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## Exercise 2.2 Page: 34

1. Find the value of the polynomial (x)=5x−4x2+3

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

Let f(x) = 5x−4x2+3

(i) When x = 0

f(0) = 5(0)-4(0)2+3

= 3

(ii) When x = -1

f(x) = 5x−4x2+3

f(−1) = 5(−1)−4(−1)2+3

= −5–4+3

= −6

(iii) When x = 2

f(x) = 5x−4x2+3

f(2) = 5(2)−4(2)2+3

= 10–16+3

= −3

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

(i) p(y)=y2−y+1

Solution:

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

Solution:

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

Solution:

p(x) = x3

∴p(0) = (0)= 0

p(1) = (1)= 1

p(2) = (2)= 8

(iv) P(x) = (x−1)(x+1)

Solution:

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

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

(i) p(x)=3x+1, x=−1/3

Solution:

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

Solution:

For, x = 4/5, p(x) = 5x–π

∴ p(4/5) = 5(4/5)- = 4-

∴ 4/5 is not a zero of p(x).

(iii) p(x)=x2−1, x=1, −1

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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).

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

(i) p(x) = x+5

Solution:

p(x) = x+5

⇒ x+5 = 0

⇒ x = −5

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

(ii) p(x) = x–5

Solution:

p(x) = x−5

⇒ x−5 = 0

⇒ x = 5

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

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

Solution:

p(x) = 2x+5

⇒ 2x+5 = 0

⇒ 2x = −5

⇒ x = -5/2

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

(iv) p(x) = 3x–2

Solution:

p(x) = 3x–2

⇒ 3x−2 = 0

⇒ 3x = 2

⇒x = 2/3

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

(v) p(x) = 3x

Solution:

p(x) = 3x

⇒ 3x = 0

⇒ x = 0

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

(vi) p(x) = ax, a0

Solution:

p(x) = ax

⇒ ax = 0

⇒ x = 0

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

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

Solution:

p(x) = cx + d

⇒ cx+d =0

⇒ x = -d/c

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

Class 9th Maths Polynomial