2006-06-22 05:06:16 · 7 answers · asked by happy 1 in Science & Mathematics ➔ Mathematics
9a*a+6a-4b*b+1
= (9a^2+6a+1)-4b^2
=(3a+1)^2-(2b)^2
=(3a+2b+1)(3a-2b+1)
2006-06-22 05:12:04 · answer #1 · answered by ag_iitkgp 7 · 1⤊ 0⤋
First of all, factorize 9a^2 +6a into 3a(3a+2).
Secondly, use the difference of two squares to factorize -4b^2+1 into (1+2b)(1-2b). Add: 3a(3a+2) + (1+2b)(1-2b)
2006-06-22 13:48:38 · answer #2 · answered by vishalarul 2 · 0⤊ 0⤋
You could also change it to:
9a^2 - 4b^2 + 6a + 1
Then factor the first two terms:
(3a + 2b)(3a - 2b) + 6a + 1
2006-06-22 12:13:49 · answer #3 · answered by KHB 2 · 0⤊ 0⤋
(9a*a+6a+1) - (4b*b)
= (3a+1)^2 - (2b)^2
= (3a+2b+1)*(3a-2b+1)
2006-06-22 14:06:40 · answer #4 · answered by akg 3 · 0⤊ 0⤋
(3a+1)^2 - (2b)^2
2006-06-22 12:13:27 · answer #5 · answered by Nate 3 · 0⤊ 0⤋
3a(3a+2)-1(2b+1)(2b-1)
2006-06-22 12:10:24 · answer #6 · answered by CRAZYDEADMOTH 3 · 0⤊ 0⤋
3a(3a+2)-4b^2+1
Other than that. there's nothing you can do.
2006-06-22 12:09:19 · answer #7 · answered by Theoretical chem guy 2 · 0⤊ 0⤋