NCERT Solutions – Polynomials Exercise 2.1 – 2.4

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.

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

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.

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.

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.

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.

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

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

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

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.

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–√.

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.

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.