We already know that a polynomial is an algebraic term with one or many terms. Zeroes of Polynomial are the real values of the variable for which the value of the polynomial becomes zero. So, real numbers, ‘m’ and ‘n’ are zeroes of polynomial p(x), if p(m) = 0 and p(n) = 0.

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## Understanding Zeroes of Polynomial

### Example 1

Let’s look at the polynomial, p(x) = 5x^{3} – 2x^{2} + 3x -2. Now, let’s find the value of the polynomial(x) at x = 1, p(1) = 5(1)^{3} – 2(1)^{2} + 3(1) – 2 = 5 – 2 + 3 – 2 = 4. Therefore, we can say that the value of the polynomial p(x) at x = 1 is 4.

Next, let’s find the value of the polynomial(x) at x = 0, p(0) = 5(0)^{3} – 2(0)^{2} + 3(0) – 2 = 0 – 0 + 0 – 2 = – 2. Therefore, we can say that the value of the polynomial p(x) at x = 0 is – 2.

**Browse more Topics under Polynomials**

- Polynomial and its Types
- Value of Polynomial and Division Algorithm
- Degree of Polynomial
- Factorisation of Polynomials
- Remainder Theorem
- Factor Theorem
- Geometrical Representation of Zeroes of a Polynomial

### Example 2

Let’s look at another polynomial now, p(x) = x – 1. Let’s find the value of the polynomial at x = 1, p(1) = 1 – 1 = 0. So, the value of the polynomial p(x) at x = 1 is 0. Therefore, 1 is the zero of the polynomial p(x).

Similarly, for the polynomial q(x) = x – 2, the zero of the polynomial is 2. To summarize, zeroes of polynomial p(x) are numbers c and d such that p(c) = 0 and p(d) = 0.

**Browse more Topics under Polynomials**

- Polynomial and its Types
- Value of Polynomial and Division Algorithm
- Degree of Polynomial
- Factorisation of Polynomials
- Remainder Theorem
- Factor Theorem
- Geometrical Representation of Zeroes of a Polynomial

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## Calculating Zeroes of polynomial

When we calculated zeroes of polynomial p(x) = x – 1, we equated it to 0, x – 1 = 0 or, x = 1. Hence, we say that p(x) = 0 is the Polynomial Equation. 1 is the Root of the Polynomial equation p(x) = 0. OR 1 is the Zero of the Polynomial equation p(x) = x – 1 = 0.

Now, let’s look at a constant polynomial ‘5’. You can write this as 5x^{0}. What is the Root of this constant polynomial? The answer is a Non-zero constant polynomial has no zero. Also, every real number is a zero of the Zero Polynomial.

Let’s look at the following linear polynomial to understand the calculation of the roots or ‘zeroes of polynomial’: p(x) = ax + b … where a ≠ 0. To find a zero, we must equate the polynomial to 0. [p(x) = 0]. Hence, ax + b = 0 … where a ≠ 0. So, ax = – b or, x = – b/a. Therefore, ‘- b/a’ is the only zero of p(x) = ax + b. We can also say that a linear polynomial has only one zero.

## Observations

- A zero of a polynomial need not be 0.
- 0 may be a zero of a polynomial.
- Every linear polynomial has one and only one zero.
- A polynomial can have more than one zero.

* Learn Degree of Polynomials in detail. *

## More Solved Examples for You

**Question:** Find p(0), p(1) and p(2) for each of the following polynomials:

- p(y) = y
^{2}– y + 1 - p(t) = 2 + t + 2t
^{2}– t^{3} - p(x) = x
^{3} - p(x) = (x – 1) (x + 1)

**Solution:**

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 = 3

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 + 8 – 8 = 4

p(x) = x^{3}

- p(0) = (0)
^{3}= 0 - p(1) = (1)
^{3}= 1 - p(2) = (2)
^{3}= 8

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

### Verify the following

**Question:** Verify whether the following are zeroes of polynomials indicated against them.

- p(x) = 3x + 1, x = – 1/3
- p(x) = 5x – π, x = 4/5
- p(x) = (x + 1) (x – 2), x = – 1, 2
- p(x) = 3x
^{2}– 1, x = – 1/√3 , 2/√3

**Solution:**

- p(x) = 3x + 1, x = – 1/3

p(-1/3) = 3(-1/3) + 1 = – 1 + 1 = 0.

Hence, x = -1/3 is a zero of polynomial 3x + 1.

- p(x) = 5x – π, x = 4/5

p(4/5) = 5(4/5) – π = 4 – π ≠ 0.

Hence, x = 4/5 is not a zero of polynomial 5x – π.

- p(x) = (x + 1) (x – 2), x = – 1, 2

p(-1) = (– 1 + 1)(- 1 – 2) = (0)(- 3) = 0.

Hence, x = – 1 is a zero of polynomial (x + 1) (x – 2).

p(2) = (2 + 1)(2 – 2) = (3)(0) = 0.

Hence, x = 2 is a zero of polynomial (x + 1) (x – 2).

- p(x) = 3x
^{2}– 1, x = – 1/√3 , 2/√3

p(- 1/√3) = 3(- 1/√3)^{2}– 1 = 3 (1/3) – 1 = 1 – 1 = 0.

Hence, x = – 1/√3 is a zero of polynomial 3x^{2}– 1.

p(2/√3) = 3(2/√3) – 1 = 3(4/3) – 1 = 4 – 1 = 3 ≠ 0.

Hence, x = 2/√3 is not a zero of polynomial 3x^{2}– 1.

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