please help. im not sure where to get help on this one.
lots of thanks to any one who can help.
2007-06-30 05:16:53 · 7 answers · asked by starcruiserGalaxy2029 2 in Science & Mathematics ➔ Mathematics
than you all, very much, 4 your answers.
You're great.
2007-06-30 05:53:38 · update #1
(x-5)^4 - (x+2)(x-5)^3
= (x - 5)³ [(x -5) - (x + 2)] removing the common factor
= (x - 5)³ [ -7]
= -7(x - 5)³
2007-06-30 05:26:13 · answer #1 · answered by fred 5 · 0⤊ 0⤋
(x-5)^4 - (x+2)(x-5)^3= -7(x- 5)^3
(x-5)^3[x-5 -(x+2)] = -7(x-5)^3
-7(x-5)^3 = -7(x-5)^3
2007-06-30 12:24:14 · answer #2 · answered by ironduke8159 7 · 1⤊ 0⤋
by taking (x-5)^3 as a common factor
(x-5)^4 - (x+2)(x-5)^3= (x-5)^3[ (x-5)-(x+2)]
= (x-5)^3[ x-5-x-2)]= -7(x- 5)^3
2007-06-30 12:24:54 · answer #3 · answered by pioneers 5 · 0⤊ 0⤋
LHS=(x-5)^3{(x-5)-(x+2)}
(x-5)^3(x-5-x-2)
(x-5)^3(-7)
-7(x-5)^3
=RHS
2007-06-30 16:46:22 · answer #4 · answered by MAHAANIM07 4 · 0⤊ 0⤋
(x-5)^4 - (x+2)(x-5)^3= -7(x- 5)^3
LHS
=(x-5)^4 - (x+2)(x-5)^3
take (x-5)^3 as commom
= (x-5)^3[ (x-5) - (x+2)]
= (x-5)^3 [x-5-x-2]
= (x-5)^3 [-7]
= -7 (x-5)^3
= RHS
2007-06-30 12:24:49 · answer #5 · answered by Sam 3 · 1⤊ 0⤋
(x-5)^4 - (x+2)(x-5)^3
factor out (x-5)^3
(x-5)^3 [ (x - 5) - (x + 2)
(x - 5)^3 (x - 5 - x - 2)
(x - 5)^3 (-7)
-7 (x - 5)^3
hope this helps
2007-06-30 12:24:51 · answer #6 · answered by 7 · 0⤊ 0⤋
Using FOIL or whatever method you learned, expand the expression on the left. Then you can re-factor the solution into a simpler form, which should be what's on the left. It's not hard, it just might take you a little while.
2007-06-30 12:21:52 · answer #7 · answered by Anonymous · 0⤊ 1⤋