CBSE - Class 9 - Maths - CH2 - Polynomials - Exercise 2.2

Carl Friedrich Gauss -
who proved fundamental
theorem of Algebra
(source:wikipedia)

ZEROS (ROOTS) of a POLYNOMIAL

1. A zero of a polynomial p(x) is a number c such that p(c) = 0.
   OR x = c (or x -c = 0 ) is called the root of the polynomial equation p(x).

2. If p(x) = ax + b, a ≠ 0, is a linear polynomial, x = -b/a, is the only zero of p(x), i.e., a linear polynomial has one and only one zero.

3. A zero of a polynomial need not be 0.

4. 0 may be a zero of a polynomial.

5. Every linear polynomial has one and only one zero.

6. A polynomial can have more than one zero.

7. Zeros of the polynomial are also called roots of the polynomial.

EXERCISE 2.2

Q1. Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0

(ii) x = –1

(iii) x = 2

Answer:

Let p(x) =  5x – 4x² + 3

(i) x = 0

p(0) = 5(0) - 4(0)² + 3 = 0 + 0 + 3  = 3

(ii) x = -1

p(-1) = 5(-1) - 4(-1)² + 3 = -5 -4 + 3 = -6

(iii) x = 2

p(2) = 5(2) - 4(2)² + 3 = 10 - 16 + 3 = -3

Q2. 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)

Answer:

(i)  p(y) = y² – y + 1

p(0) =  0² – 0 + 1 = 1

p(1) =  1² – 1 + 1 = 1

p(2) =  2² – 2 + 1= 4 -2 + 1 = 3

(ii)  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 -1 = 4

p(2) = 2 + 2 + 2(2)² – (2)³ = 2 + 2 + 2(4) -8 = 2 + 2 + 8 - 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

Q3. 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) = 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

Answer:

(i) p(x) = 3x + 1

If x = -1/3 is the zero of the polynomial, ⇒ p(-1/3) = 0

⇒ p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0

∴ x = -1/3 is the zero of the given polynomial.

(ii)  p(x) = 5x – π, x = 4/5

If x = 4/5 is the zero of the polynomial, p(4/5) must be 0.

⇒  p(4/5) = 5(4/5) - π = 4 - π ≠ 0

∴ x = 4/5 is not the zero of the given polynomial.

(iii)  p(x) = x² – 1, x = 1, –1

if x =1 and x = -1 are the zeros of the given polynomial, then p(1) = 0 and p(-1) = 0.

⇒   p(1) = (1)² – 1 = 1- 1 = 0

and p(-1) = (-1)² – 1 = 1- 1 = 0

∴  x = 1, –1 are the zeros of the given polynomial.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2

if x = -1 and x = 2 are the zeros of the polynomial, then p(-1) = 0 and p(2) = 0.

⇒   p(-1)  = (-1+1)(-1-2) = (0)(-3) = 0 

and p(2)  = (2+1)(2-2) = (3)(0) = 0

∴  x = -1, 2 are the zeros of the given polynomial.

(v) p(x) = x² , x = 0

If x = 0 is the zero of the polynomial, p(0) = 0

⇒   p(0)  = (0)²= 0

∴  x = 0 is the zero of the given polynomial.

(vi)  p(x) = lx + m, x = – m/l

if  x = -m/l is the zero of the polynomial, then p(-m/l) = 0

⇒  p(-m/l) = l(-m/l) + m = -m + m = 0

∴  x = – m/l is the zero of the given polynomial.

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

if x = 1/2 is the zero of the given polynomial, p(1/2) = 0

⇒ p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0

∴   x = 1/2is not the zero of the given polynomial.

Q4. 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.

Answer: A zero of a polynomial p(x) is a number c such that p(c) = 0. 

To find c, let us equate p(x) = 0.

(i) p(x) = x + 5 

Let p(x) = 0

⇒  p(x) = x + 5 = 0   ⇒ x = -5

p(-5) = -5 + 5 = 0

∴   x = -5 is the zero of the given polynomial.

(ii) p(x) = x – 5 

Let p(x)  = 0

⇒  p(x)= x - 5 = 0 ⇒  x = 5 

∴   x = 5 is the zero of the given polynomial.

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

Let p(x)  = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = -5/2 

For x = -5/2 the value of the polynomial is 0, ∴  x = -5/2 is the zero of the given polynomial.

(iv) p(x) = 3x – 2

To find zero of the polynomial, p(x) = 0

⇒  3x - 2 = 0

⇒ 3x = 2

⇒x = 2/3 which is the zero of the given polynomial.

(v) p(x) = 3x

p(x) = 0

⇒   3x = 0 

⇒  x = 0 is the zero of the given polynomial.

(vi) p(x) = ax, a ≠ 0

Let p(x) = 0

⇒  ax = 0

⇒  x = 0 

For x = 0, p(x) becomes zero. ∴   x = 0 is the zero of the given polynomial.

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

Let p(x) = 0

cx + d = 0

cx = -d

x = -d/c, which is the zero of the given polynomial.

Q5: 'A zero of a polynomial need not be 0.' What do you mean by this statement? 

Answer: It means a zero of the polynomial can be that non-zero value at which the value of the polynomial becomes zero.

Q6: Can a linear polynomial have more than one zeros of the polynomial.

Answer: No. The degree of the polynomial is 1. At the maximum it can have one zero of the polynomial. 

Q7: What is the Fundamental theorem of algebra? Who proposed it?

Answer: It states "Number of zeros of a polynomial ≤ the degree of the polynomial.".

Carl Friedrich Gauss proposed this theorem.

Q8: How many zeros can exist for the polynomial x² - 9?

Answer: The polynomial x²- 9 has degree = 2. It means it can have ≤2 number of zeros.

Let p(x) = 0

x²- 9 = 0

x²= 9 ⇒ x = √9 = ±3

∴ x = +3, -3 are the zeros of the polynomial  x² - 9.

Q9: Can zeros of the polynomial be identified on graph?

Answer: On X-Y graph, x values represent on x-axis and p(x) on y-axis. The line or curve when cuts the x-axis, those points are the zeros of the polynomial.

As shown below, for polynomial p(x) = 2x + 5. x = -5/2 is the zero of the polynomial.

The following graph also shows two zeros for the polynomial x²- 4