NCERT Class 9 Mathematics Textbook for Blind and Visually Impaired Students made Screen Readable by Professor T K bansal.

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

Can you find p(−1)?

Example 2

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.
\[(ii)\ q(y)\ =\ 3y^3\ −\ 4y\ +\ 11\]
at y = 2.
\[(iii)\ p(t)\ =\ 4t^4\ +\ 5t^3\ −\ t^2\ +\ 6\]
at t = a.

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

\[(ii)\ q(y)\ =\ 3y^3\ −\ 4y\ +\ √11\]
The value of the polynomial q(y) at y = 2 is given by
q(2) = 3(2)^3 − 4(2) + √11
= 24 − 8 + √11
= 16 + √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, we 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.

We observe 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 we 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.

Example 3

Check whether − 2 and 2 are zeroes of the polynomial x + 2.

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 a zero of the polynomial.

Example 4

Find a zero of the polynomial p(x) = 2x + 1.

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.

Example 5

Verify whether 2 and 0 are zeroes of the polynomial x^2 − 2x.

Solution :

Let p(x) = x^2 − 2x
Then p(2) = 2^2 − 4
= 4 − 4 = 0
and p(0) = 0 − 0 = 0

Hence, both, 2 and 0 are 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 zeros.

EXERCISE 2.2

Q1. Find the value of the following polynomial at the below mentioned values:
\[5x\ −\ 4x^2\ +\ 3\]
(i) x = 0
(ii) x = −1
(iii) x = 2

A1.
(i) 3
(ii) −6
(iii) −3

Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
\[(i)\ p(y)\ =\ y^2\ −\ y\ +\ 1\]
\[(ii)\ p(t)\ =\ 2\ +\ t\ +\ 2t^2\ −\ t^3\]
\[(iii)\ p(x)\ =\ x^3\]
\[(iv)\ p(x)\ =\ (x − 1) (x + 1)\]

A2.
(i) 1, 1, 3
(ii) 2, 4, 4
(iii) 0, 1, 8
(iv) −1, 0, 3

Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
\[(i)\ p(x)\ =\ 3x\ +\ 1; x = − \frac{1}{3} \]
\[(ii)\ p(x)\ =\ 5x\ −\ π;\ x = \frac{4}{5}\]
\[(iii)\ p(x)\ =\ x^2\ −\ 1;\ x = 1,\ −1\]
\[(iv)\ p(x)\ =\ (x\ +\ 1) (x − 2);\ x = − 1,\ 2\]
\[(v)\ p(x)\ =\ x^2;\ x = 0\]
\[(vi)\ p(x)\ =\ lx\ +\ m;\ x = − \frac{m}{l}\]
\[(vii)\ p(x)\ =\ 3x^2\ −\ 1;\ x\ =\ − \frac{1}{√3};\ \frac{2}{√3}\]
\[(viii)\ p(x)\ =\ 2x\ +\ 1;\ x =1/2\]

A3.
(i) Yes
(ii) No
(iii) Yes
(iv) Yes
(v) Yes
(vi) Yes
(vii) −(1/ √3) is a zero, but 2 / √3 is not a zero of the polynomial
(viii) No

Q4. Find the zero of the polynomial in each of the following cases:
\[(i)\ p(x)\ =\ x\ +\ 5\]
\[(ii)\ p(x)\ =\ x\ −\ 5\]
\[(iii)\ p(x)\ =\ 2x\ +\ 5\]
\[(iv)\ p(x)\ =\ 3x\ −\ 2\]
\[(v)\ p(x)\ =\ 3x\]
\[(vi)\ p(x)\ =\ ax;\ a≠0\]
\[(vii)\ p(x)\ =\ cx\ +\ d;\ c≠0\]

A4.
(i) −5
(ii) 5
(iii) −5/2
(iv) 2/3
(v) 0
(vi) 0
(vii) − (d/c)