#question on how many solutions for a function

11 messages · Page 1 of 1 (latest)

harsh nexusBOT
ionic niche
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hmm

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lets imagine

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we have the equation

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(x-3)(x+2)(x-5)=0

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this mean X can have 3 values where the equation equals zero

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x=3
x=-2
x=5

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and if you multiply all the parantheses you would get something like

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ax^3 + bx^2 + cx + d = 0

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so the number of solutions

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normally is equal to the higher exponent