x maths Polynomials

 

Exercise - 2.1 (Mathematics NCERT Class 10th)

Q.1     The graphs of y = p(x) are given in figures below for some polynomials p(x). Find the number of zeroes of p(x) , in each case.
            (i)   
          (ii)  
         (iii)
         (iv)
        (v) 
         (vi)   
Sol.
        (i) There are no zeroes as the graph does not intersect the x-axis.
        (ii) The number of zeroes is one as the graph intersects the x-axis at one point only.
        (iii) The number of zeroes is three as the graph intersects the x-axis at three points.
        (iv) The number of zeroes is two as the graph intersects the x-axis at two points.
        (v) The number of zeroes is four as the graph intersects the x-axis at four points.
        (vi) The number of zeroes is three as the graph intersects the x-axis at three points.

Polynomials : Exercise - 2.2 (Mathematics NCERT Class 10th)

Q.1       Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
             (i) x2−2x−8                      
             (ii) 4s2−4s+1
             (iii) 6x2−3−7x                                         
             (iv) 4u2+8u
             (v) t2−15                                                    
             (vi) 3x2−x−4
Sol.       (i) We have, x2−2x−8 =x2+2x−4x−8
                                                   =x(x+2)−4(x+2)
                                                   =(x+2)(x−4)
                  The value of x2−2x−8 is zero when the value of (x + 2) (x – 4) is zero, i.e.,
                  when x + 2 = 0 or x – 4 = 0 , i.e., when x = – 2 or x = 4.
                   So, The zeroes of x2−2x−8 are – 2 and 4.
                  Therefore , sum of the zeroes = (– 2) + 4 = 2
                                                             =−CoefficientofxCoefficientofx2
                  and product of zeroes = (– 2) (4) = – 8 =−81
                                                  =ConstanttermCoefficientofx2

            (ii) We have, 4s2−4s+1 =4s2−2s−2s+1
                                                    =2s(2s−1)−1(2s−1)
                                                    =(2s−1)(2s−1)
                 The value of 4s2−4s+1 is zero when the value of  
                 (2s – 1) (2s – 1) is zero, i.e., when 2s – 1 = 0 or 2s – 1 = 0,
                 i.e., when s=12ors=12.
                 So, The zeroes of 4s2−4s+1are12and12
                 Therefore, sum of the zeroes =12+12=1
                                                           =−CoefficientofsCoefficientofs2
                 and product of zeroes =(12)(12)=14
                                                 =ConstanttermCoefficientofs2
           (iii) We have, 6x2−3−7x = 6x2−7x−3
                                                   =6x2−9x+2x−3
                                                   =3x(2x−3)+1(2x−3)
                                                   =(3x+1)(2x−3)
                 The value of 6x2−3−7x is zero when the value of (3x + 1) (2x – 3) is zero, i.e., when 3x + 1 = 0 or 2x – 3 = 0, i.e, when x=−13orx=32
                So, The zeroes of 6x2−3−7xare−13and32
                Therefore, sum of the zeroes =−13+32=76
                                                           =−CoefficientofxCoefficientofx2
                 and product of zeroes =(−13)(32)=−12
                                                 =ConstanttermCoefficientofx2
           (iv)  We have, 4u2+8u = 4u (u + 2) 
                  The value of 4u2+8u is zero when the value of 4u(u + 2) is zero, i.e., when u = 0 or u + 2 = 0, i.e., when u = 0 or u = – 2.
                  So, The zeroes of 4u2+8u and 0 and – 2 
                  Therefore, sum of the zeroes = 0 + (– 2) = – 2 
                                                           =CoefficentofuCoefficientofu2
                  and , product of zeroes = (0) (–2) = 0 
                                                    =ConstanttermCoefficientofu2
            (v) We have t2−15 =(t−15−−√)(t+15−−√)
                 The value of t2−15 is zero when the value of (t−15−−√)(t+15−−√) is zero,
                 i.e., when t−15−−√=0ort+15−−√ = 0 i.e., when t=15−−√ort=−15−−√
                 So, The zeroes of t2−15are15−−√and−15−−√
                 Therefore , sum of the zeroes = 15−−√+(−15−−√)=0
                 =−CoefficientoftCoefficientoft2
                 and, product of the zeroes = (15−−√)(−15−−√)=−15
                                                    =ConstanttermCoefficientoft2  
           (vi) We have, 3x2−x−4 =  3x2+3x−4x−4
                                                  =3x(x+1)−4(x+1)
                                                  =(x+1)(3x−4)
                 The value of 3x2−x−4 is zero when the value of (x + 1) (3x – 4) is zero, i.e., when x + 1 = 0 or 3x – 4 = 0, i.e., when x = – 1 or  x=43.
                 So, The zeroes of 3x2−x−4are−1and43
                 Therefore , sum of the zeroes =−1+43=−3+43
                                                              =13=CoefficientofxCoefficientofx2
                 and, product of the zeroes =(−1)(43)=−43
                                                        =ConstanttermCoefficientofx2

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Q.2       Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
            (i) 14,−1                 
            (ii) 2–√,13                      
            (iii) 0,5–√
            (iv) 1, 1
            (v) −14,14                                                  
            (vi) 4, 1
Sol.      (i) Let the polynomial be ax2+bx+c, and its zeroes be αandβ. Then , 
                                                    α+β=14=−ba
                  and,                            αβ=−1=−44=ca
                 If a = 4,  then  b = – 1 and c = – 4.
                 Therefore, one quadratic polynomial which fits the given conditions is 4x2−x−4.

          (ii) Let the polynomial be ax2+bx+c, and its zeroes be αandβ. Then,
                                                 α+β=2–√=32√3=−ba
                and                            αβ=13=ca
                If a = 3, then b =32–√andc=1
                So, One quadratic polynomial which fits the given conditions is 3x2−32–√x+1.

           (iii) Let the polynomial  be ax2+bx+c, and its zeroes be αandβ. Then, 

                                                  α+β=0=01=−ba
                 and                           αβ=5–√=5√1=ca
                 If  a = 1, then b = 0 and c = 5–√
                 So, one quadratic polynomial which fits the given conditions is x2−0.x+5–√,i.e.,x2+5–√.

          (iv) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then,
                                                 α+β=1
                                                 =−(−1)1=−ba
and                               αβ=1 =11=ca
                If  a = 1, then b = – 1 and c = 1.
                So, one quadratic polynomial which fits the given conditions is x2−x+1.

            (v) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then
                                                   α+β=−14=−ba 
                  and                               αβ=14=ca
                  If  a = 4 then b = – 1 and c = 1. 
                  So, one quadratic polynomial which fits the given conditions is 4x2−x+1.

            (vi) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then, 
                                                   α+β=4=−ba
                                                   αβ=1=11=ca
                  If a = 1, then b = – 4 and c = 1
                  Therefore, one quadratic polynomial which fits the given conditions is x2−4x+1.

Polynomials : Exercise - 2.3 (Mathematics NCERT Class 10th)

Q.1       Divide the polynomial p(x) by the polynomial g (x) and find the quotient and remainder in each of the following : 
            (i) p(x)=x3−3x2+5x−3,g(x)=x2−2
            (ii) p(x)=x4−3x2+4x+5,g(x)=x2+1−x
            (iii) p(x)=x4−5x+6,g(x)=2−x2
Sol.      (i) We have, 

        Therefore, the quotient is x – 3 and the remainder is 7 x – 9

        (ii) Here, the dividend is already in the standard form and the divisor is also in the standard form.
         We have,

         Therefore, the quotient is x2+x−3 and the remainder is 8.

        (iii) To carry out the division, we first write divisor in the standard form.
         So, divisor = −x2+2
         We have,

         Therefore, the quotient is −x2+2
         and the remainder is – 5x + 10.

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Q.2         Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial : 
              (i) t2−3,2t4+3t3−2t2−9t−12
              (ii) x2+3x+1,3x4+5x3−7x2+2x+2
              (iii) x3−3x+1,x5−4x3+x2+3x+1
Sol.      (i) Let us divide 2t4+3t3−2t2−9t−12byt2−3.
            We have, 

        Since the remainder is zero, therefore, t2−3 is a factor of 2t4+3t3−2t2−9t−12.

        (ii) Let us divide 3x4+5x3−7x2+2x+2byx2+3x+1.
        We get,

        Since the remainder is zero, therefore x2+3x+1 is a factor of
                                                               3x4+5x3−7x2+2x+2

        (iii) Let us divide x5−4x3+x2+3x+1 by x3−3x+1
         We get,

         Here remainder is 2(≠ 0). Therefore,  x3−3x+1 is not a factor of
                                                                            x5−4x3+x2+3x+1.

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Q.3      Obtain all the zeroes of 3x4+6x3−2x2−10x−5 if two of its zeroes are 53−−√and−53−−√.
Sol.      Since two zeroes are 53−−√and−53−−√, so (x−53−−√)and(x+53−−√) are the factors of the given polynomial.
           Now, (x−53−−√)(x+53−−√)=x2−53 
           ⇒ (3x2−5) is a factor of the given polynomial. 
           Applying the division algorithm to the given polynomial and 3x2−5, we have

         Therefore, 3x4+6x3−2x2−10x−5 = (3x2−5)(x2+2x+1)
          Now, x2+2x+1 =x2+x+x+1
                                    =x(x+1)+1(x+1)
                                   =(x+1)(x+1)
          So, its other zeroes are – 1 and – 1.
          Thus,  all the zeroes of the given fourth degree polynomial are
                   53−−√,−53−−√,−1and−1. 

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Q.4       On dividing x3−3x2+x+2 by a polynomial g(x) , the quotient and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Sol.         Since on dividing x3−3x2+x+2 by a polynomial (x – 2) × g(x), the quotient and remainder were (x – 2) and (– 2x + 4) respectively, therefore,
              Therefore, Quotient × Divisor + Remainder = Dividend
          ⇒ (x−2)×g(x)+(−2x+4)  =x3−3x2+x−2
         ⇒ (x – 2) × g(x) =x3−3x2+x−2+2x−4
         ⇒ g(x)=x3−3x2+3x−2x−2                  ... (1)
         Let us divide x3−3x2+3x−2byx−2. We get

           Therefore, equation (1) gives g (x) =x2−x+1

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Q.5     Give examples of polynomials p(x), g(x) , q (x) and r (x), which satisfy the division algorithm and 
           (i) deg p(x) = deg q(x)           
           (ii) deg q(x) = deg r (x)                          
           (iii) deg r (x) = 0
Sol.     There can be several examples for each of (i), (ii) and (iii). 
           However, one example for each case may be taken as under : 
        (i) p(x)=2x2−2x+14,g(x)=2,
             q(x)=x2−x+7,r(x)=0
       (ii)  p(x)=x3+x2+x+1,g(x)=x2−1,
             q(x)=x+1,r(x)=2x+2
       (iii) p(x)=x3+2x2−x+2,
             g(x)=x2−1,q(x)=x+2,r(x)=4

Polynomials : Exercise - 2.4 Optional (Mathematics NCERT Class 10th)

Q.1        Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
              (i) 2x3+x2−5x+2;12,1,−2 
              (ii) x3−4x2+5x−2;2,1,1
Sol.      (i) Comparing the given polynomial with ax3+bx2+cx+d, we get
            a = 2 , b = 1, c = – 5 and d = 2. 
                   p(12)=2(12)3+(12)2−5(12)+2
                    =14+14−52+2
                    =1+1−10+84=04=0
                    p(1)=2(1)3+(1)2−5(1)+2
                    =2+1−5+2=0
                    p(−2)=2(−2)3+(−2)2−5(−2)+2
                    =2(−8)+4+10+2
                    =−16+16=0
            Therefore, 12, 1 and – 2 are the zeroes of 2x3+x2−5x+2.
            So, α=12,β=1andγ=−2.
            Therefore, α+β+γ=12+1+(−2)=1+2−42
                     =−12=−ba
                     αβ+βγ+γα=(12)(1)+(1)(−2)+(−2)(12)
                     =12−2−1
                     =1−4−22=−52=ca
           and     αβγ=12×1×−2=1=−22=−dα

           (ii) Comparing the given polynomial with ax3+bx2+cx+d, we get
                                           a = 1, b = – 4, c = 5 and d = – 2. 
                     p(2)=(2)3−4(2)2+5(2)−2=8−16+10−2=0
                     p(1)=(1)3−4(1)2+5(1)−2=1−4+5−2=0
              Therefore , 2 , 1 and 1 are the zeros of  x3−4x2+5x−2
           Thus,     α=2,β=1andγ=1.
            Now α+β+γ=2+1+1=4=−(−4)1=−ba
                           αβ+βγ+γα=(2)(1)+(1)(1)+(1)(2)
                           =2+1+2=5=51=ca
           and           αβγ=(2)(1)(1)=2=−(−2)1=−da

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Q.2       Find a cubic polynomial with the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively. 
Sol.     Let zeroes be α,βandγ
We have given α+β+γ=2
αβ+βγ+γα=−7
and αβγ=−14
Cubic polynomial is given by (x−α)(x−β)(x−γ)=0
⇒x3−x2(α+β+γ)+x(αβ+βγ+γα)−αβγ=0
⇒x3−2x2−7x+14=0
αβγ=−14=−141=−da
             If a = 1, then b = – 2, c = – 7 and d = 14
             So, one cubic polynomial which fits the given conditions will be x3−2x2−7x+14.

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Q.3       If the zeroes of the polynomial x3−3x2+x+1 are a – b, a, a + b , find a and b
Sol.       Since (a – b) , a (a + b) are the zeroes of the polynomial x3−3x2+x+1,
             Therefore, (a−b)+a+(a+b)=−(−3)1=3 
              ⇒ 3a = 3 ⇒ a = 1 
               (a – b) a + a(a + b) + (a + b) (a – b) =11=1
              ⇒ a2−ab+a2+ab+a2−b2=1
              ⇒ 3a2−b2=1
              ⇒ 3(1)2−b2=1 [Since a = 1]
              ⇒ 3−b2=1
              ⇒ b2=2
              ⇒ b=±2–√
            Hence , a = 1 and b=±2–√.

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Q.4       If two zeroes of the polynomial x4−6x3−26x2+138x−35are2±3–√ , find other zeroes. 
Sol.       Since 2±3–√, are two zeroes of the polynomial p(x)=x4−6x3−26x2+138x−35
             Let x=2±3–√ ⇒ x−2=±3–√
             Squaring we get x2−4x+4=3
              ⇒ x2−4x+1=0
             Let us divide p(x) by x2−4x+1 to obtain other zeroes.

          Therefore, p(x)=x4−6x3−26x2+138x−35
                                 =(x2−4x+1)(x2−2x−35)
                                 =(x2−4x+1)(x2−7x+5x−35)
                                 =(x2−4x+1)[x(x−7)+5(x−7)]
                                 =(x2−4x+1)(x+5)(x−7)
          ⇒ (x + 5) and (x – 7) are other factors of p(x).
          Therefore,  – 5 and 7 are other zeroes of the given polynomial.

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Q.5       If the polynomial x4−6x3+16x2−25x+10 is divided by another polynomial x2−2x+k, the remainder comes out to be x + a find k and a. 
Sol.       Let us divide x4−6x3+16x2−25x+10 by x2−2x+k

            So, remainder = (2 k – 9) x – (8 – k) k + 10 
            But the remainder is given as x + a. 
            On comparing their coefficients , we have  2 k – 9 = 1 
            ⇒ 2 k = 10 
            ⇒ k = 5 
            and                             – (8 – k)k + 10 = a 
            ⇒ a = – (8 – 5)5 + 10  = – 3 × 5 + 10 = – 15 + 10 = – 5
            Hence, k = 5 and a = – 5.