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Help with factoring quadratics?
I asked this earlier, but I don't think I'm going to get any more answers because this forum moves so quickly. Anyway, what I'm having trouble with, is how to simplify this:
(x-1)[x^2(x-4)-4(x-4)]=0
into this:
(x-1)(x-4)(x^2-4)=0
?
I keep thinking there should be an extra (x-4) in there, it seems like it just disappears. I have a feeling this is something easy that I'm missing, but I really need help.
2 Answers
- ?Lv 49 years agoFavorite Answer
(x-1) [x^2(x-4) - 4(x-4)] = 0
x^2(x-4) - 4(x-4) <--- here, each term, x^2(x-4) and - 4(x-4), have (x-4) as their factor.
We factor (x-4) out.
x^2(x-4) divided by (x-4) is x^2.
-4(x-4) divided by (x-4) is -4.
After factoring out (x-4), terms left are: x^2 and -4. As 1 expression, what's left is (x^2 - 4).
so
x^2(x-4) - 4(x-4) = (x-4)(x^2-4)
Attach (x-1) and you get (x-1)(x-4)(x^2-4).
hope I helped :D
- 9 years ago
Think like this: let M = (x - 4)
so you have (x-1) (x^2 * M - 4 * M) = (x - 1)[(x^2 - 4) * M]
So the answer should be (x-1)(x-2)(x+2)(x-4)
where for any a, b: a^2 - b^2 = (a-b)(a+b) .
