Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.
BSEB Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2
Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p(x) = 5x – 4x² + 3
(i) At x = 0: p(0) = 5(0) – 4(0)² + 3
= 0 – 0 + 3 = 3
(ii) At x = – 1 : p(- 1) = 5(- 1) – 4(- 1)² + 3
= – 5 – 4 + 3 = – 6.
(iii) At x = 2 : p(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 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)(» + 1)
Solution:
(i) We have, p(y) = y² – y + 1
p(0) = (0)² – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)² – 1 + 1 = 1 – 1 + 1 = 1,
and p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 3
(ii) We have, p(t) = 2 + t + 2t² – t³
∴ p(0) = 2 + 0 + 2(0)² – (0)³
= 2 + 0 + 0 – 0 = 2,
p(1) = 2 + 1 + 2(1)² – (1)³
= 2 + 1 + 2 – 3 = 5 – 3 = 2,
and p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4
(iii) We have, p(x) = x³
∴ p( 0) = (0)³ = 0,
p(1) = (1)³ = 1,
and p(2) = (2)³ = 8
(iv) We have, p(x) = (x – 1)(x + 1)
p(0) = (0 – 1)(0 + 1) = (- 1)(1) = – 1,
p(1) = (1 – 1)(1 + 1) = (0)(2) = 0
and p(2) = (2 – 1)(2 + 1) = (1)(3) = 3
Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them,
(i) p(x) = 3x + 1, x = – \(\frac { 1 }{ 3 }\)
(ii) p(x) = 5x – π, x = \(\frac { 4 }{ 5 }\)
(iii) 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 = – \(\frac { m }{ l }\)
(vii) p(x) = 3x² – 1, x = \(-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
(viii) p(x) = 2x + 1, x = \(\frac { 1 }{ 2 }\)
Solution:
(i) We have, p(x) = 3x + 1
At x = – \(\frac { 1 }{ 3 }\), P(-\(\frac { 1 }{ 3 }\)) = 3(-\(\frac { 1 }{ 3 }\)) + 1 = – 1 + 1 = 0
∴ – \(\frac { 1 }{ 3 }\) is a zero of p(x).
(i) We have, p(x) = 5x – π
At x = – \(\frac { 4 }{ 5 }\), P(\(\frac { 4 }{ 5 }\)) = 5(\(\frac { 4 }{ 5 }\)) – 4 = 4 – 4 = 0
∴ \(\frac { 4 }{ 5 }\) is a zero of p(x).
(iii) We have, p(x) = x³ – 1
At x = 1, p(1) = (1)² – 1 = 1 – 1 = 0
∴ 1 is a zero of p(x).
Also, at x = – 1 p(- 1) = (- 1)² – 1 = 1 – 1 = 0
∴ – 1 is a zero of p(x).
(iv) We have, p(x) = (x + 1)(x – 2)
At x = – 1, p(- 1) = (- 1 + 1)(- 1 – 2) = (0)(- 3) = 0
∴ – 1 is a zero of p(x).
Also, at x = 2, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0
∴ 2 is zero of p(x).
(v) We have, p(x) = x²
At x = 0, p(0) = (0)² = 0
∴ 0 is a zero of p(x).
(vi) We have p(x) = lx + m
At x = – \(\frac { m }{ l }\), p(-\(\frac { m }{ l }\)) = l(\(\frac { -m }{ l }\)) + m = – m + m = 0
∴ \(\frac { -m }{ l }\) is a zero of p(x)
(viii) We have, p(x) = 2x + 1
At x = \(\frac { 1 }{ 2 }\), p(\(\frac { 1 }{ 2 }\)) = 2(\(\frac { 1 }{ 2 }\)) + 1 = 1 + 1 = 2.
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(xr) = 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:
(i) We have to solve p(x) = 0
⇒ x + 5 = 0 ⇒ x = – 5
∴ – 5 is a zero of the polynomial x + 5.
(ii) We have to solve p(x) = 0
⇒ x – 5 = 0 ⇒ x = 5
∴ 5 is a zero of the polynomial x – 5.
(iii) We have to solve p(x) = 0
2x + 5 = 0 ⇒ x = – \(\frac { 5 }{ 2 }\)
∴ – \(\frac { 5 }{ 2 }\) is a zero of the polynomial 2x + 5.
(iv) We have to solve p(x) = 0
3x – 2 – 0 ⇒ x = \(\frac { 2 }{ 3 }\)
∴\(\frac { 2 }{ 3 }\) is a zero of the polynomial 3x – 2.
(v) We have to solve p(x) = 0
⇒ 3x = 0 ⇒ x = 0
∴ 0 is a zero of the polynomial 3x.
(vi) We have to solve p(x) = ax, a ≠ 0
⇒ ax = 0 ⇒ x = 0
∴ 0 is a zero of the polynomial ax.
(vii) We have to solve p(x) = 0, c ≠ 0
⇒ cx + d = 0 ⇒ x = – \(\frac { d }{ c }\)
∴ – \(\frac { d }{ c }\) is a zero of the polynomial cx + d.