NCERT Class 9 Maths Solutions Chapter - 2 Polynomials, Ex 2.2

Ex - 2.2

Question 1.  Find the value of the polynomial 5x - 4x² +3  at 
(i) x = 0    (ii) x = - 1    (iii) x = 2

Solution

Concept Insight: Given Polynomial is p(x) to find the  value of given polynomial at any particular value of x replace the variable x with its corresponding value. Remember for odd power of negative number the negative sign remains while for even power of negative numbers the negative sign vanishes.  Also check for  calculation errors; there are chances of making calculation mistake while computing square, cubes and higher powers of numbers.

Question 2.  Find p(0), p(1) and p(2) for each of the following polynomials:

(i)     p(y) = y² - y + 1    (ii)     p(t) = 2 + t + 2t² - t³    
(iii)    p(x) = x³            (iv)     p(x) = (x - 1) (x + 1)

Solution

(i)       p(y) = y² - 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 + 2t² - t³
          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) = x³   
          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

Concept Insight: Replace the variable with 0, 1 or  2 in the given polynomials to obtain the required value.  Be careful about the calculations, there are chances of making calculation mistake while computing square, cubes and higher powers of numbers. Carefully apply the properties of addition, subtraction and multiplication of numbers. While multiplying two binomials multiply each term of the binomial to each term of the other binomial.

Question 3.  Verify whether the following are zeroes of the polynomial, indicated against them.

Solution

(i)    If  is a zero of given polynomial p(x) = 3x + 1, then

        

        So,  is a zero of given polynomial

(ii)    If  is a zero of polynomial p(x) = 5x -  ,                                          

        then  should be 0

        

         So,  is not a zero of given polynomial

(iii)    If x = 1 and x = - 1 are zeroes of polynomial p(x) = x² - 1 then p(1) and p(- 1) 

         should be 0
         Now, p(1) = (1)² - 1 = 0
         p(- 1) = (- 1)² - 1 = 0
         Hence x = 1 and - 1 are zeroes of polynomial.

(iv)    If x = - 1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x - 2), then p(- 1) and

         p(2)should be 0.
         Now, p(- 1) = (- 1 + 1) (- 1 - 2) = 0 (-3) = 0
         p(2) = (2 + 1) (2 -  2 ) = 3 (0) = 0
         So, x = - 1 and x = 2 are zeroes of given polynomial.

(v)     If x = 0 is a zero of polynomial p(x) = x² then p(0) should be zero.
         Now, p(0) = (0)² = 0
         Hence x = 0 is a zero of given polynomial

(vi)    If  is a zero of polynomial p(x) = lx + m, thenis 0.

(vii)    If  and  are zeroes of polynomial p(x) = 3x² - 1, then

      Hence,  is a zero of given polynomial but  is not a zero of given

            polynomial.

(viii)      If    is a zero of polynomial p(x) = 2x + 1 then  should be 0.

  So,  is not a zero of polynomial.

Concept Insight: Key idea here is  Zero of the polynomial is not the real number zero but it is that value of the variable which makes the  value of the polynomial equal to zero. A polynomial can have more than one zeroes.

Question 4.  Find the zero of the polynomial in each of the following cases:

Solution

Zero of a polynomial is that value of variable at which value of polynomial comes to 0.
(i)     p(x) = x + 5
        p(x) = 0
        x + 5 = 0
        x = - 5
        So, for x = - 5, value of polynomial is 0 and hence x = - 5 is a zero of polynomial.

(ii)    p(x) = x - 5
        p(x) = 0
        x - 5 = 0
        x = 5
        So, for x = 5 value of polynomial is 0 and hence x = 5 is a zero of polynomial.

(iii)   p(x) = 2x + 5
        p(x) = 0
        2x + 5 = 0
        2x = - 5
         
        So, for   value of polynomial is 0 and hence   is a zero of polynomial.

(iv)   p(x) = 3x - 2
        p(x) = 0
        3x - 2 = 0
         
        So, for  , value of polynomial is 0 and hence   is a zero of polynomial.

(v)    p(x) = 3x
        p(x) = 0
        3x = 0
        x = 0
        So, for x = 0 value of polynomial is 0 and hence x = 0 is a zero of polynomial.

(vi)   p(x) = ax
        p(x) = 0
        ax = 0
        x = 0
        So, for x = 0, value of polynomial is 0. Hence x = 0 is a zero of polynomial.

(vii)   p(x) = cx + d
        p(x) = 0
        cx+ d = 0
        

        So, for , value of polynomial is 0. Hence  is a zero of polynomial.

Concept Insight: Equate the polynomial to zero and solve the corresponding linear equation to get the value of variable. Be careful while transposing the terms to the other side. For verification substitute the value of the variable obtained in the polynomial.

Ex-2.3