Today was the first day back in school and it's been along time since I've had Algebra. So I was wondering if someone could help me factor this problem, and show complete detail. Thank you.
Factor Completely:
4(x-2)^3 (x+1) + 6(x+1)^2 (x-2)^2
2006-08-24 12:20:59 · 5 answers · asked by livingall_4_god 2 in Education & Reference ➔ Higher Education (University +)
Thanks for your answers. And by the way, I'm in college.
2006-08-24 13:53:52 · update #1
4(x - 2)^3 (x + 1) + 6(x + 1)^2 (x - 2)^2
You take out their GCF, ending with the following:
2(x - 2)^2 (x + 1) [ 2(x - 2) + 3(x + 1) ]
Simplify the equation within the brackets:
2(x - 2)^2 (x + 1) (2x - 4 + 3x + 3)
2 (x - 2)^2 (x+ 1) (5x - 1)
Since it can't be simplified any further, the answer is
2 (x - 2)^2 (x+ 1) (5x - 1)
2006-08-24 12:52:39 · answer #1 · answered by lemons 3 · 0⤊ 0⤋
=16+17+7=14=12=144 aha
2006-08-24 19:25:56 · answer #2 · answered by lins 4 · 0⤊ 0⤋
factorized it is (x-2)^2(x+1) [4(x-2)+6(x+1)]
i got a d in calculus.
2006-08-24 19:35:43 · answer #3 · answered by I SLEEP TO DIE. 3 · 0⤊ 0⤋
4(x-2)^3 (x+1) + 6(x+1)^2 (x-2)^2
[[4(x-2)4(x-2)4(x-2)](x+1) + [6(x+1)6(x+1)][(x-2)(x-2)]]
[[(4x-8)(4x-8)(4x-8)](x+1) + [(6x+6)(6x+6)](x^2-4x+4)]
[[(16x^2-64x+64)(4x^2-4x-8)] + [(36x^2+72x+36)(x^2-4x-4)]]
[(64x^4-256x^3+256x^2-64x^3+256x^2-256x-128x^2+512x-512) + (36x^4+72x^3+36x^2-144x^3-288x^2-144-144x^2-288x-144)]
[(64x^4-320x^3+384x^2-256x-512) + (36x^4-72x^3-396x^2-288x-288)]
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100x^4-392x^3-12x^2-544x-800
that's the answer i got...lol...it toOk a long time. so, you're welcome.
2006-08-24 19:51:05 · answer #4 · answered by xretrospective 2 · 0⤊ 0⤋
haha..not in school yet...
2006-08-24 19:22:42 · answer #5 · answered by * Flave' * 3 · 0⤊ 0⤋