Math, asked by kodgegundamma, 9 months ago

PD p(x) x-1 is a factor of x^3-1​

Answers

Answered by adinarayanakanta
0

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)

Answered by ksonakshi70
0

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)

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