2007-05-20 16:00:41 · 6 answers · asked by colin l 1 in Science & Mathematics ➔ Mathematics
It's for calculus so I know the basics. Just need it set up like (?x+?)(?x+?)=0
2007-05-20 16:05:48 · update #1
12x^2-64
=12x^2-8^2
= (12x-8)^2
=[4(3x-2)]^2
if u wan (?x-?)(?x-?)
=12x^2-64
=(sq.root12 x-8)(sq.root 12 +8)
this is because of the diff of square rule
hope that helped :)
2007-05-20 16:16:26 · answer #1 · answered by veena_dracks84 2 · 0⤊ 0⤋
well.. firs tthing i looked a is different of perfect squares.. but it doesn't quite makessense in this case..
take 4 common and u get
4 (3x^2 - 16)
if u want to further factor (3x^2 - 16) u can get
(x + sqrt(16/3)) (x-sqrt(16/3)) (4)
but then I don't think that is necessary..
first step should be more than enough
2007-05-20 16:05:19 · answer #2 · answered by sudhi_kandi 3 · 0⤊ 0⤋
12x^2 - 64
4(3x^2 - 16)
2007-05-20 16:03:59 · answer #3 · answered by misshahila 2 · 0⤊ 1⤋
since 12 does not have a complete square you do it this way
4(3x^2 -16)
2007-05-20 16:02:55 · answer #4 · answered by jay gal 3 · 0⤊ 1⤋
= 4.(3x² - 16)
= 4.(â3.x - 4).(â3.x + 4)
2007-05-20 19:41:42 · answer #5 · answered by Como 7 · 0⤊ 0⤋
4(3x^2-16)
2007-05-20 16:05:50 · answer #6 · answered by dwinbaycity 5 · 0⤊ 0⤋