can someone solve this...?

0 ⤊

it says that you have to group...

b^2-9b-b+9=0...

2007-09-12 17:31:37 · 4 answers · asked by Jay-j 2 in Science & Mathematics ➔ Mathematics

Do you mean like this?

(b-9)(b-1)=0

b=9 and b=1

2007-09-12 17:38:34 · answer #1 · answered by morgan_becca07 2 · 0⤊ 0⤋

The form suggests the factorization:

(b-9)(b-1) = 0

b=9 â¨ b=1

2007-09-13 00:37:11 · answer #2 · answered by Pascal 7 · 0⤊ 0⤋

b^2-9b-b+9=0
b^2-b-9b+9=0
b(b-1)-9(b-1)=0
(b-9)(b-1)=0
b-9 = 0 or b-1=0
b = 9 or 1

2007-09-13 00:38:28 · answer #3 · answered by james w 5 · 0⤊ 0⤋

uh... i'm in 10th grade math, maybe i could help.
b^2-10b=0
b^2=10b
___ ___
b b
b=10
i think th grouping is just combining like terms.

2007-09-13 00:42:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋