Polynomials - Class 9 NCERT Solutions

1. NCERT Solutions for Class 9 MATHS – Polynomials 1. Write the degree of each of the following polynomials (i) 5 4 7 x x x + + + (ii) The degree of 4 y − − t − 2 3 2 4 (iii) 5 7 y (ii) + is 2. (iv) 3 3 2 Sol. (i) The degree of is 3. 5 4 7 x x x 2 − = is 0. − 1 (iii) The degree of (iv) The degree of Classify the following as linear, quadratic and cubic polynomials: (i) x + (ii) x − 5 3 3x = 7 5 7 is 1. t t 0 2. + + 2 2x 3 + 4 y y (iii) (iv) 1 x x 2 3 r (v) 3t (vi) (vii) 7x Sol. + Quadratic polynomial. 3 − Cubic polynomial. 4 y y + + Quadratic polynomial. x + Linear polynomial. 2x x (i) (ii) (iii) (iv) 1 (v) 3t Linear polynomial. (vi) r Quadratic polynomial. (vii) 7x Cubic polynomial. x x 2 2 3 3. − + at 2 Find the value of the polynomial (i) 0 x = Sol.(i) Putting x = (0) 5 0 4(0) p =  − (ii) Putting x = − , we get ( 1) 5 ( 1) 4( 1) p − =  − − (iii) Putting x = (2) 5 2 4(2) p =  − 5 4 3 x 1 x x = − x = (ii) (iii) 2 , we get 3 3 + = 1 − 2 , we get 3 10 16 3 + = 0 2 + =− − + =− 2 3 5 4 3 6 − + =− 2 3

2. 4. + + 1 2 + is divided by 3 2 Find the remainder when 3 3 1 x x x (iv) x  + x − x + (i) (ii) (iii) x 1 + ( ) p x 1 x + (v) 5 2x = + + + 3 2 Sol. Let 3 3 1 x = − x x x (i) Putting Using remainder theorem, when = − + − 1 3 3 1 = − + − + = 0 1 2 1 2 we get x + = 1 0, 1 p − = + + + is divided by x + 3 2 remainder is given by ( 1) ( ) p x 3 3 1 x x x 1, + 3( 1) 1 − + 3 2 ( 1) 3( 1) x − (ii) 1 2 x − = x = Putting we get, 0, p      1, 2 1 2 = + + + divided by x− 3 2 Using remainder theorem, when remainder is given by ( ) p x 3 3 1 x x x 3 2       1 8 1 6 12 8 8 27 8 (iii) x Putting Using remainder theorem, when = + + 0 1 = + 0 = (iv) x  + Putting 0, x  + = Using remainder theorem, when   = − + −    =− + −             1 2 1 2 1 2 = + + + 3 3 1 1 4 1 2 = +  +  + 3 3 1 + + + = = x = we get 0, = + + + is divided by , x remainder is given by (0) 3 2 ( ) p x 3 3 1 p x x x + 3 3 (0) 3(0) 3(0) 1 we get, x =−  p  − = + + + is divided by x  + 3 2 remainder is given by ( ) ( ) p x 3 3 1 , x x x + + ) 1  − + 3 2 ( ) 3( ) 3( 3 2 3 3 1

3. (v) 5 + 2x 5 2 x = − + = Putting 5 we get 2 0, x p    5 2 = + + + is divided by 5 − + 3 2 Using remainder theorem, when remainder is given by ( ) p x 3 3 1 x x x 2 , x   3 2                   5 2 5 2 5 2 = − + − + − + 3 3 1 125 8 125 150 60 8 8 27 8 25 4 − 5 2 = − +  −  + 3 3 1 − + + = = − 5. Sol. Let − . − + x a − 3 2 Find the remainder when ( ) p x x = Putting x a − Using remainder theorem, when = − + = − + − 5a = is divided by x a 6 x ax − = + x a − 3 2 6 ax 0, = we get x a = − + x a − is divided by x a − 3 2 remainder is given by ( ) p a ( ) p x 6 , x ax − 3 2 ( ) a ( ) a a 6( ) a a 3 3 6 a a a a 6. x − is a factor of (x) Find the value of k, if = in each of the following cases: p 1 + + 2 (i) ( ) p x x x k = + + 2 ( ) p x 2 2 x kx (ii) = − + 2 ( ) p x 2 1 kx x (iii) = − + 2 (iv) ( ) p x x− 0 = 3 kx x k = + + 2 Sol. (i) If ( 1) ( ) p x , is a factor of then x x k (1) p   + + = = 2 (1) 1 1 + + 1 k 0 k  =− 2 k 0

4. (ii) If ( (1) p = + + 2 x− 0 p( ) 2 2, x x kx is a factor of then 1) =  + + =  + + = 2 2(1) (1) 2 0 2 2 0 k k  = − + (2 2) k = − + then 2 x− 0 = (iii) If ( (1) p ( ) p x 2 1, kx x is a factor of 1) − 2(1) 1 0 + =  − 2 1 0 + = 2  (1) k k  (iv) If ( (1) p  k = 2 1 − 1) x− 0 = − = − + 2 is a factor of then ( ) p x 3 , kx x k  − + = + = 2 3 0 (1) 3(1) 0 k k k k 3 2 =  = 2 3 k k 7. Factorise the following using appropriate identities: 2 y x − + + − + 2 2 2 2 (i) (ii) (iii) 9 6 4 4 1 x xy y y y 100 + + = + + − − + 2 2 2 2 Sol. (i) 9 6 (3 ) 2(3 )( ) ( ) x y x xy y x y = + = + 2 (3 ) (3 + = )(3 ) x y − x y x y + 2 2 2 (ii) = 4 4 1 (2 ) − 2(2 )(1) (1) y y y = y − 2 (2 1) (2 1)(2 1) y y y 2 2      y y − = − 2 2 (iii) ( ) x x 100    10       y y = − + x x 10 10 Factorise each of the following: (i) 27 125 y z + + = + = + − 8. − 3 3 3 3 64 343 = (ii) m n + 3 3 3 3 Sol. (i) 27 125 (3 ) − (5 ) y z y z + 2 2 (3 5 )[(3 ) z (3y)(5z) (5z) ] + − y y − 2 2 (3 5 )(9y z 15 = 25 ) z y yz 3 3 3 3 (ii) 64 343 (4 ) m (7 ) m n n

5. = = Factorise: (i) 27 − − + + 2 2 (4 7 )[(4 ) n (4 )(7 ) (7 ) ] m n + m m n + 2 2 (4 7 )[16m n 28 49 ] m mn n 9. + + − 3 3 3 9 x + + + y + + + z − xyz = + + + − − + − 3 3 3 3 3 3 Sol. 27 = = If 9 (3 ) + + 3(3x)(y)(z) − − x y z xyz x y z − 2 2 2 (3 )[(3 ) (3 ) (3 )] z x x y z x y z x y − yz 2 2 2 (3 ) (9 3 3 ) zx x y z x y z xy yz + + = + + = 3 3 3 10. Sol. We know that, 3 x + = = 0 x  show that 0, 3 , x y z x y z xyz + + − + = + + + + + = − − − 3 3 2 2 2 3 ( − )( ) y z xyz − x y − z x y + z xy yz zx 2 2 2 0( ) ( 0 given) x y z xy yz zx x y z + + = 3 3 3 Hence proved. 3 y z xyz For complete NCERT Solutions visit www.ntseguru.in & take a free demo. Or Download NTSE GURU Android App for free from Google Playstore.