Question

Determine the zeros of f(x) = x3 – 3x2 – 16x + 48.

Answer 1

factor it

Answer 2

alright thanks

Answer 3

solve by factoring and making f(x) 0
0=x^2-3x^2-16x+48 take out x^2 frm first 2 terms and -16 from last 2
0=x^2(x-3)-16(x-3) combine the x-3 to make
0=(x^2-16)(x-3) factor the perfect square
0=(x-4)(x+4)(x-3) make equal to 0
x-4=0 x+4=0 x-3=0 solve
x=4 x=-4 x=3 zeros are
(4,0)(-4,0)(3,0)

Answer 4

thanks