Posted by rachel on Saturday, August 23, 2008 at 11:36am.
I'm not exactly sure how to factor cubed roots:
8x^3 - 1 = (2x)^3 - 1
that's what i get; can it be simplified anymore?
- math - rachel, Saturday, August 23, 2008 at 11:43am
i'm looking at another one, and i don't know where to start:
x^3 - 2x^2 - 4x +8
hmm, cause u can't pull and x out, no quadratic . . .???
- math - Reiny, Saturday, August 23, 2008 at 12:14pm
your first one fits the pattern for the difference of cubes
A^3 - B^3 = (A-B)(A^2 + AB + B^2)
so (2x)^3 - 1 = (2x-1)(4x^2 + 2x + 1)
test it by expanding my answer
for you second one, try grouping
I will get you started
x^3 - 2x^2 - 4x +8
= x^2(x-2) - 4(x-2)
- math - rachel, Saturday, August 23, 2008 at 12:16pm
for function operations:
If f(x) = x^2 -4 and g(x) = [square root of] (2x+4)
g(f(a+2)) ---> then
x^2 - 4 +2
x^2 - x
[square root of] (2(x^2 - 2) +4)
= x [square root of] 2
wher does 'a' come from?, but is that right?
- math - Reiny, Saturday, August 23, 2008 at 12:29pm
the x has been replaced by a+2
so since f(x) = x^2 -4 and g(x) = √ (2x+4)
g(f(x)) = √(2(x^2-4) + 4)
= √(2x^2 -4)
= √(2(a+2)^2 - 4)
=√(2(a^2 + 4a + 4) - 4)
= √(2a^2 + 8a + 4)
check my algebra, sometimes without writing it down first, I tend to make errors while just working it out only on the screen.
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