# Class 9 NCERT Solutions- Chapter 2 Polynomials – Exercise 2.2

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

^{2}+ 3

**(i) x = 0**

**(ii) x = –1**

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**(iii) x = 2**

**Solution:**

Given equation: 5x − 4x

^{2}+ 3Therefore, let f(x) = 5x – 4x

^{2 }+ 3

(i)When x = 0f(0) = 5(0)-4(0)

^{2}+3= 3

(ii)When x = -1f(x) = 5x−4x

^{2}+3f(−1) = 5(−1)−4(−1)

^{2}+3= −5–4+3

= −6

(iii)When x = 2f(x) = 5x−4x

^{2}+3f(2) = 5(2)−4(2)

^{2}+3= 10–16+3

= −3

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

**(i) p(y) = y ^{2}−y+1**

**(ii) p(t) = 2+t+2t ^{2}−t^{3}**

**(iii) p(x) = x ^{3}**

**(iv) P(x) = (x−1)(x+1)**

**Solution:**

(i)p(y) = y^{2 }– y + 1Given equation: p(y) = y

^{2}–y+1Therefore, p(0) = (0)

^{2}−(0)+1 = 1p(1) = (1)

^{2}–(1)+1 = 1p(2) = (2)

^{2}–(2)+1 = 3Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y

^{2}–y+1

(ii)p(t) = 2 + t + 2t^{2 }− t^{3}Given equation: p(t) = 2+t+2t

^{2}−t^{3}Therefore, p(0) = 2+0+2(0)

^{2}–(0)^{3}= 2p(1) = 2+1+2(1)

^{2}–(1)^{3}= 2+1+2–1 = 4p(2) = 2+2+2(2)

^{2}–(2)^{3}= 2+2+8–8 = 4Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t

^{2}−t^{3}

(iii)p(x) = x^{3}Given equation: p(x) = x

^{3}Therefore, p(0) = (0)

^{3}= 0p(1) = (1)

^{3}= 1p(2) = (2)

^{3}= 8Hence, p(0) = 0,p(1) = 1, p(2) = 8, for the equation p(x)=x

^{3}

(iv)p(x) = (x−1)(x+1)Given equation: p(x) = (x–1)(x+1)

Therefore, 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

Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x+1)

**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) = x ^{2}−1, x=1, −1**

**(iv) p(x) = (x+1)(x–2), x =−1, 2**

**(v) p(x) = x ^{2}, x = 0**

**(vi) p(x) = lx+m, x = −m/l**

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

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

**Solution:**

(i)p(x)=3x+1, x=−1/3Given: p(x)=3x+1 and x=−1/3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/3

p(−1/3) = 3(-1/3)+1

= −1+1

= 0

Hence, p(x) of -1/3 = 0

(ii)p(x)=5x–π, x = 4/5Given: p(x)=5x–π and x = 4/5

Therefore, substituting the value of x in equation p(x), we get.

For, x = 4/5

p(4/5) = 5(4/5)–π

= 4–π

Hence, p(x) of 4/5 ≠ 0

(iii)p(x)=x^{2}−1, x=1, −1Given: p(x)=x

^{2}−1 and x=1, −1Therefore, substituting the value of x in equation p(x), we get.

For x = 1

p(1) = 1

^{2}−1=1−1

= 0

For, x = -1

p(−1) = (-1)

^{2}−1= 1−1

= 0

Hence, p(x) of 1 and -1 = 0

(iv)p(x) = (x+1)(x–2), x =−1, 2Given: p(x) = (x+1)(x–2) and x =−1, 2

Therefore, substituting the value of x in equation p(x), we get.

For, x = −1

p(−1) = (−1+1)(−1–2)

= (0)(−3)

= 0

For, x = 2

p(2) = (2+1)(2–2)

= (3)(0)

= 0

Hence, p(x) of −1, 2 = 0

(v)p(x) = x^{2}, x = 0Given: p(x) = x

^{2 }and x = 0Therefore, substituting the value of x in equation p(x), we get.

For, x = 0

p(0) = 0

^{2}= 0Hence, p(x) of 0 = 0

(vi)p(x) = lx+m, x = −m/lGiven: p(x) = lx+m and x = −m/l

Therefore, substituting the value of x in equation p(x), we get.

For, x = −m/l

p(-m/l)= l(-m/l)+m

= −m+m

= 0

Hence, p(x) of -m/l = 0

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

^{2}−1 and x = -1/√3 , 2/√3Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/√3

p(-1/√3) = 3(-1/√3)

^{2 }-1= 3(1/3)-1

= 1-1

= 0

For, x = 2/√3

p(2/√3) = 3(2/√3)

^{2 }-1= 3(4/3)-1

= 4−1

=3 ≠ 0

Hence, p(x) of -1/√3 = 0

but, p(x) of 2/√3 ≠ 0

(viii)p(x) =2x+1, x = 1/2Given: p(x) =2x+1 and x = 1/2

Therefore, substituting the value of x in equation p(x), we get.

For, x = 1/2

p(1/2) = 2(1/2)+1

= 1+1

= 2≠0

Hence, p(x) of 1/2 ≠ 0

**Question 4: Find the zero of the polynomials 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 **

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

**Solution:**

(i)p(x) = x+5Given: p(x) = x+5

To find the zero, let p(x) = 0

p(x) = x+5

0 = x+5

x = −5

Therefore, the zero of the polynomial p(x) = x+5 is when x = -5

(ii)p(x) = x–5Given: p(x) = x–5

p(x) = x−5

x−5 = 0

x = 5

Therefore, the zero of the polynomial p(x) = x–5 is when x = 5

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

p(x) = 2x+5

2x+5 = 0

2x = −5

x = -5/2

Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2

(iv)p(x) = 3x–2Given: p(x) = 3x–2

p(x) = 3x–2

3x−2 = 0

3x = 2

x = 2/3

Therefore, the zero of the polynomial p(x) = 3x–2 is when x = 2/3

(v)p(x) = 3xGiven: p(x) = 3x

p(x) = 3x

3x = 0

x = 0

Therefore, the zero of the polynomial p(x) = 3x is when x = 0

(vi)p(x) = ax, a0Given: p(x) = ax, a≠ 0

p(x) = ax

ax = 0

x = 0

Therefore, the zero of the polynomial p(x) = ax is when x = 0

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

p(x) = cx + d

cx+d =0

x = -d/c

Therefore, the zero of the polynomial p(x) = cx+d is when x = -d/c