Factoring...
\[x^2-6x+8\]
@EndersWorld take your best shot at it.
Okay. Gimme a few.
a=1 b=-6 c=8
ac=8
-2-4=-6 -2*-4=8
\[x^2-(2+4)x+8\]
\[x^2-2x-4x+8\]
\[x(x-2)-4(x-2)\]
Great work so far.
\[(x-2)(x-4)\]
I can’t remember if that’s its or not...
The Disburtitive Property?
If I were you though, I would try my best to avoid as much as possible using negative numbers for m and n. Instead do this: \(m + n = |b|\) \(m \times n = ac\)
Factoring.
That way you don't get hung up on the negative numbers.
You factored correctly though.
Okay... good... I found a trick,
In this case, when finding \(m\) and \(n\) you should write: \(2 + 4 = 6\) \(2 \times 4 = 8\)
When you do it this way, you know to substitute \(-6 = -(2 + 4)\) flawlessly with no confusion. But I think your confusion may be finally cleared up. I told you that you would get this eventually. Congrats.
Two shots of fireball helped.
Two `shots` of `fireball`???
...
Cinnamon whiskey.
Cinnamon Whiskey? How old are you?
Not old enough. Moving on.
You shouldn't do that.
Never Drink or do Drugs. They will kill You.
Slowly
..
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