JEE Mathematics Algebraic Expressions Notes Set 01

Algebraic Expression and Its Terms

An algebraic expression is built from numbers and variables joined by +, -, x and division. In \( ax^2 + bx + c \) the terms are \( ax^2 \), \( bx \) and \( c \); here \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \), which does not depend on \( x \), is the absolute (constant) term. The letter \( x \) is a variable because it can take many values.

Factors

When parts are multiplied together to form a term or an expression, each part is a factor. So \( x \) and \( y \) are factors of \( xy \), and \( (p+q) \) and \( (r+s) \) are factors of \( (p+q)(r+s) \).

Kinds of Expression

1. Monomial: only one term, e.g. \( 8a^2x \), \( \frac{u}{v} \).
2. Binomial: two terms, e.g. \( p^2 + q^2 \), \( \frac{3}{n} - \frac{2}{p} \).
3. Trinomial: three terms, e.g. \( p^2 + 2pq + q^2 \), \( \frac{3a}{b} + \frac{3c}{d} + \frac{3e}{f} \).
4. Polynomial (multinomial): an expression with more than one term.
5. Homogeneous expression: every term has the same total degree. For example \( 2x^3 + 5y^3 - 5x^2z + 3xyz \) is homogeneous of the third degree in \( x, y, z \), because the powers in each term add up to 3.

Equation and Identity

An equation is true only for particular values of the variable. For \( 4x + 2 = 10 \) we get \( 4x = 8 \), so \( x = 2 \) is the only solution (the root). To solve an equation means to find all such values.
An identity is true for every value of the variable. For example \( 5x(x-2) + 2(5x+3) = 5x^2 + 6 \) holds for all \( x \), because the left side simplifies to the right side.

Like Terms and Terms of the Same Type

Terms with exactly the same variable part are like terms and can be added or subtracted: \( 5p^2q \) and \( 2p^2q \); \( \frac{xy}{z} \) and \( \frac{5xy}{z} \).
Terms are of the same type when one can be obtained from the other by interchanging the variables. In two variables \( x, y \): the type \( x^2 \) contains \( x^2, y^2 \); the type \( x^2y \) contains \( x^2y, xy^2 \). In three variables: the type \( xy \) contains \( xy, yz, zx \); the type \( x^2yz \) contains \( x^2yz,\; y^2zx,\; z^2xy \).

Degree

The degree of a term is the sum of the powers of its variables: \( 5x^2y \) and \( 6xy^2 \) are both of degree 3. Terms of the same type always have the same degree, but terms of the same degree need not be of the same type (\( x^2 \) and \( xy \) are both of degree 2 yet not of the same type).
The degree of a polynomial is the degree of its highest-degree term. A polynomial is homogeneous when all its terms have the same degree.
Examples: the degree of \( 2x^3 - 3x^2 - 5x + 7 \) is 3; the degree of \( ax^2y + bxy^2 + cx^3y \) is 4; \( 2x^2 + 3xy - 5y^2 \) is homogeneous of degree 2 in \( x, y \).

Example 1. Find the degrees of (i) \( xy^3 - 4x^2y^2 + x^3y^2 - xy \); (ii) \( 7x - y + xy - x^2y + y^2x \).
Answer:
(i) The degrees of the terms are 4, 4, 5, 2, so the highest is 5: degree 5.
(ii) The degrees are 1, 1, 2, 3, 3, so the degree is 3.

Example 2. Which of x = 1, x = 2, x = 3 make \( x^2 - 4x + 3 \) zero?
Answer:
Let \( f(x) = x^2 - 4x + 3 \). \( f(1) = 0 \), \( f(2) = 4 - 8 + 3 = -1 \), \( f(3) = 9 - 12 + 3 = 0 \).
So x = 1 and x = 3 make it zero; they are the roots of \( x^2 - 4x + 3 = 0 \).

Brackets

In \( 2(x+5) \) the number 5 is added to \( x \) and the result is multiplied by 2. A bar written over terms (a vinculum) groups them like a bracket, so \( 2\overline{x+3} = 2(x+3) \).
Distributive law: \( a(x + y) = ax + ay \), and more generally \( a(p+q+r) = ap + aq + ar \). This is visible as the area of a rectangle of height \( a \) and base \( x + y \), split into two rectangles of areas \( ax \) and \( ay \).

Removing Brackets

A bracket preceded by + keeps the signs of its terms; a bracket preceded by - changes every sign inside:
\( a + (b - c) = a + b - c \)
\( a - (b + c) = a - b - c \)
\( a - (b - c) = a - b + c \)
With nested brackets, remove the innermost bracket first and work outwards.

Example 3. Simplify \( 1 - [\,1 - \{\,1 - (1 - \overline{1+1})\,\}\,] \).
Answer:
Innermost: \( \overline{1+1} = 2 \), so \( 1 - 2 = -1 \).
Then \( 1 - (-1) = 2 \), so the curly bracket is 2.
Then \( 1 - 2 = -1 \) is the square bracket; finally \( 1 - (-1) = \)2.

Example 4. Simplify \( [\,x + \{x - (x - 2)\}\,] - [\,x - \{x - (x + 2)\}\,] \).
Answer:
First bracket: \( x - (x-2) = 2 \), so \( x + 2 \).
Second bracket: \( x - (x+2) = -2 \), so \( x - (-2) = x + 2 \).
Difference: \( (x+2) - (x+2) = \)0.

Example 5. Simplify \( a - b - [\,a + b - \{\,a - b + (a + b - \overline{a - b})\,\}\,] \).
Answer:
\( a + b - \overline{a-b} = a + b - a + b = 2b \).
\( a - b + 2b = a + b \); then \( a + b - (a+b) = 0 \).
Result: \( a - b - 0 = \)a - b.