__Exercise 2.2 Page: 34__

__Exercise 2.2 Page: 34__

**1. Find the value of the polynomial (x)=5x−4x ^{2}+3 **

**(i) x = 0**

**(ii) x = – 1**

**(iii) x = 2**

Solution:

Let f(x) = 5x−4x^{2}+3

(i) When x = 0

f(0) = 5(0)-4(0)^{2}+3

= 3

(ii) When x = -1

f(x) = 5x−4x^{2}+3

f(−1) = 5(−1)−4(−1)^{2}+3

= −5–4+3

= −6

(iii) When x = 2

f(x) = 5x−4x^{2}+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)=y ^{2}−y+1**

Solution:

p(y) = y^{2}–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+2t ^{2}−t^{3}**

Solution:

p(t) = 2+t+2t^{2}−t^{3}

∴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)=x ^{3}**

Solution:

p(x) = x^{3}

∴p(0) = (0)^{3 }= 0

p(1) = (1)^{3 }= 1

p(2) = (2)^{3 }= 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)=x ^{2}−1, x=1, −1**

Solution:

For, x = 1, −1;

p(x) = x^{2}−1

∴p(1)=1^{2}−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) = x ^{2}, x = 0**

Solution:

For, x = 0 p(x) = x^{2}

p(0) = 0^{2 }= 0

∴ 0 is a zero of p(x).

**(vi) p(x) = lx+m, x = −m/l**

Solution:

For, x = -m/*l *; p(x) = *l*x+m

∴ p(-m/*l)*= *l*(-m/*l*)+m = −m+m = 0

∴-m/*l* is a zero of p(x).

**(vii) p(x) = 3x ^{2}−1, x = -1/√3 , 2/√3**

Solution:

For, x = -1/√3 , 2/√3 ; p(x) = 3x^{2}−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