PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.2

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at (i) x = 0, (ii) x = – 1 and (iii) x = 2.
Answer:
Here, p (x) 5x – 4x² + 3
(i) The value of polynomial p (x) at x = 0 is given by
p(0) = 5(0) – 4(0)² + 3
= 3

(ii) The value of polynomial p (x) at x = – 1 is given by
p(- 1) = 5(- 1) – 4 (- 1)² + 3
= – 5 – 4 + 3
= – 6

(iii) The value of polynomial p (x) at x = 2 is given by
p(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) = y² – y + 1
Answer:
p(y) = y² – y + 1
∴ p(0) = (0)² – (0) + 1 = 1
∴ P(1) = (1)² – (1) + 1 = 1 – 1 + 1 = 1
∴ p(2) = (2)² – (2) + 1 = 4 – 2 + 1 = 3

(ii) p (t) = 2 + t + 2t² – t³
Answer:
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
Answer:
p(x) = x³
∴ p (0) = (0)³ = 0
∴ p ( 1) = (1)³ = 1
∴ p (2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)
Answer:
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

Question 3.
Verify whether the following are zeros of the polynomial, indicated against them:
(i) p(x) = 3x + 1, x = – \(\frac{1}{3}\)
Answer:
Here, p (x) = 3x + 1
Then, p\(\left(-\frac{1}{3}\right)\) = 3\(\left(-\frac{1}{3}\right)\) + 1 = – 1 + 1 = 0
Hence, – \(\frac{1}{3}\) is a zero of polynomial
p(x) = 3x + 1

(ii) p(x) = 5x – π, x = \(\frac{4}{5}\)
Answer:
Here, p(x) = 5x – π,
Then, p\(\left(\frac{4}{5}\right)\) = 5\(\left(\frac{4}{5}\right)\) – π = 4 – π ≠ 0
Hence, \(\frac{4}{5}\) is not a zero of polynomial
p(x) = 5x – π

(iii) p(x) = x² – 1, x = 1, – 1
Answer:
Here, p(x) = x² – 1
Then, p(1) = (1)² – 1 = 1 – 1 = 0 and
p(- 1) = (- 1)² – 1 = 1 – 1 = 0.
Hence, 1 and – 1 both are zeroes of polynomial p(x) = x² – 1.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
Answer:
Here, p (x) = (x + 1) (x – 2)
Then, p(- 1) = (- 1 + 1) (- 1 – 2) = 0 × (-3)= 0
and p (2) = (2 + 1) (2 – 2) = 3 × O = O.
Hence, – 1 and 2 both are zeros of polynomial p(x) = (x + 1) (x – 2).

(v) p(x) = x², x = 0
Answer:
Here, p(x) = x²
Then, p(0) = (0)² = 0
Hence, 0 is a zero of polynomial p (x) = x².

(vi) p(x) = lx + m, x = –\(\frac{n}{l}\)
Answer:
Here, p (x) = lx + m
Then, p \(\left(-\frac{m}{l}\right)\) = l\(\left(-\frac{m}{l}\right)\) + m = – m + m = 0
Hence, \(-\frac{m}{l}\) is a zero of polynomial
p(x) = lx + m.

(vii) p(x) = 3x² – 1, x = – \(\frac{1}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)
Answer:
Here, p(x) = 3x² – 1
Then, p\(\left(-\frac{1}{\sqrt{3}}\right)\) = 3\(\left(-\frac{1}{\sqrt{3}}\right)^{2}\) – 1
= 3\(\left(\frac{1}{3}\right)\) – 1 = 1 – 1 = 0
and p\(\left(\frac{2}{\sqrt{3}}\right)\) = 3\(\left(\frac{2}{\sqrt{3}}\right)^{2}\) – 1
= 3\(\left(\frac{4}{3}\right)\) – 1 = 4 – 1 = 3 ≠ 0
Hence, –\(\frac{1}{\sqrt{3}}\) is a zero of polynomial p (x) = 3x² – 1, but \(\frac{2}{\sqrt{3}}\) is not a zero of polynomial p(x) = 3x² – 1.

(viii) p (x) = 2x + 1, x = \(\frac{1}{2}\)
Answer:
Here, p(x) = 2x + 1
Then, p\(\left(\frac{1}{2}\right)\) = 2\(\left(\frac{1}{2}\right)\) + 1 = 1 + 1 = 2 ≠ 0
Hence, \(\frac{1}{2}\) is not a zero of polynomial
p(x) = 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer:
To find the zero of polynomial p (x) = x + 5,
we solve the equation p (x) = 0.
∴ x + 5 = 0
∴ x = – 5
Thus, – 5 is the zero of polynomial
p(x) = x + 5.

(ii) p(x) = x – 5
Answer:
To find the zero of polynomial p(x) = x – 5,
we solve the equation p (x) = 0.
∴ x – 5 = 0
∴ x = 5
Thus, 5 is the zero of polynomial
p(x) = x – 5.

(iii) p (x) = 2x + 5
Answer:
To find the zero of polynomial p(x) = 2x + 5,
we solve the equation p (x) = 0.
∴ 2x + 5 = 0
∴ 2x = – 5
Thus, –\(\frac{5}{2}\) is the zero of polynomial
p(x) = 2x + 5.

(iv) p (x) = 3x – 2
Answer:
To find the zero of polynomial p (x) = 3x – 2,
we solve the equation p (x) = 0.
∴ 3x – 2 = 0
∴ 3x = 2
Thus, \(\frac{2}{3}\) is the zero of polynomial
p(x) = 3x – 2.

(v) p(x) = 3x
Answer:
To find the zero of polynomial p (x) = 3x.
we solve the equation p (x) = 0.
∴ 3x = 0
∴ x = 0
Thus, 0 is the zero of polynomial P(x) = 3x.

(vi) p(x) = ax, a ≠ 0
Answer:
To find the zero of polynomial p (x) = ax,
a ≠ 0, we solve the equation p (x) = 0.
∴ ax = 0
∴ x = 0 (∵ a ≠ 0)
Thus, 0 is the zero of polynomial
p(x) = ax, a ≠ 0.

(vii) p(x) = cx + d, c ≠ 0, C, d are real numbers.
Answer:
To find the zero of polynomial
p(x) = cx + d, c ≠ 0, c, d are real numbers, we solve the equation p (x) = 0.
∴ cx + d = 0
∴ cx = – d
∴ x = – \(\frac{d}{c}\)
Thus, – \(\frac{d}{c}\) is the zero of polynomial p (x) = cx + d, c ≠ 0, c, d are real numbers.