Question

find p(0),p(1) and p(2) for each of the following polynomials.p(x)=(x-1)(x+1)​

Answer 1

Step-by-step explanation:

ANSWER:-

How to Attempt this question?

• Given the value, place the value in x
• Then solve the polynomial

First we will make IT quadratic.

(x-1)(x+1)

= x²-0x-1

From Identity

${x}^{2} - {y}^{2} = (x + y)(x - y)$

For p(0)'

${(0)}^{2} + 0 \times 0 - 1$

$= - 1$

For p(1)-

${(1)}^{2} + (1) \times 0 - 1$

$= 0$

For p(2)-

${(2)}^{2} + 2 \times 0 - 1$

$= 4 - 1$

$= 3$

These are the values.

Note:-

• Whenever squaring is done, then negative changes to positive
• Positive remained the same
• Place the value of p(c) in place of x to get desired results
• For verification it must match with the output provided in question.

Answer 2

Answer:

p(0)=-1  , p(1)=0 , p(2)=3     , subscribe to my youtube channel #smtuitions

Step-by-step explanation:

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

now using algebraic identity (a-b)(a+b) = a² - b²

p(x)=x²-1²

p(0)=0²-1²

p(0)=-1

p(x)=x²-1²

p(1)=1²-1²

p(1)=0

p(x)=x²-1²

p(2)=2²-1²

p(2)=4-1

p(2)=3