(x - 16) / (x^2 + x - 2)
2006-12-12 00:43:44 · 4 answers · asked by jonesin_am 1 in Science & Mathematics ➔ Mathematics
LET A N B BE THE COEFFICIENTS OF 1/(X-2) N 1/(X+1)
THEN
A*(X+1)+B*(X-2)=X-16
A+B=1
A-2*B=-16
A=-14/3
B=17/3
2006-12-12 00:51:57 · answer #1 · answered by well thts it...... 3 · 0⤊ 1⤋
(x - 16) / (x^2 + x - 2) =
(x - 16) / (x + 2)(x - 1) =
A / (x + 2) + B / (x - 1)
(x + 2)(x - 1)[(x - 16) / (x^2 + x - 2)] = (x + 2)(x - 1)[A / (x + 2) + B / (x - 1)]
x - 16 = A(x - 1) + B(x + 2) = x(A + B) + (-A + 2B)
1 = A + B
-16 = -A + 2B
A = 6, B = -5
6/(x + 2) - 5/(x - 1)
2006-12-12 00:59:50 · answer #2 · answered by bayou64 4 · 0⤊ 0⤋
(x - 16) / (x^2 + x - 2)
=(x-16)/[(x+2)(x-1)] = A/(x+2) + B/(x-1)
= [A(x-1) + B(x+2)]/[(x-1)(x+2)]
So numerators of original fraction and new fraction now equal
Ax-A +Bx+2B =(A+B)x +(A+2B) = new numerator
A+B =1
A+2B=-16
So B= -17 and A = 18
Thus (x-16)/[(x+2)(x-1)] = 18/((x+2) - 17/(x-1)
2006-12-12 01:07:14 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋
do your damn homework yourself.
2006-12-12 00:45:39 · answer #4 · answered by jhstha 4 · 0⤊ 1⤋