NCERT Class 9 Maths Solutions Chapter - 2 Polynomials, Ex 2.4

Ex - 2.4

Question 1.  Determine which of the following polynomials has (x + 1) as a factor:

Solution

(i)    If (x + 1) is a factor of p(x) = x³ + x² + x + 1, p (- 1) must be zero.

        Here, p(x) = x³ + x² + x + 1  
               p(-1) = (- 1)³ + (- 1)² + (- 1) + 1  
                       = - 1 + 1 - 1 + 1 = 0
        Hence, x + 1 is a factor of this polynomial

(ii)    If (x + 1) is a factor of p(x) = x⁴ + x³ + x² + x + 1, p (- 1) must be zero.
      
        Here, p(x) = x⁴ + x³ + x² + x + 1  
              p( -1) = (- 1)⁴ + (- 1)³ + (- 1)² + (- 1) + 1
                       = 1 - 1 + 1 -1 + 1 = 1
        As, 

        So, x + 1 is not a factor of this polynomial

(iii)    If (x + 1) is a factor of polynomial p(x) = x⁴ + 3x³ + 3x² + x + 1, p(- 1) must be 0.

         p(- 1) = (- 1)⁴ + 3(- 1)³ + 3(- 1)² + (- 1) + 1
                  = 1 - 3 + 3 - 1 + 1 = 1
         As, 

         So, x + 1 is not a factor of this polynomial

(iv)    If (x + 1) is a factor of polynomial
         p(x) =  ,  p(- 1) must be 0.

                

   As, 

         So, (x + 1) is not a factor of this polynomial.

Concept Insight: A linear polynomial 'x-a' is a factor of the polynomial p(x) iff p(a) = 0. Note that 'a' is a zero of polynomials x-a and  p(x) . Be careful while squaring and cubing the numbers.

Question 2.  Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

    (i)    p(x) = 2x³ + x² - 2x - 1, g(x) = x + 1
    (ii)   p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
    (iii)    p(x) = x³ - 4 x² + x + 6, g(x) = x - 3

Solution

(i)    If g(x) = x + 1 is a factor of given polynomial p(x), p(- 1)  must be zero.
        p(x) = 2x³ + x² - 2x - 1
        p(- 1) = 2(- 1)³ + (- 1)² - 2(- 1) - 1
                 = 2(- 1) + 1 + 2 - 1 = 0
        Hence, g(x) = x + 1 is a factor of given polynomial.

(ii)    If g(x) = x + 2 is a factor of given polynomial p(x), p(- 2) must be 0.
        p(x) = x³ +3x² + 3x + 1
        p(- 2) = (- 2)³ + 3(- 2)² + 3(- 2) + 1
                 = - 8 + 12 - 6 + 1
                 = - 1
       

        Hence g(x) = x + 2 is not a factor of given polynomial.

(iii)    If g(x) = x - 3 is a factor of given polynomial p(x), p(3) must be 0.
            p(x) = x³ - 4 x² + x + 6
            p(3) = (3)³ - 4(3)² + 3 + 6
                   = 27 - 36 + 9 = 0
        So, g(x) = x - 3 is a factor of given polynomial.

Concept Insight: The problem is a direct application of Factor theorem. g(x) will be the factor of the polynomial p(x) iff the zero of the linear polynomial g(x) when put in place of the variable of polynomial results to zero. Be careful while squaring and cubing the numbers.

Question 3.  Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:

Solution

If x - 1 is a factor of polynomial p(x), then p(1) = 0

(i)    p(x) = x² + x + k
       p(1) = 0
        (1)² + 1 + k = 0
        2 + k = 0
        k = - 2
        So, value of k is - 2.

Concept Insight: x-1 is a factor of the given polynomial p(x) iff p(1) = 0 thus equating p(1) to zero will give the required value of constant k. Be careful with arithmetic simplifications.

Question 4.  

Solution

(i)    12x² - 7x + 1  
       The two numbers such that pq = 12  1 = 12 and p + q = - 7. They are p = - 4 and

       q = - 3
       Now, 12x² - 7x + 1 = 12x² - 4x - 3x + 1  
                                   = 4x (3x - 1) - 1 (3x - 1)
                                   = (3x - 1) (4x - 1)

(ii)    2x² + 7x + 3
        The two numbers such that pq = 2  3 = 6 and p + q = 7.
        They are p = 6 and q = 1
        Now, 2x² + 7x + 3 = 2x² + 6x + x + 3  
                             = 2x (x + 3) + 1 (x + 3)
                             = (x + 3) (2x+ 1)

(iii)    6x² + 5x - 6  
        The two numbers such that pq = - 36 and p + q = 5.
        They are p = 9 and q = - 4
        Now,
        6x² + 5x - 6 = 6x² + 9x - 4x - 6  
                    = 3x (2x + 3) - 2 (2x + 3)
                    = (2x + 3) (3x - 2)

(iv)    3x² - x - 4  
         The two numbers such that pq = 3  (- 4) = - 12

         and p + q = - 1.
         They are p = - 4 and q = 3.
         Now,
         3x² - x - 4 = 3x² - 4x + 3x - 4  
                  = x (3x - 4) + 1 (3x - 4)
                  = (3x - 4) (x + 1)

Concept Insight: To factorise the polynomial ax²+bx+c, by splitting the middle term, 
b is expressed as the sum of two numbers whose product is ac.
Do not forget to consider the sign of the terms while splitting.
Remember

ac>0	b>0	b =(p+q) where p>0,q>0
ac>0	b<0	b =(p+q) where p<0,q<0
ac<0	b>0	b =(p+q) where p > q then p>0 and q<0
ac<0	b<0	b =(p+q) where p > q then p<0 and q>0

Question 5.  Factorise:

(i)   x³ - 2x² - x + 2                         (ii)  x³ - 3x² - 9x - 5

(iii) x³ + 13x² + 32x + 20                (iv) 2y³ + y² - 2y - 1

Solution

(i)    Let p(x) = x³ - 2x² - x + 2
       Factors of 2 are � 1, � 2.
       By hit and trial method
       p(2) = (2)³ - 2(2)² - 2 + 2
           = 8 - 8 - 2 + 2 = 0
       So, (x - 2) is factor of polynomial p(x)
   
       By long division

       

       Now,     Dividend = Divisor  Quotient + Remainder
     x³ - 2x² - x + 2 = (x + 1) (x² - 3x + 2) + 0
                  = (x + 1) [x² - 2x - x + 2]
                  = (x + 1) [x (x - 2) - 1 (x - 2)]
                  = (x + 1) (x - 1) (x - 2)
                  = (x - 2) (x - 1) (x + 1)

(ii)    Let p(x) = x³ - 3x² - 9x - 5
        Factors of 5 are �1, � 5.
        By hit and trial method
        p(- 1) = (- 1)³ - 3(- 1)² - 9(- 1) - 5
           = - 1 - 3 + 9 - 5 = 0
        So x + 1 is a factor of this polynomial
        Let us find the quotient while dividing x³ + 3x² - 9x - 5 by x + 1
        By long division

       

        Now, Dividend = Divisor  Quotient + Remainder
         x³ - 3x² - 9x - 5 = (x + 1) (x² - 4 x - 5) + 0
                                     = (x + 1) (x² - 5 x + x - 5)
                                     = (x + 1) [(x (x - 5) +1 (x - 5)]
                                     = (x + 1) (x - 5) (x + 1)
                                     = (x - 5) (x + 1) (x + 1)

(iii)    Let p(x) = x³ + 13x² + 32x + 20
         The factors of 20 are �1, � 2, � 4, � 5 ... ...
         By hit and trial method
         p(- 1) = (- 1)³ + 13(- 1)² + 32(- 1) + 20
                   = - 1 + 13 - 32 + 20
                   = 33 - 33 = 0
         As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).

         Let us find the quotient while dividing x³ + 13x² + 32x + 20 by (x + 1)
          By long division

         

         We know that
         Dividend = Divisor  Quotient + Remainder
         x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20) + 0
                                         = (x + 1) (x² + 10x + 2x + 20)
                                         = (x + 1) [x (x + 10) + 2 (x + 10)]
                                         = (x + 1) (x + 10) (x + 2)
                                         = (x + 1) (x + 2) (x + 10)

(iv)    Let p(y) = 2y³ + y² - 2y - 1
         By hit and trial method
         p(1) = 2 ( 1)³ + (1)² - 2( 1) - 1
                = 2 + 1 - 2 - 1= 0
         So, y - 1 is a factor of this polynomial
         By long division method,

         

          p(y) = 2y³ + y² - 2y - 1
                 = (y - 1) (2y² +3y + 1)
                 = (y - 1) (2y² +2y + y +1)
                 = (y - 1) [2y (y + 1) + 1 (y + 1)]
                 = (y - 1) (y + 1) (2y + 1)

Concept Insight: To factorise p(x) when its degree is greater than or equal to 3 note down all the factors of constant term considering both negative and positive sign.
Check the obtained factors for the possible zeroes of the polynomial p(x)  Using Factor theorem one zero can be obtained continue the process till all the zeroes are obtained or use long division method. To obtain the other quadratic factor use long division to determine the other factors. The degree of the polynomial is  less than or  equal to the number of real factors the polynomial.

Ex-2.5