factoring

PLAYNOISE

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Joined
Aug 16, 2005
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I have three more factoring problems that i have been stuck on... i just dont know what to do.. and is the first one i did right?

1.)6x^2+3x=0
2x^2+x=0
x(2x-1)=0
2x=-1
x=0
x=-1/2_(<was_that_right?)

2.)x^2-2x-8=0

3.)3+5x-2x^2=0

4.)x^2+4x=12

Edit: Irritating formatting removed.
 
1) If 2x - 1 = 0, how did you get "2x = -1"?

2) Apply the quadratic formula, or find factors of -8 that are two units apart.

3) Rearrange in descending order. Divide through by -1 to change all the signs. Then find factors of (2)(-3) = -6 that are five units apart.

4) Subtract the 12 to the left-hand side. Find factors of -12 that are four units apart.

Eliz.
 
1.)6x^2+3x=0
2x^2+x=0
x(2x-1)=0
2x=-1
x=0
x=-1/2_(<was_that_right?)

I think someone else already explained this...

2.)x^2-2x-8=0

just think what times what will get you -8, but added together give you -2?
so just go through the factors...
-1 and 8
-8 and 1
-2 and 4
-4 and 2
which one works? if you're still not sure, just take one of those and put it into the equation, and check...
(x-1) (x+8)
= x^2 + 8x - x - 8
= x^2 + 7x - 8
does this work? no? keep going then...

3.)3+5x-2x^2=0

for this, you might want to switch everything over to the other side first...
2x - 5x - 3 = 0
Then since you know that nothing that multiply to make -3 and add together to make -5 exist...you know that that 2 in 2x is going to come in somewhere, so just use -1 and 3 and -3 and 1, and just switch them around until it fits:

(2x-1) (x+3)
(2x+1) (x-3)
(2x+3) (x-1)
(2x-3) (x+1)

4.)x^2+4x=12

So first step is to get everything on one side.
x^2 + 4x - 12 = 0
then solve the same way as #2
 
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