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Determine the zeros of f(x) = x3 – 3x2 – 16x + 48. (1 point)
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you can factor this
gcf of x^3,3x^2 is x^2 gcf of 16x,48 is 16
so factor the first two terms by writing x^2(something) minus so factor the second two terms by writing 16(something) so we have x^2(something) - 16(something) =(something)(x^2-16) =(something(x-4)(x+4)
x2(x-3)-16(x-3)=(x2-16)(x-3) x2=16 x_1,_2=+/-4 x_3=3
ok, but the answers I got were x=-4 x=3 and x=4 So I did get them right?
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thats what he has
Awesome, thanks guys, I checked it and yeah its tht c:
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