NCERT Class 9 Mathematics Chapter 2 Exercise 2.2

 Exercise 2.2 of NCERT Class 9 Mathematics Chapter 2, "Polynomials," focuses on evaluating polynomials at given values and verifying potential zeroes. Below are the solutions to the exercise questions:

1. Evaluate the polynomial $f(x) = 5x - 4x^2 + 3$ at:

• (i) $x = 0$
  

  Substitute $x = 0$into $f(x):$

  $f(0) = 5(0) - 4(0)^2 + 3 = 0 - 0 + 3 = 3$

• (ii) $x = -1$
  

  Substitute $x = -1$ into $f(x):$

  $f(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = -6$

• (iii) $x=2$

  Substitute $x = 2$into $f(x):$

  $f(2) = 5(2) - 4(2)^2 + 3 = 10 - 4(4) + 3 = 10 - 16 + 3 = -3$
  

2. For each polynomial, find $p(0)$, $p(1)$, and $p(2)$:

• (i) $p(y) = y^2 - y + 1$
  

  • $p(0) = (0)^2 - (0) + 1 = 1$

  • $p(1)=(1{)}^2-(1)+1=1$

  • $p(2) = (2)^2 - (2) + 1 = 3$
    

• (ii) $p(t)=2+t+2t^2-t^3$

  • $p(0)=2+(0)+2(0{)}^2-(0{)}^3=2$

  • $p(1) = 2 + (1) + 2(1)^2 - (1)^3 = 2 + 1 + 2 - 1 = 4$
    

  • $p(2) = 2 + (2) + 2(2)^2 - (2)^3 = 2 + 2 + 8 - 8 = 4$
    

• (iii) $p(x) = x^3$
  

  • $p(0)=(0{)}^3=0$

  • $p(1)=(1{)}^3=1$

  • $p(2) = (2)^3 = 8$
    

• (iv) $p(x)=(x-1)(x+1)$

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

• $p(1) = (1 - 1)(1 + 1) = (0)(2) = 0$
  

• $p(2) = (2 - 1)(2 + 1) = (1)(3) = 3$
  

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