Question

class 9th ch 2 polynomial question (ex-2.2)​

Answer 1

Answer:

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

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

= -1×1

=-1

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

=0×2

=0

p(2)=(2-1) (2+1)

= 1 × 3

=3

Answer 2

Solution:-

━━━━━━━━━━━━━━━━━━━━━━━━━

$\large\to{\underbrace{\underline{\sf{Understanding\:the\:concept:-}}}}$

$\implies$ Here it is given in the question that we have to find p(0), p(1) and p(2) for the given polynomials. Here as you asked we will solve for the last part that is p(x) = (x - 1)(x + 1). Follow the steps to get the answer.

⛤ HOW TO DO:-

$\to$ Here to solve this we will one by one put the values given for finding the zero of the polynomial and by doing this we will get the answer. As 3 values of p are given in the question we will have to put three different values.

----------------------

ANSWER:-

• p(0) = -1
• p(1) = 0
• p(2) = 3

GIVEN:-

• 1st value = 0
• 2nd value = 1
• 3rd value = 2

TO FIND:-

• We have to find the p(0), p(1) and p(2) for the asked polynomial.

SOLUTION:-

• We will put the values and solve.

• 1st CASE p(0) :-

➤ p(x) = (x - 1)(x + 1)

➤ p(0) = (0 - 1)(0 + 1)

➤ -1 × 1

➤ -1

----------------------

• 2nd CASE p(1) :-

➤ p(x) = (x - 1)(x + 1)

➤ p(1) = (1 - 1)(1 + 1)

➤ 0 × 2

➤ 0

----------------------

• 3rd CASE p(2) :-

➤ p(x) = (x - 1)(x + 1)

➤ p(2) = (2 - 1)(2 + 1)

➤ 1 × 3

➤ 3

----------------------

━━━━━━━━━━━━━━━━━━━━━━━━━