HELP!
2007-01-25 16:09:32 · 4 answers · asked by savannah_banana03 1 in Science & Mathematics ➔ Mathematics
9x^2 + 12xy + 4y^2 - 25
= (3x)^2 + 2(3x)(2y) + (2y)^2 - 25
= (3x + 2y)^2 - 25
= (3x + 2y + 5) (3x + 2y -5)
by difference of perfect squares.
2007-01-25 16:20:06 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
9x^2+12xy+4y^2-25
= (3x)^2 + 2*(3x)*(2y) + (2y)^2 - 5^2
now using [ a^2+2*a*b + b^2 = (a+b)^2 ] where a = 3x , b = 2y
we get
= (3x + 2y )^2 - 5^2
now using [ c^2 - d^2 = (c+d)(c-d) ]
c = (3x + 2y ) , d = 5
= (3x+ 2y + 5 )(3x+ 2y - 5 )
hence factors are 1, (3x+ 2y + 5 ) and (3x+ 2y - 5 )
2007-01-26 00:26:28 · answer #2 · answered by rajeev_iit2 3 · 0⤊ 0⤋
(9 x^2 + 12 xy + 4 y^2) - 25 = (3x + 2y)^2 - 5^2
= (3x + 2y + 5)(3x + 2y - 5)
2007-01-26 00:18:53 · answer #3 · answered by wild_turkey_willie 5 · 1⤊ 0⤋
(3x + 2y)^2 - 25
= (3x + 2y -5)(3x + 2y +5)
2007-01-26 00:19:25 · answer #4 · answered by Anonymous · 0⤊ 0⤋