2007-12-04 16:11:16 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x² - 9
a = 1
b = 0
c = - 9
2007-12-05 06:20:38 · answer #1 · answered by Como 7 · 2⤊ 0⤋
I think tyou mean ax^2+bx +c
(x-3)(x+3) = x^2-9 so
a = 1 b = 0 and c = -9
b = 0
2007-12-05 00:14:34 · answer #2 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
if (x-3)(x+3) is written in ax^+ bx + c form, what is the value of b?
(x-3)(x+3)
no answer cos short of one more equations..
2007-12-05 00:21:17 · answer #3 · answered by Lachinos 3 · 0⤊ 1⤋
b=0
in ax^2 + bx +c form
(x-3)(x+3)
= x^2 -3x +3x -9
=x^2 -9
there is no bx, because they cancel when the brackets are expanded
2007-12-05 00:22:33 · answer #4 · answered by Anonymous · 0⤊ 0⤋
0, SINCE (x-3)(X+3)=X^2-9
2007-12-05 00:15:12 · answer #5 · answered by Patricia F 1 · 0⤊ 0⤋
b= 0 because the -3x and the 3x cancel out when you multiply and distribute.
2007-12-05 00:18:40 · answer #6 · answered by myself 3 · 0⤊ 0⤋
0
these are conjugates
2007-12-05 00:13:39 · answer #7 · answered by Dan L 2 · 0⤊ 0⤋