**Zeros Of A Polynomial Function**

If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x).

**For Example:**

(i) For polynomial p(x) = x – 2; p(2) = 2 – 2 = 0

∴ x = 2 or simply 2 is a zero of the polynomial

p(x) = x – 2.

(ii) For the polynomial g(u) = u^{2} – 5u + 6;

g(3) = (3)^{2} – 5 × 3 + 6 = 9 – 15 + 6 = 0

∴ 3 is a zero of the polynomial g(u)

= u2 – 5u + 6.

Also, g(2) = (2)^{2} – 5 × 2 + 6 = 4 – 10 + 6 = 0

∴ 2 is also a zero of the polynomial

g(u) = u^{2} – 5u + 6

(a) Every linear polynomial has one and only one zero.

(b) A given polynomial may have more than one zeroes.

(c) If the degree of a polynomial is n; the largest number of zeroes it can have is also n.

**For Example**:

If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a

polynomial is 8; largest number of zeroes it can have is 8.

(d) A zero of a polynomial need not be 0.

**For Example**: If f(x) = x^{2} – 4,

then f(2) = (2)2 – 4 = 4 – 4 = 0

Here, zero of the polynomial f(x) = x^{2} – 4 is 2 which itself is not 0.

(e) 0 may be a zero of a polynomial.

**For Example**: If f(x) = x^{2} – x,

then f(0) = 0^{2} – 0 = 0

Here 0 is the zero of polynomial

f(x) = x^{2} – x.

**Zeros Of A Polynomial Function With Examples**

**Example 1:** Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :

(i) p(x) = 3x + 1, x = \(-\frac{1}{3}\)

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

(iii) p(x) = x^{2}, x = 0

(iv) p(x) = lx + m, x = \(-\frac{m}{l}\)

(v) p(x) = 2x + 1, x = \(\frac{1}{2}\)

**Sol.**

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

\(\Rightarrow p\left( -\frac{1}{3} \right)=3\times -\frac{1}{3}+1=-1+1=0\)

∴ x = \(-\frac{1}{3}\) is a zero of p(x) = 3x + 1.

(ii) p(x) = (x + 1) (x – 2)

⇒ p(–1) = (–1 + 1) (–1 – 2) = 0 × –3 = 0

and, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0

∴ x = –1 and x = 2 are zeroes of the given polynomial.

(iii) p(x) = x^{2 }

⇒ p(0) = 0^{2} = 0

∴ x = 0 is a zero of the given polynomial

(iv) p(x) = lx + m

\(\Rightarrow p\left( -\frac{m}{l} \right)=l\left( -\frac{m}{l} \right)+m\)

= – m + m = 0

∴ x = \(-\frac{m}{l}\) is a zero of the given polynomial.

(v) p(x) = 2x + 1

\(\Rightarrow p\left( \frac{1}{2} \right)=2\times \frac{1}{2}+1\)

= 1 + 1 = 2 ≠ 0

∴ x = \(\frac{1}{2}\) is not a zero of the given polynomial.

**Example 2:** Find the zero of the polynomial in each of the following cases :

(i) p(x) = x + 5

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

(iii) p(x) = 3x – 2

**Sol.**

To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.

(i) For the zero of polynomial p(x) = x + 5

p(x) = 0 ⇒ x + 5 = 0 ⇒ x = –5

∴ x = –5 is a zero of the polynomial.

p(x) = x + 5.

(ii) p(x) = 0 ⇒ 2x + 5 = 0

⇒ 2x = –5 and x = \(-\frac{5}{2}\)

∴ \(-\frac{5}{2}\) is a zero of p(x) = 2x + 5.

(iii) p(x) = 0 ⇒ 3x – 2 = 0

⇒ 3x = 2 and x = \(\frac{2}{3}\).

∴ x = \(\frac{2}{3}\) is zero of p(x) = 3x – 2.