Asked by tasu
if x^2+2x-15=(x+r)(x+s) for all values of x one possible value of for r-s is?
Answers
Answered by
Reiny
x^2+2x-15=(x+r)(x+s)
(x+5)(x-3) = (x+r)(x+s)
so r = 5 and s = -3
r - s
= 5 - (-3)
= 8
(x+5)(x-3) = (x+r)(x+s)
so r = 5 and s = -3
r - s
= 5 - (-3)
= 8
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