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 = (AB)(A^2 + AB + B^2)
so (2x)^3  1 = (2x1)(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(x2)  4(x2)
= ......

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the x has been replaced by a+2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^24) + 4)
= √(2x^2 4)
then g(f(a+2))
= √(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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