Question

How do I find all the rational zeros of p(x)=x3−12x−16?

Answer 1

hi,

constant = 16

possible factors of 16 are +1, -1, +2, -2, +4, -4, +8, -8.

by hit and trial method,

p(x) =x^3 -12x - 16

• at x = 1

p(1) = (1)^3 -12(1) - 16

= 1 - 12 - 16

= 1 - 28

= -27

1 is not a zero of p(x).

• at x= -2

p(x) = x^3 - 12x - 16

p(-2) = (-2)^3 - 12(-2) - 16

= -8 + 24 - 16

= +16 - 16

= 0

-2 is a zero of p(x).

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