Question

if p(t)=(t-1)(t+1) then find the values of p(1), p(0).?

Answer 1

$Given \: p(t) = ( t - 1 )( t + 1 )$

$i ) \red{Value \: of \: p(1) }$

$= ( 1 - 1 )( 1 + 1 )$

$= 0 \times 2$

$\green { = 0 }$

$ii ) \red{Value \: of \: p(0) }$

$= ( 0 - 1 )( 0 + 1 )$

$= (-1) \times 1$

$\green { = -1 }$

Therefore.,

$\red{ Value \: of \: p(1) } \green { = 0}$

$\red{ Value \: of \: p(0) } \green { = -1}$

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Answer 2

Step-by-step explanation:

p(1)=(1-1) (1+1)

= 0×2

= 0

p(0)=(0-1) (0+1)

= -1×1

= -1

hope it's help you