Question

Find $p\left(0\right),p\left(1\right),p\left(2\right)$ for $p\left(x\right)=(x-1)(x+1)$

• A. $p\left(0\right)=-1$
• B. $p\left(1\right)=0$
• C. $p\left(2\right)=3$
• D. None of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A $p\left(1\right)=0$
B $p\left(0\right)=-1$
C $p\left(2\right)=3$

Here, $p\left(x\right)=(x-1)(x+1)=x^2-1$

$\Rightarrow$ $p\left(0\right)=0^2-1=-1$

$\Rightarrow$ $p\left(1\right)=1^2-1=1-1=0$

$\Rightarrow$ $p\left(2\right)=2^2-1=4-1=3$