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2006-11-25 22:17:02 · 14 answers · asked by gina s 1 in Science & Mathematics Mathematics

14 answers

take 2(x-2) as common factor

= 2(x-2)((x+3) + (x-3))
= 2(x-2)(2x)
=4x^2-8x

2006-11-25 22:20:41 · answer #1 · answered by Mein Hoon Na 7 · 0 0

It would be useful to tell exactly what you are trying to do... but I assume you need to simplify.
First, get rid of the stuff in parenthesis...
You do a fun little process call 'foil' First, outside, inside, last.
(x+3)(x-2) becomes x^2+3x-2x-6 = x^2 + x -6
(x-2)(x-3) becomes x^2 -2x -3x +6 = x^2 -5x +6
notice how first you multiply the first terms in each of the parenthesis, then the middle terms, then the outsides, then the last.
now, both of those answers were multiplied by 2, soo
2(x^2 + x -6) = 2x^2 + 2x - 12
2(x^2 -5x +6) = 2x^2 -10x + 12
now just add the two parts:
= 4x^2 - 8x

now, idk what kinda math your doing, probably alegbra and that's what you want, but you can still simplify this...
factor out a 4 and an x
4x(x-2)

Hope that helps.

2006-11-25 22:26:57 · answer #2 · answered by Anonymous · 0 0

Does there could desire to be in basic terms 5 steps? that's what I have been given: 10(2x-3)-2(3x+4)=4(x-one million)+sixteen {unique question} 20x-30-6x-8=4x-4+sixteen {use distributive regulation to do away with the brackets-multiply in> 10 circumstances 2x equals 20x and 10 circumstances 3 equals 30 etc.} 14x-38=4x+12 {integrate like words> 20x-6x=14x etc.} 10x-38=12 {flow the x's to a minimum of one part} 10x=38+12 {isolate the x}10x+50 {divide via 10 on the two aspects to isolate x some extra} x=5 desire this helped!!! :D

2016-12-29 12:21:45 · answer #3 · answered by calvete 3 · 0 0

if you need it to be in the simplest form then observe:
2(x+3)(x-2)+2(x-2)(x-3) divide by (x-2)
= 2(x+3)+2(x-3) open the brakets by multipling the(2)by them
= 2x+6+2x-6 notice ( -6+6=0)
= 2x+2x add
= 4x

2006-11-26 03:01:23 · answer #4 · answered by Anonymous · 0 0

2(x + 3)(x - 2) + 2( x - 2)(x - 3) =
2(x - 2)(x + 3 + x - 3) =
4x(x - 2) =
4x^2 - 8x

2006-11-25 22:51:27 · answer #5 · answered by Helmut 7 · 0 0

2(x+3)(x-2) + 2(x-2)(x-3)=
(2x+6)(x-2) + (2x-4)(x-3)=
(2xˇ2 + 6x -4x - 12) +
+ (2xˇ2 - 4x- 6x +12)=
4xˇ2 - 8x

2006-11-25 22:27:15 · answer #6 · answered by Andreja K 3 · 0 0

= 2(x+3)(x-2)+2(x-2)(x-3)
-> 2(x-2) {(x+3) + (x-3)}
-> 2(x-2) 2x
-> 4x(x-2)
ans 4x(x-2)

2006-11-25 22:25:00 · answer #7 · answered by kamal_sheru 1 · 0 0

It is called order of operations referred to as PEMDAS followed by the accronym Please Excuse My Dear Aunt Sally. Other words to solve problem Parrenthsis, Exponents, Multiplication, Division, Addition, and Subtraction. The rest is up to you that is all the help you get. I don't do peoples work for them life and all of its complexities doesn't get handed to you on a silver platter you do have to work for it!

2006-11-25 22:21:59 · answer #8 · answered by foxy 3 · 0 0

2(x+3)+2(x-2)(x-3)

= (2x+ 6)+(2x-4)(x-3)
= 2x+6+(2xx-6x-4x+12)
= 2x-10x+18+2xx

= 2xx-8x+18

2006-11-25 22:26:21 · answer #9 · answered by zymzyv 3 · 0 0

the answer is -24. do the brackets first and you get -12 + 12

2006-11-25 22:32:59 · answer #10 · answered by Lillian 2 · 0 0

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