﻿ Class 9 NCERT Math Solution
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TOPICS
Exercise - 2.2

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

Solution :-
```(i) At x = 0
p(x) = 5x – 4x² + 3
p(0) = 5 x 0 - 4 x 0 + 3
= 0 - 0 + 3
= 3
```
```(ii) At x = -1
p(x) = 5x – 4x² + 3
p(-1) = 5 x (-1) – 4 x (-1)² + 3
= -5 - 4 x 1 + 3
= -5 - 4 + 3
= -9 + 3
= -6
```
```(iii) At x = 2
p(x) = 5x – 4x² + 3
p(2) = 5 x 2 - 4 x 2² + 3
= 10 - 4 x 4 + 3
= 13 - 16
= -3
```

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 + t + 2t² – t³,
(iii) p(x) = x³,
(iv) p(x) = (x – 1) (x + 1).

Solution :-
```(i) p(y) = y² – y + 1 at

p(0) = 0² – 0 + 1
= 0 - 0 + 1
= 1

p(1) = 1² – 1 + 1
= 1 - 1 + 1
= 2 - 1
= 1

p(2) = 2² – 2 + 1
= 4 - 2 + 1
= 5 - 2
= 3
```
```
(ii) p(t) = 2 + t + 2t² – t³

p(0) = 2 + 0 + 2 x 0² – 0³
= 2 + 0
= 2

p(1) = 2 + 1 + 2 x 1² – 1³
= 3 + 2 - 1
= 5 - 1
= 4

p(2) = 2 + 2 + 2 x 2² – 2³
= 4 + 2 x 4 - 8
= 4 + 8 - 8
= 12 - 8
= 4
```
```
(iii) p(x) = x³

p(0) = 0³ = 0

p(1) = 1³ = 1

p(2) = 2³ = 8
```
```(iv) p(x) = (x – 1) (x + 1)

p(0) = (0 – 1) (0 + 1)
= -1 x 1
= -1

p(1) = (1 – 1) (1 + 1)
= 0 x 2
= 0

p(2) = (2 – 1) (2 + 1)
= 1 x 3
= 3
```

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) = 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 = -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
p(-1/3) = 3 x (-1/3) + 1
= -1 + 1
= 0
Hence, x = -1/3 is a zero of polynomial.
```
```(ii) p(x) = 5x - π, x = 4/5
p(4/5) = 5 x 4/5 - π
= 4 - π
Hence, x = 4/5 is not a zero of polynomial.
```
```(iii) p(x) = x² – 1, x = 1, –1
p(1) = (1)² – 1
= 1 - 1
= 0
p(-1) = (-1)² – 1
= 1 - 1
= 0
Hence, x = 1, -1 are zeroes of polynomial.
```
```(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
p(-1) = (-1 + 1)(-1 - 2)
= 0 x (-3)
= 0
p(2) = (2 + 1)(2 - 2)
= 3 x 0
= 0
Hence, x = -1, 2 are zeroes of polynomial.
```
```(v) p(x) = x², x = 0
p(0) = 0² = 0
Hence, x = 0 is a zero of polynomial.
```
```(vi) p(x) = lx + m, x = -m/l
p(-m/l) = l x (-m/l) + m
= -m + m
= 0
Hence, x = -m/l is a zero of polynomial.
```
```(vii) p(x) = 3x² – 1, x = -1/√3, 2/√3
p(-1/√3) = 3 x (-1/√3)² – 1
= 3 x 1/3 - 1
= 1 - 1
= 0
p(2/√3) = 3 x (2/√3)² – 1
= 3 x 4/3 - 1
= 4 - 1
= 3
Hence, x = -1/√3 is a zero of polynomial.
But x = 2/√3 is not a zero of polynomial.
```
```(viii) p(x) = 2x + 1, x = 1/2
p(1/2) = 2 x 1/2 + 1
= 1 + 1
= 2
Hence, x = 2 is not a zero of polynomial.
```

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, c, d are real numbers.

Solution :-
```    Zero of a polynomial is that value of the variable at which
the value of the polynomial is obtained as 0. So, p(x) = 0.

(i) p(x) = x + 5
p(x) = 0
x + 5 = 0
x = -5
Therefore, x = -5 is a zero of polynomial p(x) = x + 5.
```
```(ii) p(x) = x – 5
p(x) = 0
x - 5 = 0
x = 5
Therefore, x = 5 is a zero of polynomial p(x) = x - 5.
```
```(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = -5
x = -5/2
Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5.
```
```(iv) p(x) = 3x – 2
p(x) = 0
3x - 2 = 0
3x = 2
x = 2/3
Therefore, x = 2/3 is a zero of polynomial p(x) = 3x - 2.
```
```(v) p(x) = 3x
p(x) = 0
3x = 0
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = 3x.
```
```(vi) p(x) = ax, a ≠ 0
p(x) = 0
ax = 0
x = 0/a
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = ax.
```
```(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
p(x) = 0
cx + d = 0
cx = -d
x = -d/c
Therefore, x = -d/c is a zero of polynomial p(x) = cx + d.
```
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