How do you find zeros of #y=5x^3-5x#?

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Answer 1 · 2016-10-09T09:41:57

#x=-1, 0, 1#

Let's start with the original:

#y=5x^3-5x#

We can factor out a #5x# from both terms, so we get:

#y=5x(x^2-1)#

And now let's factor the expression within the brackets:

#y=5x(x+1)(x-1)#

We set it all equal to 0:

#0=5x(x+1)(x-1)#

And we can now find the zeros:

#5x=0#

#x=0#

~~~

#x+1=0#

#x=-1#

~~~

#x-1=0#

#x=1#

So #x=-1, 0, 1#