+8
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January 6th 2009, 04:17 PM

(5x - 4)(x - 2)


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Now I'm a stranger in your eyes.
   
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January 6th 2009, 04:26 PM

(5x-4)(x-2)
   
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January 6th 2009, 04:32 PM

Hii, you made me brush up on math skills! Oy! (:

So, typically, when one has these problems to factor out (which is basically reversing what has been done with the polynomial...expanding - reverse expansion...heh), the leading coefficient is one.
This leads us to think:

Hmm...what adds/subtracts to get the constant at the end but adds/subtracts to fulfill the secondary coefficient (the one in the middle)?

It all works well enough until...curveball! We're thrown a not-one as a leading coefficient - which you've got here.

There's one way to do this (not using a calculator) which I will illustrate here: *Think of these things as a puzzle...
5x-14x+8

What can equal 40x (5*8)x through addition/subtraction and also equal the middle coefficient?

Well, 4x and 10x can equal 14x if added together, and, if multiplied, it also equals 40x...yay. (:

So, just re-write as such:

5x-10x-4x+8

(5x-10x)(-4x+8) *Now take out what is common...

Do you see how -10x-4x=-14x? It's simply another way of writing 14x to fulfill all of the restrictions

Now, the biggest rule in factoring, pull out what is common (and this is so neat, 'cause it works every time, nearly)...
5x(x-2)-4(x-2)
Take what's left over, and what is common, and throw them in parentheses --
(5x-4)(x-2)

Alternate way (the cheap way) with a calculator *DO NOT READ UNLESS PRIOR KNOWLEDGE OF QUADRATIC FUNCTIONS CAN BE APPLIED* Don't get frazzled if you don't get the cheap way...heh:
You can use the Quadratic Formula because this is a quadratic equation!
Find the roots (two of them) using a program on your Ti-83 Plus (you can go to their website and hand-program it into your's, it's not bad...)
If you understand this, you can understand the factoring. Factoring also shows you the roots of any function, which is a nice thing to have in graphing them, I think. (That's the purpose of factoring - finally a reason, right?) (:
For the program:
A=5
B=-14
C=8
Enter=> Solution
These would be your two roots...but with the 0.8, there's a trick to it. Remember, 0.8=8/10 which reduces to 4/5. That means that it's going to be (5x-4) and then the two (other solution/root) would be (x-2). If you don't get the second way, don't worry. It's using knowledge of solutions/roots and quadratics, which really have nothing to do with factoring, in the short-term, but once you learn them, things make sooo much more ultimate sense. So, if you have the knowledge of functions, use it!
PM if you have any questions? (:

Last edited by Simply Complicado; January 6th 2009 at 04:49 PM.
   
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January 6th 2009, 05:52 PM

I am SimplyComplicado, so just PM me if you need any help. Switched names, yeah.
   
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January 6th 2009, 06:07 PM

It's in the form of ax(squared)+bx+c

To factorise, first you find; a times c (in this case 5x8=40)
Then think what numbers do you add to give -14 and multiply to give you 40?
-10 and -4

Then re-write as
5x(squared)-10x-4x+8. Seperate...
5x(squared)-10x -4x+8 see what is common for both.
5x(x-2) -4(x-2) Becomes...
(5x-4)(x-2)

Basically what Simply Complicado but a bit less confusing (I hope )

PM me if you need anymore help (:
   
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January 6th 2009, 07:21 PM

Better put as what Marionette has.
How ironic with my username.
;p ha.
x
   
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