Class 9

### Topic Covered

♦ Zeroes of a Polynomial

### Zeroes of a Polynomial

Consider the polynomial p(x) = 5x^3 – 2x^2 + 3x – 2.

If we replace x by 1 everywhere in p(x), we get

p(1) = 5 × (1)^3 – 2 × (1)^2 + 3 × (1) – 2

= 5 – 2 + 3 –2

= 4

So, we say that the value of p(x) at x = 1 is 4.

Similarly, p(0) = 5(0)^3 – 2(0)^2 + 3(0) –2

= –2
Q 3240545413

Find the value of each of the following polynomials at the indicated value of variables:

(i) p(x) = 5x^2 – 3x + 7 at x = 1.

(ii) q(y) = 3y^3 – 4y + sqrt (11) at y = 2.

(iii) p(t) = 4t^4 + 5t^3 – t^2 + 6 at t = a.
Class 9 Chapter 2 Example 2
Solution:

(i) p(x) = 5x^2 – 3x + 7

The value of the polynomial p(x) at x = 1 is given by

p(1) = 5(1)^2 – 3(1) + 7

= 5 – 3 + 7 = 9

(ii) q(y) = 3y^3 – 4y + sqrt 11

The value of the polynomial q(y) at y = 2 is given by

q(2) = 3(2)^3 – 4(2) + sqrt 11 = 24 – 8 + sqrt 11 = 16 + sqrt 11

(iii) p(t) = 4t^4 + 5t^3 – t^2 + 6

The value of the polynomial p(t) at t = a is given by

p(a) = 4a^4 + 5a^3 – a^2 + 6

Now, consider the polynomial p(x) = x – 1.

What is p(1)? Note that : p(1) = 1 – 1 = 0.

As p(1) = 0, we say that 1 is a zero of the polynomial p(x).

Similarly, you can check that 2 is a zero of q(x), where q(x) = x – 2.

In general, we say that a zero of a polynomial p(x) is a number c such that p(c) = 0.

You must have observed that the zero of the polynomial x – 1 is obtained by
equating it to 0, i.e., x – 1 = 0, which gives x = 1. We say p(x) = 0 is a polynomial
equation and 1 is the root of the polynomial equation p(x) = 0. So we say 1 is the zero
of the polynomial x – 1, or a root of the polynomial equation x – 1 = 0.

Now, consider the constant polynomial 5. Can you tell what its zero is? It has no
zero because replacing x by any number in 5x^0 still gives us 5. In fact, a non-zero
constant polynomial has no zero. What about the zeroes of the zero polynomial? By
convention, every real number is a zero of the zero polynomial.
Q 3250545414

Check whether –2 and 2 are zeroes of the polynomial x + 2.
Class 10 Chapter 2 Example 3
Solution:

Let p(x) = x + 2.
Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0
Therefore, –2 is a zero of the polynomial x + 2, but 2 is not.
Q 3200545418

Find a zero of the polynomial p(x) = 2x + 1.
Class 9 Chapter 2 Example 4
Solution:

Finding a zero of p(x), is the same as solving the equation

p(x) = 0

Now, 2x + 1 = 0 gives us x = - 1/2

So , -1/2 is a zero of the polynomial 2x + 1.

Now, if p(x) = ax + b, a ≠ 0, is a linear polynomial, how can we find a zero of
p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x),
amounts to solving the polynomial equation p(x) = 0.

Now, p(x) = 0 means ax + b = 0, a ≠ 0

So, ax = –b

i.e., x = - b/a .

So, x = - b/a is the only zero of p(x), i.e., a linear polynomial has one and only one zero.

Now we can say that 1 is the zero of x – 1, and –2 is the zero of x + 2.
Q 3210645510

Verify whether 2 and 0 are zeroes of the polynomial x^2 – 2x.
Class 9 Chapter 2 Example 5
Solution:

Let p(x) = x^2 – 2x

Then p(2) = 2^2 – 4 = 4 – 4 = 0

and p(0) = 0 – 0 = 0

Hence, 2 and 0 are both zeroes of the polynomial x^2 – 2x.

Let us now list our observations:

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

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

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

(iv) A polynomial can have more than one zero.