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NCERT Solutions for Class 9 Math Chapter 2.2 POLYNOMIALS ALL SOLUTION

CLASS :- 9, EX :- 2.2



1.    Find the value of the polynomial 5x-4x2+3 at

(i)    x=0

(ii)    x=-1

(iii)    x=2

solution :

(i)    x=0

    Let,

    p(x)=5x-4x2+3

    The value of the polynomial p(x) at x=0 is given

    p(0)=5×(0)-4×(0)2+3

    =0-0+3

    =3

(ii)    x=-1

    Let,

    p(x)=5x-4x2+3

    The value of the polynomial p(x) at x=-1 is given

    p(-1)=5×(-1)-4×(-1)2+3

    =(-5)-4+3

    =-5-4+3

    =-9+3

    =-6

(iii)    x=2

    Let,

    p(x)=5x-4x2+3

    The value of the polynomial p(x) at x=2 is given

    p(2)=5×(2)-4×(2)2+3

    =10-(4×4)+3

    =10-16+3

    =-6+3

    =-3

2.    Find p(0), p(1) and p(2) for each of the following polynomials:

(i)    p(y)=y2-y=1

(ii)    p(t)=2+t+2t2-t3

(iii)    p(x)=x3

(iv)    p(x)=(x-1)(x+1)

solution :

(i)    p(y)=y2-y=1

    p(y)=y2-y=1

    When,  p(0)

    p(0)=02-0+1

    =0-0+1

    =1

    When,  p(1)

    p(1)=12-1+1

    = 1 − 1 + 1

    =0+1

    =1

    When,  p(2)

    p(2)=22-2+1

    =4-2+1

    2+1

    =3

(ii)    p(t)=2+t+2t2−t3

    p(t)=2+t+2t2−t3

    When,  p(0)

    p(0)=2+0+2×(0)2−03

    =2+0−0

    =2

    When,  p(1)

    p(1)=2+1+2×(1)2−13

    =3+2-1

    =5-1

    =4

    When,  p(2)

    p(2)=2+2+2×(2)2-23

    =4+8-8

    =4+0

    =4

(iii)    p(x)=x3

    p(x)=x3

    When,  p(0)

    P(0)=03

    =0

    When,  p(1)

    p(1)=13

    =1

    When,  P(2)

    p(2)=23

    =8

(iv)    P(x)=(x−1)(x+1)

    P ( x ) = ( x − 1 ) ( x + 1 )

    When,  p(0)

    p(0)=(0-1)(0+1)

    =-1×1

    =−1

    When,  p(1)

    p(1)=(1-1)(1+1)

    =0×2

    =0

    When,  p(2)

    p(2)=(2-1)(2+1)

    =1×3

    =3

3.    Verify whether the following are zeroes of the polynomial, indicated against them.

(i)    p(x)=3x+1,x=-13

(ii)    p(x)=5x-Ï€,x=45

(iii)    p(x)=x2-1,x=1,-1

(iv)    p(x)=(x+1)(x-2), x=-1,2

(v)    p(x)=x2,x=0

(vi)    p(x)=lx+m, x=-ml

(vii)    p(x)=3x2-1, x=-1√3,2√3

(viii)    p(x)=2x+1,x=12

solution :

(i)    p(x)=3x+1,x=-13

    For, x=-13,

    p(x)=3x+1

    p(-13)=3(-13)+1

    =−1+1

    =0

    So, -13 is a zero of p(x).

(ii)    p(x)=5x-Ï€,x=45

    For, x=45,

    p(x)=5x-Ï€

    p(45)=5×(45)-Ï€

    =4−π

    So, 45 is not a zero of p(x).

(iii)    p(x)=x2-1,x=1,-1

    For, x=1,

    p(x)=x2-1,

    p(1)=12-1

    =1−1

    =0

    For, x=-1,

    p(x)=x2-1,

    p(-1)=-12-1

    =1−1

    =0

    So, 1 and -1 is a zero of p(x).

(iv)    p(x)=(x+1)(x-2), x=-1,2

    For, x=-1,

    p(x)=(x+1)(x-2),

    p(-1)=(-1+1)(-1–2)

    =0×(-3)

    =0

    For, x=2,

    p(x)=(x+1)(x-2),

    p(2)=(2+1)(2–2)

    =3×0

    =0

    So, -1 and 2 is a zero of p(x).

(v)    p(x)=x2,x=0

    For, x=0,

    p(x)=x2

    p(0)=02

    =0

    So, 0 is a zero of p(x).

(vi)    p(x)=lx+m,x=-ml

    For, x=-ml

    p(x)=lx+m

    p(-ml)=l(-ml)+m

    =-m+m

    =0

    So, -ml is a zero of p(x).

(vii)    p(x)=3x2-1 , x=-1√3,2√3

    For, x=-1√3

    p(x)=3x2-1

    p(-1√3)=3(-1√3)2-1

    =3×(13)-1

    =1-1

    =0

    For, x=2√3

    p(x)=3x2-1

    p(2√3)=3(2√3)2-1

    =3×(43)-1

    =4-1

    =3

    =3≠0

    So, -1√3 is a zero of p(x) but (2√3) is not a zero of p(x).

(viii)    p(x)=2x+1,x=12

    For, x=12,

    p(x)=2x+1

    p(12)=2(12)+1

    =1+1

    =2

    =2≠0

    So, 12 is not a zero of p(x).

4.    Find the zero of the polynomial in each of the following cases:

(i)    p(x)=x+5

(ii)    p(x)=x–5

(iii)    p(x)=2x+5

(iv)    p(x)=3x-2

(v)    p(x)=3x

(vi)    p(x)=ax, a≠0

(vii)    p(x)=cx+d, c≠0,c,d are real numbers.

solution :

(i)    p(x)=x+5

    â‡’x+5=0

    â‡’x=−5

    So, -5 is a zero of the polynomial p(x).

(ii)    p(x)=x–5

    p(x)=x–5

    â‡’x-5=0

    â‡’x=5

    So, 5 is a zero of the polynomial p(x).

(iii)    p(x)=2x+5

    â‡’2x+5=0

    â‡’2x=-5

    â‡’x=-52

    So, -52 is a zero of the polynomial p(x).

(iv)    p(x)=3x-2

    â‡’3x-2=0

    â‡’3x=2

    â‡’x=23

    So, 23 is a zero of the polynomial p(x).

(v)    p(x)=3x

    â‡’3x=0

    â‡’x=03

    â‡’x=0

    So, 0 is a zero of the polynomial p(x).

(vi)    p(x)=ax, a≠0

    â‡’ax=0

    â‡’x=0a

    â‡’x=0

    So, 0 is a zero of the polynomial p(x).

(vii)    p(x)=cx+d, c≠0,c,d are real numbers.

    â‡’cx+d=0

    â‡’cx=−d

    â‡’x=-dc

    So, -dc is a zero of the polynomial p(x).



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