Bunuel wrote:
HinaTabassum wrote:
I reached this part.
x^2(x+2)(x-3)=0
From there, I solved for x=-2 and x=3
How is x=0 calculated?
x^2(x+2)(x-3)=0
From there, I solved for x=-2 and x=3
How is x=0 calculated?
The product of number of multiples is 0 when at least one of the multiples is 0. So, x^2*(x + 2)*(x - 3) to be 0, either one of x^2, x + 2, or x - 3 should be 0:
x^2 = 0 --> x = 0;
x + 2 = 0 --> x = -2;
x - 3 = 0 --> x = 3.
Does this makesense?
Put that way, it does. And looks simple too. Thanks Bunuel.
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