Exercise 2.1
1. Find the degrees of the following polynomials:
(i) 2×2
– 5x + 3
(ii) y3
+ 2y – 1
(iii) – 9
(iv) 4z – 3
2. Write polynomials of degrees 1, 2 and 3.
3. What are the coefficients of x2
and x3
in the polynomial
x4
– 3×3
+ 6×2
– 2x + 7?
4. What is the coefficient of z in the polynomial 4z3 + 5z2
– 11?
5. What is the constant term of the polynomial 9×3
+ 5×2
– 8x –10?
Recall that polynomials of degree 1 are called linear polynomials.
In this chapter, we shall study linear polynomials.
2.2 Linear Polynomials
We begin with some examples involving linear polynomials.
Example 4: The perimeter of a square of side x is 4x, which is a
linear polynomial in the variable x.
Think and Reflect
Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and
3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Example 5: A chess club charges a joining fee of `200 plus `50 for
every match played. The following table shows the amount a player will
have to pay as the number of matches varies.
Hence, if m is the number of matches played, the total cost will be
`(200 + 50m). Observe that 200 + 50m is a linear polynomial in the
variable m. The amount paid increases by the constant value of `50 for
every additional match played.
Think and Reflect
If a player paid `750, how many matches did he play?
The examples given above highlight a characteristic feature of linear
polynomials — that the difference between the successive values at
integers is constant. In Example 4, the perimeters increase by 2 cm each
time the side of the square increases by 0.5 cm. Similarly in Example 5,the amount paid by a player increases by `50 for every additional match
played. Such patterns are called linear patterns.
When we equate a linear polynomial in one variable to a constant,
we get a linear equation. Let us consider the following example.
Example 6: The sum of two numbers is 64. One of the numbers is 10
more than the other. What are the two numbers?
Let the smaller number be x. Then the larger number must be
x + 10. Since their sum is 64, we have the linear equation x + (x + 10) = 64.
This implies that 2x + 10 = 64. Note that 2x + 10 is a linear polynomial.
By equating it to 64 we get a linear equation 2x = 54 or x = 27. The numbers
are therefore, 27 and 37.
Polynomials can also be thought of as input-output processes.
For instance, consider the linear polynomial 2x + 3. For every x, there is
a corresponding value of the polynomial 2x + 3. For instance, if x = 4, we
substitute 4 in the expression to get 2 × 4 + 3 = 11.
If x = –6, then we substitute –6 in the expression to get 2 × –6 + 3 = –9.
Fig. 2.3 shows this process as an input-output machine where the
input is the value of x and the output is the value of 2x + 3. This process can
be referred to as a function where the expression 2x + 3 is a function of
the variable x. You will learn more about functions in later grades.
Think and Reflect
We have learnt that to evaluate the value of an algebraic expression,
we substitute a value of the variable in the given expression. Consider
Example 3, where the wire is bent to form a rectangle. Here, the area
of the rectangle, 10x – x2
, is a function of x. Can you interpret this as
an input-output process? What value does the expression take when
x = 6 cm?
Note that 2x + 3 is a linear function, whereas 10x – x2
is a quadratic function.