In the previous section we saw some examples for the formation of algebraic expressions. In this section we will see a few more examples.
Solved example 20.1
In rectangles with one side 1 cm shorter than the other, take the length of the shorter side as 'x' cm.
(i) Taking perimeters as p(x), write the relation between p(x) and (x) as an equation
(ii) Taking their areas as a(x), write the relation between a(x) and (x) as an equation
(iii) Calculate p(1), p(2), p(3), p(4) and p(5)
(iv) Calculate a(1), a(2), a(3), a(4) and a(5)
Solution:
1. Length of the shorter side (b) = x cm
2. So length of the longer side (l) = (x+1)
3. Perimeter (p) = 2(l+b) = 2(x+1+x) = 2(2x+1) = 4x+2
4. Area (a) = l×b = x(x+1) = x2 + x
5. So the algebraic expression showing the 'relation between perimeter and x' for the given rectangle is: p(x) = 4x+2. This is solution of part (i)
6. Also, the algebraic expression showing the 'relation between area and x' for the given rectangle is:
a(x) = x2 + x. This is solution of part (ii)
7. To find p(1), put x = 1 in p(x)
• From (5), we have p(x) = 4x+2.
• So p(1) = 4×1 + 2 = 6
• Similarly, p(2) = 4×2 + 2 = 10
• p(3) = 4×3 + 2 = 14
• p(4) = 4×4 + 2 = 18
• p(5) = 4×5 + 2 = 22
This is solution of part (iii)
8. To find a(1), put x = 1 in a(x)
From (6), we have a(x) = x2 + x
So a(1) = 12+ 1 = 1 + 1 = 2
a(2) = 22+ 2 = 4 + 2 = 6
a(3) = 32+ 3 = 9 + 3 = 12
a(4) = 42+ 4 = 16 + 4 = 20
a(5) = 52+ 5 = 25 + 5 = 30
This is solution of part (iv)
Solved example 20.2
From the four corners of a rectangle 7 × 5 cm (fig.20.4.a), equal squares are cut off (fig.20.4.b), and the edges are folded up to make a box (fig.20.4.c).
(i) Taking sides of the squares as 'x' cm, write the dimensions of the box in terms of 'x'
(ii) Taking volume of the box as v(x) cm3, write the relation between v(x) and x as an equation
(iii) Calculate v(1⁄2), v(1) and v(11⁄2)
Solution:
1. When squares of side x cm are cut off,
• the remaining length becomes (7-2x) cm
• the remaining width becomes (5-2x) cm
• When these remaining edges are folded to make the box, the heigth of the box is x cm. So
♦ length of box (l) = (7-2x) cm
♦ width of box (b) = (5-2x) cm
♦ height of box (h) = x cm
This is solution of part (i)
2. Thus volume of box (v) = l×b×h = (7-2x)(5-2x)x
= (35-10x-14x+4x2)x = (35-24x+4x2)x = 35x-24x2+4x3
3. So the algebraic expression showing the 'relation between volume and x' for the given box is:
v(x) = 35x-24x2+4x3. This is solution of part (ii)
4. For calculating v(1⁄2), it is more convenient to use the initial expression in (2)
• We have v(x) = (7-2x)(5-2x)x.
• Put x = 1⁄2 . We get: v(1⁄2) = [(7-(2×1⁄2)][(5-(2×1⁄2)]×1⁄2
⇒ v(1⁄2) = [(7-1)][(5-1)]×1⁄2 = 6 × 4 × 1⁄2 = 12
■ For calculating v(1), we can use the final expression in (2)
• We have v(x) = 35x-24x2+4x3
• Put x = 1. We get: v(1) = (35 × 1) - (24 × 12) + (4 × 13)
⇒ v(1) = 35 - 24 + 4 = 15
■ For calculating v(11⁄2), it is more convenient to use the initial expression in (2)
We have v(x) = (7-2x)(5-2x)x.
Put x = 3⁄2 . We get: v(3⁄2) = [(7-(2×3⁄2)][(5-(2×3⁄2)]×3⁄2.
⇒ v(3⁄2) = [(7-3)][(5-3)]×3⁄2 = 4 × 2 × 3⁄2 = 12
This is solution of part (iii)
So far we have seen a number of algebraic expressions.
Solved example 20.1
In rectangles with one side 1 cm shorter than the other, take the length of the shorter side as 'x' cm.
(i) Taking perimeters as p(x), write the relation between p(x) and (x) as an equation
(ii) Taking their areas as a(x), write the relation between a(x) and (x) as an equation
(iii) Calculate p(1), p(2), p(3), p(4) and p(5)
(iv) Calculate a(1), a(2), a(3), a(4) and a(5)
Solution:
1. Length of the shorter side (b) = x cm
2. So length of the longer side (l) = (x+1)
3. Perimeter (p) = 2(l+b) = 2(x+1+x) = 2(2x+1) = 4x+2
4. Area (a) = l×b = x(x+1) = x2 + x
5. So the algebraic expression showing the 'relation between perimeter and x' for the given rectangle is: p(x) = 4x+2. This is solution of part (i)
6. Also, the algebraic expression showing the 'relation between area and x' for the given rectangle is:
a(x) = x2 + x. This is solution of part (ii)
7. To find p(1), put x = 1 in p(x)
• From (5), we have p(x) = 4x+2.
• So p(1) = 4×1 + 2 = 6
• Similarly, p(2) = 4×2 + 2 = 10
• p(3) = 4×3 + 2 = 14
• p(4) = 4×4 + 2 = 18
• p(5) = 4×5 + 2 = 22
This is solution of part (iii)
8. To find a(1), put x = 1 in a(x)
From (6), we have a(x) = x2 + x
So a(1) = 12+ 1 = 1 + 1 = 2
a(2) = 22+ 2 = 4 + 2 = 6
a(3) = 32+ 3 = 9 + 3 = 12
a(4) = 42+ 4 = 16 + 4 = 20
a(5) = 52+ 5 = 25 + 5 = 30
This is solution of part (iv)
Solved example 20.2
From the four corners of a rectangle 7 × 5 cm (fig.20.4.a), equal squares are cut off (fig.20.4.b), and the edges are folded up to make a box (fig.20.4.c).
 |
| Fig.20.4 |
(ii) Taking volume of the box as v(x) cm3, write the relation between v(x) and x as an equation
(iii) Calculate v(1⁄2), v(1) and v(11⁄2)
Solution:
1. When squares of side x cm are cut off,
• the remaining length becomes (7-2x) cm
• the remaining width becomes (5-2x) cm
• When these remaining edges are folded to make the box, the heigth of the box is x cm. So
♦ length of box (l) = (7-2x) cm
♦ width of box (b) = (5-2x) cm
♦ height of box (h) = x cm
This is solution of part (i)
2. Thus volume of box (v) = l×b×h = (7-2x)(5-2x)x
= (35-10x-14x+4x2)x = (35-24x+4x2)x = 35x-24x2+4x3
3. So the algebraic expression showing the 'relation between volume and x' for the given box is:
v(x) = 35x-24x2+4x3. This is solution of part (ii)
4. For calculating v(1⁄2), it is more convenient to use the initial expression in (2)
• We have v(x) = (7-2x)(5-2x)x.
• Put x = 1⁄2 . We get: v(1⁄2) = [(7-(2×1⁄2)][(5-(2×1⁄2)]×1⁄2
⇒ v(1⁄2) = [(7-1)][(5-1)]×1⁄2 = 6 × 4 × 1⁄2 = 12
■ For calculating v(1), we can use the final expression in (2)
• We have v(x) = 35x-24x2+4x3
• Put x = 1. We get: v(1) = (35 × 1) - (24 × 12) + (4 × 13)
⇒ v(1) = 35 - 24 + 4 = 15
■ For calculating v(11⁄2), it is more convenient to use the initial expression in (2)
We have v(x) = (7-2x)(5-2x)x.
Put x = 3⁄2 . We get: v(3⁄2) = [(7-(2×3⁄2)][(5-(2×3⁄2)]×3⁄2.
⇒ v(3⁄2) = [(7-3)][(5-3)]×3⁄2 = 4 × 2 × 3⁄2 = 12
This is solution of part (iii)
Algebraic expressions like:
p(x) = 10 + 4x
a(x) = 6 + 5x + x2
v(x) = x3 + 6x2 + 11x + 6
are called polynomials. But
they must satisfy some conditions. Then only they will be
called as polynomials. Let us analyse the various properties first:
• In p(x) = 10 + 4x, there are
two components on the right side of the '=' sign
• Each component is called a term. So one term is '10', and the other term is '4x'
• Every term can be split into
two parts: A coefficient and a variable part. In '4x', 4 is the
coefficient, and x is the variable part.
• '10' does not seem to have a
variable part. But in reality, it also has a variable part. This can
be explained as follows:
♦ 10 can be written as 10 × 1
♦ Now, this 1 is same as x0
♦ So 10 = 10x0
♦ Thus, in the term '10', 10 is
the coefficient, and x0 is the variable part.
• Variables are usually denoted
by letters x, y or z
• In a polynomial, the variables
can take different values. But there are some terms that will not
change values
• For example, in the polynomial
'p(x) = 10 + 4x' , what ever be the value of x, values of 10 and 4
does not change. Such values are called constants.
• Constants are usually denoted
by letters a, b, c etc.,
■ Let us take the next
polynomial:
• a(x) = 6 + 5x + x2
• Terms: 6, 5x and x2
• Splitting the terms:
♦ 6: coefficient is 6, variable part is x0
♦ 5x: coefficient is 5, variable
part is x
♦ x2: coefficient is 1 (∵ x2 = 1x2), variable part is x2
• Constants: 6 and 5 are the
constants
• This polynomial has only one
variable 'x'
■ Let us take the next
polynomial:
• v(x) = x3 + 6x2 + 11x + 6
• Terms: x3, 6x2, 11x and 6
• Splitting the terms:
♦ x3: coefficient is 1, variable part is x3
♦ 6x2: coefficient is 6, variable
part is x2
♦ 11x: coefficient is 11,
variable part is x
• Constants: 6, 11 and 6 are the
constants
• This polynomial has only one
variable 'x'
■ Let us consider one more
example:
• r(y) = 2y2 – 5y – 3
• Terms: 2y2, 5y, and 3
• Splitting the terms:
♦ 2y2: coefficient is 2, variable part is y2
♦ 5y: coefficient is 5, variable
part is y
♦ 3: coefficient is 3, variable
part is y0
• Constants: 2, 5 and 3 are the
constants
• This polynomial has only one
variable 'y'
Number of terms in a polynomial
■ Polynomials having only one
term are called monomials ('mono' means one)
Examples:
• a(x) = 5x
• t(u) = u2
• p(x) = x3
■ Polynomials having only two
terms are called binomials ('bi' means two)
Examples:
• p(y) = y2+2
• a(x) =
5x2+2x
• r(x) = x12 - 1
■ Polynomials having only three
terms are called trinomials ('tri' means three)
Examples:
• r(x) = x2-x+2
• p(y) = y4+y2-1
• t(y) = y32-2y+7
■ A polynomial can have a large
number of terms. An example with 151 terms is given below:
t(x) = x150 + x149 + x148 +
..... + x2 + x + 1
Conditions for an algebraic expression to be called as a polynomial
• We have seen so many
polynomials above. All of them have a definite number of terms. Look
at the sign between those terms. We find that the sign will either be
'+' or '-'.
• That means, there are only two possible operations between
the terms of a polynomial: addition or subtraction.
• If any other
operation is involved, it is not a polynomial.
Condition 2:
• We have seen that every term
in a polynomial can be split into two parts. A coefficient and a
variable part.
• Consider the variable part.
Every variable will be having an exponent. (If there is no exponent
written, it is understood that, the exponent is '1').
• For an algebraic expression
to be a polynomial, the exponent in every term should be a whole
number. We have seen whole numbers here. So the exponent should
belong to the list: 0, 1, 2, 3. . .
♦ The exponent cannot be a
negative number.
♦ The exponent cannot be a
fraction
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