## EX 2.2 QUESTION 1.

Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:

Let p(x) = 5x – 4x2 + 3

(i) p(0)

= 5(0) – 4(0)2 + 3

= 0 – 0 + 3 = 3

(ii) p(-1)

= 5(-1) – 4(-1)2 + 3
= – 5x – 4x2 + 3

= -9 + 3 = -6

(iii) p(2)

= 5(2) – 4(2)2 + 3

= 10 – 4(4) + 3

= 10 – 16 + 3 = -3

## EX 2.2 QUESTION 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 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
Solution:

(i) p(y) = y2–y+1

∴p(0) = (0)2−(0)+1=1

p(1) = (1)2–(1)+1=1

p(2) = (2)2–(2)+1=3

(ii) p(t)=2+t+2t2−t3:

p(t) = 2+t+2t2−t3

∴p(0) = 2+0+2(0)2–(0)3=2

p(1) = 2+1+2(1)2–(1)3=2+1+2–1=4

p(2) = 2+2+2(2)2–(2)3=2+2+8–8=4.

(iii) p(x)=x3

p(x) = x3

∴p(0) = (0)= 0

p(1) = (1)= 1

p(2) = (2)= 8

(iv) P(x) = (x−1)(x+1)

p(x) = (x–1)(x+1)

∴p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3

## EX 2.2 QUESTION 3.

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –1/3
(ii) p (x) = 5x – π, x = 4/5
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = -1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = -m/l
(vii) P (x) = 3x2 – 1, x =  -1/√3 , 2/√3
(viii) p (x) = 2x + 1, x = 1/2
Solution:

(i) p(x)=3x+1,x=−1/3

For, x = -1/3,

p(x) = 3x+1

∴p(−1/3) = 3(-1/3)+1

= −1+1 = 0

∴ -1/3 is a zero of p(x).

(ii) p(x)=5x–π,x = 4/5

For, x = 4/5,

p(x) = 5x–π

∴ p(4/5) = 5(4/5) –π = 4 –π

∴ 4/5 is not a zero of p(x).

(iii) p(x)=x2−1,x=1, −1

For, x = 1, −1

p(x) = x2−1

∴p(1)=12−1=1−1 = 0

p(−1)=(-1)2−1 = 1−1 = 0

∴1, −1 are zeros of p(x).

(iv) p(x) = (x+1)(x–2), x =−1, 2

For, x = −1,2;

p(x) = (x+1)(x–2)

∴p(−1) = (−1+1)(−1–2)

= (0)(−3) = 0

p(2) = (2+1)(2–2) = (3)(0) = 0

∴−1,2 are zeros of p(x).

(v) p(x) = x2, x = 0

For, x = 0 p(x) = x2

p(0) = 0= 0

∴ 0 is a zero of p(x).

(vi) p(x) = lx+m, x = −m/l

For, x = -m/; p(x) = lx+m

∴ p(-m/l)l(-m/l)+m = −m+m = 0

∴-m/l is a zero of p(x).

(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3

For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1

∴p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0

∴p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0

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

(viii) p(x) =2x+1, x = 1/2

For, x = 1/2 p(x) = 2x+1

∴ p(1/2)=2(1/2)+1 = 1+1 = 2≠0

∴1/2 is not a zero of p(x).

## EX 2.2 QUESTION 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 where c and d are real numbers.
Solution

(i) p(x) = x+5

⇒ x+5 = 0

⇒ x = −5

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

(ii)p(x) = x−5

⇒ x−5 = 0

⇒ x = 5

∴ 5 is a zero of the polynomial p(x).

(iii) p(x) = 2x+5

p(x) = 2x+5

⇒ 2x+5 = 0

⇒ 2x = −5

⇒ x = -5/2

∴x = -5/2 is a zero  of the polynomial p(x).

(iv) p(x) = 3x–2

p(x) = 3x–2

⇒ 3x−2 = 0

⇒ 3x = 2

⇒x = 2/3

∴x = 2/3  is a zero of the polynomial p(x).

(v) p(x) = 3x

p(x) = 3x

⇒ 3x = 0

⇒ x = 0

∴0 is a zero of the polynomial p(x).

(vi) p(x) = ax, a0

p(x) = ax

⇒ ax = 0

⇒ x = 0

∴x = 0 is a zero of the polynomial p(x).

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

p(x) = cx + d

⇒ cx+d =0

⇒ x = -d/c

∴ x = -d/c is a zero of the polynomial p(x).

Categories: class 9NCERT NOTES

error: Content is protected !!