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 – 4x**

^{2}+ 3 at**(i) x = 0**

**(ii) x = –1**

**(iii) x = 2**

Answer:

Let p(x) = 5x – 4x

^{2}+ 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) = y**

^{2}– y + 1**(ii) p(t) = 2 + t + 2t**

^{2}– t^{3}**(iii) p(x) = x**

^{3}**(iv) p(x) = (x – 1) (x + 1)**

Answer:

(i) p(y) = y

^{2}– y + 1
p(0) = 0

^{2}– 0 + 1 = 1
p(1) = 1

^{2}– 1 + 1 = 1
p(2) = 2

^{2}– 2 + 1= 4 -2 + 1 = 3
(ii) p(t) = 2 + t + 2t

^{2}– t^{3}
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) = x

^{3}
p(0) = 0

^{3}=**0**
p(1) = 1

^{3}=**1**
p(2) = 2

^{3}=**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) = x**

^{2}– 1, x = 1, –1**(iv) p(x) = (x + 1) (x – 2), x = – 1, 2**

**(v) p(x) = x**

^{2}, x = 0**(vi) p(x) = lx + m, x = – m/l**

**(vii) p(x) = 3x**

^{2}– 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) = x

^{2}– 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.

^{2}, 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

(i) p(x) = x + 5

**(ii) p(x) = x – 5**

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

(iv) p(x) = 3x – 2

(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.

(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 x**

^{2}- 9?
Answer: The polynomial x

^{2}- 9 has degree = 2. It means it can have ≤2 number of zeros.
Let p(x) = 0

x

^{2}- 9 = 0
x

^{2}= 9 ⇒ x = √9 = ±3
∴ x = +3, -3 are the zeros of the polynomial x

^{2}- 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.

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