How do you solve $(x - 3) (x - 5) (x + 1) = 0$ $(x-3)(x-5)(x+1)=0$ ?

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Answer 1 · 2018-01-25T09:10:21

$x = 3, 5, - 1$ $color(blue)(x = 3, 5, -1)$

$(x - 3) \cdot (x - 5) \cdot (x + 1) = 0$ $(x-3) * (x-5) * (x+1) = 0$

i.e. $x - 3 = 0 or x - 5 = 0 or x + 1 = 0$ $x-3 =0 or x-5 = 0 or x+1 = 0$

Case 1 : $x - 3 = 0, x = 3$ $x - 3 = 0, color(green)(x = 3)$

Case 2 : $x - 5 = 0, x = 5$ $x - 5 = 0, color(green)(x = 5)$

Case 3 : $x + 1 = 0, x = - 1$ $x + 1 = 0, color(green)(x = -1)$

How do you solve (x-3)(x-5)(x+1)=0(x−3)(x−5)(x+1)=0(x-3)(x-5)(x+1)=0?