Please Wait... While Loading Full Video#### Class 9 chapter - 2 POLYNOMIALS

♦ 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`

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

(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

(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

Class 10 Chapter 2 Example 3

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

Class 9 Chapter 2 Example 4

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

Class 9 Chapter 2 Example 5

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.