walker wrote:

C

x(x-5)(x+2) = 0 --> x is {-5,0,2}

(1) x^2 - 7x does not eual 0 --> {-2,5} remain --> insufficient

(2) x^2 - 2x - 15 does not equal 0 --> {-2,0} remain --> insufficient

(1)&(2) {-2} remains --> sufficient.

how did you derive from x(x-5) (x+2) = 0 x is {-5,0,2}

Where did you get the -2 and 5 from in (1)

and where did you get the {-2,0} from in (2) when I factorised I got (x+3) (x-5)

One last question, how do you go about simplifying an equation like

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

Sorry to be a pain but I'm struggling to understand this

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