Ex 2.4

                                                  Maths NCERT Book Solutions

                                                                  Chapter 2

                                                                   Ex.2.4

Q: 1

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

Answer

(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. Q: 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   Answer (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. Q: 3 Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:          Answer 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.   Q: 4   Answer (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 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 Q: 5 Factorise: (i)   x3 - 2x2 - x + 2                         (ii)  x3 - 3x2 - 9x - 5 (iii) x3 + 13x2 + 32x + 20                (iv) 2y3 + y2 - 2y - 1   Answer   (i)    Let p(x) = x3 - 2x2 - x + 2        Factors of 2 are Â± 1, Â± 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, Â± 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, Â± 2, Â± 4, Â± 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.