Question

Find P(0) of the polynomial P(x)=(x-1)(x+1).​

Answer 1

Answer:

Answer:p(y) = y2 – y + 1

Answer:p(y) = y2 – y + 1 Plug y = 0 we get

Answer:p(y) = y2 – y + 1 Plug y = 0 we get =>p(0) = (0)2 – 0 + 1

Answer:p(y) = y2 – y + 1 Plug y = 0 we get =>p(0) = (0)2 – 0 + 1=>p(0) = 0 – 0 + 1

Answer:p(y) = y2 – y + 1 Plug y = 0 we get =>p(0) = (0)2 – 0 + 1=>p(0) = 0 – 0 + 1 => 1

Answer 2

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$☞ \: \: p(x) = (x - 1)(x + 1) \\ \\ ☞ \: \: p(0) = (0 - 1)(0 + 1) \\ \\ ☞ \: \: \: \: \: \: \: \: \: \: \: \: = ( - 1)(1) \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ☞ \: \: \: \: \: \: \: \: \: \: \: \: = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

I hope it helps you . ✔️❤️