Find the zeros of x^3-4x^2-5x=0 please help!!!
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first take out the common factor x: x(x^2 - 4x - 5) = 0
Ok thaks what do I do next?
now can you factor x^2 - 4x - 5 ?
No. I don't think so.
ok it will factor to 2 binomials of the form (x + a)(x + b) in this case a times b will = -5 and a+ b will = -4
so can you figure out what a and b are?
hint: a = -5
(x-5) (x-4)?
no because -4 * -4 = -20 not -5
* -5 * -4 = 20 i meant!
why would you do -5 * -4 though?
the last 2 numbers are multiplied to get the last number (x - 5)(x - 4) = x*x + x*-4 - 5* x -5*-4 = x^2 - 4x - 5x + 20 = x^2 - 9x + 20 which is nothing like what we require
first x is distributed over the (x - 4) then -5 is distributed over (x - 4)
Oh oh so you FOIL'd it
yes - amounts to same thing the 2 numbers you require are - 5 and +1 because -5*+1 = -5 and -5+1 = -4
Okay okay thank you. Is that all you do to find the zeros?
well you need to finish the factoring first: we have x(x - 5)(x + 1) = 0 so can u find the zeros now?
Would x=5 and x=-1?
thats 2 zeros what about the other?
I'm not sure what to do from there.
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