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2007-01-21 15:33:57 · 2 answers · asked by saravanan mathivadhani suryakala 1 in Science & Mathematics ➔ Alternative ➔ Other - Alternative
Multibly each term in the first binomial by each term in the second, and combine like terms.
(a-x)(b-x) = ab - ax -bx +x^2 = x^2 - (a+b)x +ab
2007-01-21 15:43:06 · answer #1 · answered by Steve P 2 · 0⤊ 0⤋
(a-x) (b-x)
= a(b-x) - x(b-x)
= ab - ax - bx + (x * x)
= ab - ax - bx + x ^2
x ^ 2 means x to the power 2
2007-01-21 15:48:38 · answer #2 · answered by Sushmita N 2 · 0⤊ 0⤋
3?
2007-01-24 09:39:16 · answer #3 · answered by Talonn 1 · 1⤊ 0⤋
ab - ax - bx + x ^2
2007-01-23 09:03:23 · answer #4 · answered by neon_matrix5000 2 · 0⤊ 0⤋
ab-ax-bx+x^2
2007-01-23 00:25:25 · answer #5 · answered by bldudas 4 · 0⤊ 0⤋
if your question consists of all letters of alphabet that is (a-x)(b-x)(c-x)..........(z-a) then the answer will be 0 because it has a factor (x-x)=0 and 0 multiplied by anything is 0
2007-01-23 18:09:45 · answer #6 · answered by romeo 1 · 1⤊ 0⤋
(a-x)(b-x)
ab-ax-bx-x^2
ab-x(a+b)-x^2
but man, i think ur question was something different... but it was unclear.
2007-01-22 08:40:35 · answer #7 · answered by FRIENDS FOREVER 2 · 0⤊ 0⤋
ab-ax-bx+xsquared
2007-01-21 15:41:49 · answer #8 · answered by David K 3 · 0⤊ 1⤋
(a-x)(b-x)= x^-x(a+b)-ab
2007-01-21 16:14:19 · answer #9 · answered by nikhil k 1 · 0⤊ 0⤋