Exercise 2.2

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+1Sol.      (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) = 0Sol.     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

5

√√E