1

The graphs of y = p (x) are given in the figure given below, for some polynomials p (x). Find the number of zeroes of p (x), in each case.

Answer

(i) The given graph is parallel to x-axis. It does not intersect the x-axis.
∴ It has no zeroes.
(ii) The given graph intersects the x-axis at one point only.
∴It has one zero.
(iii) The given graph intersects the x-axis at three points.
∴ It has three zeroes.
(iv) The given graph intersects the x-axis at two points.
∴It has two zeroes.
(v) The given graph intersects the x-axis at four points.
∴ It has four zeroes.
(vi) The given graph meets the x-axis at three points.
∴ It has three zeroes.

Exercise 2.1 Page Number 28

1(i)

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² - 2x - 8

Answer

(i) x² - 2x - 8
We have p (x) = x² - 2x - 8
= x² + 2x - 4x - 8 = x (x + 2) - 4 (x + 2)
= (x - 4) (x + 2)
For p (x) = 0, we have
(x - 4) (x + 2) = 0
Either x - 4 = 0 ⇒ x = 4
or x + 2 = 0 ⇒ x = - 2
∴ The zeroes of x² - 2x - 8 are 4 and - 2.

Thus, relationship between zeroes and the coefficients in x² - 2x - 8 is verified.

Exercise 2.2 Page Number 33

1(ii)

(ii) 4s² - 4s + 1

Answer

(ii) 4s²- 4s + 1
We have p (s) = 4s² - 4s + 1
= 4s² - 2s - 2s + 1 = 2s (2s - 1) - 1 (2s - 1)
= (2s - 1) (2s - 1)
For p (s) = 0, we have,
(2s - 1) = 0 ⇒ s = 1/2
∴ The zeroes of 4s² - 4s + 1 are 1/2 and 1/2
Now,

Thus, the relationship between the zeroes and coefficients in the polynomial 4s² - 4s + 1 is verified.

Exercise 2.2 Page Number 33

1(iii)

(iii) 6x² - 3 - 7x

Answer

(iii) 6x² - 3 - 7x
We have
p (x) = 6x² - 3 - 7x
= 6x² - 7x - 3
= 6x² - 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (3x + 1) (2x - 3)
For p (x) = 0, we have,

Either (3x+1) = 0

Thus, the relationship between the zeroes and the coefficients in the polynomial 6x² - 3 - 7x is verified.

Exercise 2.2 Page Number 33

1(iv)

(iv) 4u² + 8u

Answer

(iv) 4u² + 8u
We have, f(u) = 4u² + 8u = 4u (u + 2)
For f (u) = 0,
Either 4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = - 2
∴ The zeroes of 4u² + 8u are 0 and - 2.
Now, 4u² + 8u can be written as 4u² + 8u + 0.

Thus, the relationship between the zeroes and the coefficients in the polynomial 4u² + 8u is verified.

Exercise 2.2 Page Number 33

1(v)

(v) t² - 15

Answer

(v) t² - 15
We have,
f (t) = t² - 15 = (t)² - (√15)²
( t+√15) = (t- √15)
For f (t) = 0, we have
Either ( t+√15) = 0 ⇒ t = -√15
or (t- √15) = 0 ⇒ t = √15
∴ The zeroes of t² - 15 are -√15 and √15.
Now, we can write t² - 15 as t²+ 0t - 15.

Thus, the relationship between the zeroes and the coefficients in the polynomial t² - 15 is verified.

Exercise 2.2 Page Number 33

1(vi)

(vi) 3x² - x - 4

Answer

(vi) 3x² – x – 4
We have, f (x) = 3x² - x - 4 = 3x²+ 3x - 4x - 4
= 3x (x + 1) - 4 (x + 1)
= (x + 1) (3x - 4)
For f (x) = 0 ⇒ (x + 1) (3x - 4) = 0
Either (x + 1) = 0 ⇒ x = - 1
or 3x - 4 = 0 ⇒ x = 4/3
∴The zeroes of 3x²- x - 4 are 1 and
Now,

Thus, the relationship between the zeroes and the coefficients in 3x² - x - 4 is verified.

Exercise 2.2 Page Number 33

2(i)

Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively:

(i) 1/4, -1

Answer

Note: A quadratic polynomial whose zeroes are α and β is given by
p (x) = {x² -(α + β) x + αβ}
i.e., p (x) = {x² - (sum of the zeroes) x + (product of the zeroes)}

(i) Sum of the zeroes, (α + β) =
Product of the zeroes, ab = - 1
∴ The required quadratic polynomial is
x² - (α + β) x + ab

Exercise 2.2 Page Number 33

2(ii)

(ii) √2, 1/3

Answer

(ii) Sum of the zeroes, (α + β) = √2
Product of zeroes, αβ = 1/3
∴The required quadratic polynomial is

Exercise 2.2 Page Number 33

2(iii)

(iii) 0, √5

Answer

(iii) Since, sum of zeroes, (α + β) = 0
Product of zeroes, αβ = √5
∴ The required quadratic polynomial is
x² - (α + β) x + αβ
= x² - (0) x + √5
= x² + √5

Exercise 2.2 Page Number 33

2(iv)

(iv) 1 , 1

Answer

(iv) Since, sum of the zeroes, (α + β) = 1
Product of the zeroes = 1
∴The required quadratic polynomial is
x² - (α + β) x + αβ
= x² - (1) x + 1
= x²- x + 1

Exercise 2.2 Page Number 33

2(v)

(v) -1/4, 1/4

Answer

(v) Since, sum of zeroes, (α+β) = -1/4
Product of zeroes, αβ = 1/4
∴ The required quadratic polynomial

Exercise 2.2 Page Number 33

2(vi)

(vi) 4, 1

Answer

(vi) Since, sum of the zeroes, (α + β) = 4
Product of the zeroes, αβ = 1
∴ The required quadratic polynomial is
= x² - (α + β) x + αβ
= x² - (4) x + 1
= x² - 4x + 1

Exercise 2.2 Page Number 33

1

Divide the polynomial p (x) by the polynomial g (x) and find the quotient and remainder in each of the following :
(i) p (x) = x³ - 3x² + 5x - 3, g (x) = x² - 2
(ii) p (x) = x⁴ - 3x² + 4x + 5, g (x) = x² + 1 - x
(iii) p (x) = x⁴ - 5x + 6, g (x) = 2 - x²

Answer

(i) Here, dividend p (x) = x³ - 3x² + 5x - 3
divisor g (x) = x² - 2
∴ We have

Thus, the quotient = (x - 3) and remainder = (7x - 9)

(ii) Here, dividend p (x) = x⁴- 3x² + 4x + 5 = x⁴+ 0x³ - 3x² + 4x + 5
and divisor g (x) = x² + 1 - x = x² - x + 1
∴ We have

Thus, the quotient is (x² + x - 3) and remainder = 8

(iii) Here, dividend, p (x) = x⁴ - 5x + 6 = x⁴ + 0x³ +0x²- 5x + 6
and divisor, g (x) = 2 - x² = - x² + 2
∴ We have

Thus, the quotient = –x² – 2 and remainder = –5x + 10

Exercise 2.3 Page Number 36

2

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12

(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Answer

(i) Dividing 2t⁴ + 3t³ - 2t² - 9t - 12 by t² - 3, we have:

∵ Remainder = 0
∴ (t² - 3) is a factor of 2t⁴ + 3t³ - 2t² - 9t - 12.

(ii) Dividing 3x⁴ + 5x³ - 7x² + 2x + 2 by x² + 3x + 1, we have:

∵ Remainder = 0.
∴ x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x²+ 2x + 2.

(iii) Dividing x⁵ - 4x³ + x² + 3x + 1 by x³ - 3x + 1, we get:

∴ The remainder = 2, i.e., remainder ≠ 0
∴ x³ - 3x + 1 is not a factor of x⁵- 4x³ + x² + 3x + 1.

Exercise 2.3 Page Number 36

3

Obtain all other zeroes of 3x⁴ + 6x³ - 2x² - 10x - 5, if two of its zeroes

Answer

We have p (x) = 3x⁴ + 6x³- 2x² - 10x - 5

Exercise 2.3 Page Number 36

4

On dividing x³ - 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x).

Answer

Here,
Dividend p (x) = x³ - 3x² + x + 2
Divisor = g (x)
Quotient = (x - 2)
Remainder = (-2x + 4)
Since,
(Quotient × Divisor) + Remainder = Dividend

∴ [(x − 2) × g (x)] + [(−2x + 4)] = x³ − 3x² + x + 2
⇒ (x − 2) × g (x)= x³ − 3x² + x + 2 − (−2x + 4)
= x³ − 3x² + x + 2 + 2x −4

= x³ - 3x² + 3x - 2

Exercise 2.3 Page Number 36

5

Give example of polynomials p (x), g (x), q (x) and r (x), which satisfy the division algorithm and
(i) deg p (x) = deg q (x)
(ii) deg q (x) = deg r (x)
(iii) deg r (x) = 0

Answer

We have
(i) p (x) = 3x² - 6x + 27
g (x) = 3
q (x) = x² - 2x + 9
r (x) = 0

p (x)= q (x) × g (x) + r (x).

(ii) p (x)= 2x³ − 2x² + 2x + 3
g (x) = 2x² − 1
q (x) = x − 1
r (x) = 3x + 2

⇒ p (x)= q (x) × g (x) + r (x)

(iii) p (x) = 2x³ − 4x² + x + 4
g (x) = 2x² + 1
q (x)= x − 2
r (x) = 6

⇒ p (x)= q (x) × g (x) + r (x)

Exercise 2.3 Page Number 36

1(i)

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³ + x² - 5x + 2; 1/2, 1 , -2

Answer

Let p (x) = 2x³ + x² - 5x + 2

Comparing with ax³ + bx² +cx+d

a = 2, b = 1, c = -5, d = 2

Again,
p (1) = 2 (1)³ + (1)² - 5 (1) + 2
= 2 + 1 - 5 + 2
= (2 + 2 + 1) - 5
= 5 - 5 = 0
⇒ 1 is a zero of p (x).
Also p (- 2) = 2 (- 2)³ + (- 2)² - 5 (- 2) + 2
= 2(- 8) + (4) + 10 + 2
= - 16 + 4 + 10 + 2
= - 16 + 16 = 0
⇒ -2 is a zero of p (x).
Relationship
∵ p (x) = 2x³ + x² - 5x + 2
∴ Comparing it with ax³ + bx² + cx + d, we have :
a = 2, b = 1, c = - 5 and d = 2
Also 1/2 , 1 and - 2 are the zeroes of p (x)
Let α = 1/2 , β = 1 and γ = - 2

Thus, the relationship between the co-efficients and the zeroes of p (x) is verified.

Exercise 2.4 Page Number 36