Question

PD p(x) x-1 is a factor of x^3-1​

Answer 1

Answer:

yes

Step-by-step explanation:

because

g(x)=x^3-1

g(1)=1^3-1=1-1=0

g(1)=0

where 1 is the zero of p(x)

Answer 2

Answer:

$by \: the \: factor \: theorem \: we \: know \\ that \: g(x) =( x - 1) \: is \: \: factor \: of \: \\ p(x) = x {}^{3} - 1 \: then \: p(1) \: = 0 \\ p(1) = 1 {}^{3} - 1 \\ \: \: \: \: \: \: \: \: \: = 1 - 1 = 0$

g(x) is a factor of p(x)