Quantitative Aptitude Elementary Algebra Study Material

Algebriac Expressions (Polynomials)

Some useful terms defined :

Constants : A symbol having a fixed numerical value is called a constant, such as 4, -5, 2/5 ,π etc.

Variables : A symbol which takes on various numerical values is known as a variable.
We know that the circumference of a circle is given by the formula C = 2 πr where r is the radius of the circle. Here, 2 and π are constant, whereas C and r are variables.

Algebraic Expressions : A combination of constants and variables, connected by the signs of fundamental operations (i.e. +, -, x, /) is known as an algebraic expression.

The several parts of the expression in separated by + or – sign are called the terms of the expression.

Polynomials : An algebraic expression in which the variables involved have only non-negative integral powers, is called a polynomial.

Degree of Polynomial in One Variable : In case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial.
Ex.: (i) 2x – 3 is a polynomial in x of degree 1.
(ii) 5x⁴ – 3x² + 5 is a polynomial in x of degree 4.

Degree of Polynomial in Two or More Variables: In case of polynomials in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is known as the degree of the polynomial.
Ex. (i) 7x⁴y – 5x³y + 9y² is polynomial of degree 5 in x and y.
(ii) 3x³ – 5x⁴y³ + 9y² is a polynomial of degree 7 in x and y.

Polynomials of Various Degrees

• A polynomial of degree 1 is called a linear polynomial, e.g., 3 + 5x. 5 – 3y
• A polynomial of degree 2 is called a quadratic poly­nomial.e.g. 5x² + 4x + 3, 3y²-2y+1

• A polynomial of degree 3 is called a cubic polyno­mial.e.g. x³ + 3x² + 5x + 3
• A polynomial of degree 4 is called a biquadratic poly­nomial e.g. 8x⁴ + 3x² + 5x² + 4

Number of Terms in a Polynomial: A polynomial is said to be a monomial, a binomial or a trinomial according as it contains one term, two terms or three terms respectively.

Ex. (i) 4,7x, 9xv, 5xyz   —Monomials.

• 4 + 3x, 5x – 3y. 3x² + 2y — Binomials.
• 5x³ + 2x + 3. xv + yz + zx —Trinomials.

Constant Polynomials : A polynomial having one term consisting of a constant, is called a constant polynomial.

Ex. 3,5, – 5,7 etc.

Terms of a Polynomial in Ascending or Descending or­der : The terms of a polynomial are said to be in ascending or descending order according as they increase or decrease in degrees respectively.
Ex. (i) 3 – 5x + 7x² + 9x³ — In ascending order (ii) 9x³ + 7x² – 5x + 3 — In descending order

Like Terms : Terms with same variables and their same exponents are called like or similar terms, otherwise they are called unlike or dissimilar terms.
Ex. (i) 3x²,5x², – 7x² — Like terms.
(ii) 5x²y, 3xy²,7x⁴ — Unlike terms.
Rule : Two or more polynomials can be added by arrang­ing their terms and combining the like terms.

Subtraction :
Rule : Rearrange the terms of the two polynomials so that like terms occur in the same vertical columns. Now change the sign of each term of the polynomials to be subtracted and add the two polynomials.

Remember: Sign convention.

Case -1: (+) + (+) = (Add with + ve sign)
Case – II: (+) + (-) = (Subtract with the sign of greater number)
Case – III: (-) + (-) = (Add with negative sign).

Example: 5 + 3 = 8 (Case I)
5- 6 = 5 + (-6) = – 1 (Case II.)
(Since 6 has negative sign and it is greater than 5)
6-5= 1 (Case II)
-3 – 2 = – 5 (Case III)

Multiplication of Polynomials
Remember :

• The product of two factors with like signs is positive and those with unlike signs is negative, e.g., 5×4 = 20 like signs

• 5 x – 4 = 20 like signs
• 5 x 4 = – 20 unlike signs
• If x is any variable and m, n are positive integers, then x^m x xⁿ = x^m+n.

Thus, x⁵ x x⁷ = x⁵⁺⁷ = x¹¹

Rule : The co-efficient of the product of two monomials is equal to the product of their coefficients and the vari­able part in the product is the product of variables in the given monomials. In case of two polynomials, multiply each term of the multiplicand by each term of the multiplier and take the algebraic sum of these products,
e.g., 7x³ x – 5x⁴ = – (7x – 5) (x³ + x⁴)
= – 35x⁷ – 9x² x – 7x³ = (9 x -7) (x² + x³) = 63x⁵ etc.

Division of Polynomial by a Polynomial.

We may proceed according to the steps given below

(a) Arrange the terms of the dividend and divisor in descending order.
(b) Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
(c) Multiply all the terms of the divisor by the first term of the quotient and subtract the result from the dividend.
(d) Consider the remainder (if any) as a new dividend and proceed as before.
(e) Repeat this process till we obtain a remainder which is either 0 or polynomial of degree less than that of the divisor.

Remember: Like the numbers, the result dividend = divisor x quotient + remainder also hold for polynomials.
Finding remainder by Division method takes much time. We can easily find remainder by using remainder theorem.
Remainder Theorem – It gives a method for finding the remainder without actual division. When a polynomial P (x) with degree > 1 is divided by a binomial of the form (x -a), then the remainder is P (a).

Corollary : (i) if P (x) be divided by (x + a), the remainder is P (-a).
[ x + a = x-(-a)]
ii) If P (x) be divided by ax + b, the remainder is P (-b/a)
[ ax + b = 0; ax = – b, x = -b/a ]
Ex. 1. Without actual division, find the remainder when x⁴– 3x³ + 4x² – 6x + 7 is divided by x – 1.
Sol.: Here, the divisor is x – 1
So, a = 1
.-. The remainder on division by (x -1) = P (1)
.-. P(l) = (l)⁴ – 3(1)³ + 4(1)² – 6(1) + 7
= l-3 + 4-6 + 7 = 3

Ex.2. If the expressions Px³ + 3x²-3 and 2x³ – 5x + P when divided by x – 4 leave same remainder. Find the value of P.
Sol. Remainders are
R, = f (4) = P (4)³ + 3 (4)² – 3 = 64P + 45 R₂ = f (4) = 2 (4)³ -5 (4) + P = P + 108 Since, R_} = R₂, Hence .
64P + 45 = P + 108 or, 63 P = 63 => P = 1

Factorization of Polynomials

: The factor theorem based on remainder theorem is useful in the factorization of the polynomials.
Factor Theorem : If there is no remainder it is to be natu­rally concluded that the given polynomial is completely di­visible the given divisor.
Statement: Let P (x) be a polynomial and a be a real num­ber. Then (x-a) is a factor of P(x) iffP(a) = 0

1. , (i) if (x-a) is a factor of P (x), then P (a) = 0

• If P (a) = 0, then (x-a) is factor of P (x).

Ex.: Let f (x) = x³ + 2x² – 5x – 6. Find out whether x – 2, x + 3 and x -4 are factors of f (x).

Sol. We know that if (x – a) is factor of f (x), then f (a) = 0

• f (2) = 2³ + 2(2)² – 5(2) -6 = 0 Therefore, (x – 2) is a factor of f (x)
• (ii) f (- 3) = (- 3)³ + 2 (-3)² – 5 (- 3) – 6 = 0 Therefore, x + 3 is factor of f (x)
• f (4) = 4³ + 2(4)² – 5(4) – 6 = 70

Since f (4) ≠0, therefore, (x – 4) is not a factor of f (x).

Conditions of Divisibility

1. xⁿ + aⁿ is exactly divisible by (x + a) only when n is odd. e.g., a⁷ + b⁷ is exactly divisible by (a + b).

2. xⁿ + aⁿ is not exactly divisible by (x + a) when n is even. e.g., a¹⁰ + b¹⁰ is not exactly divisible by a + b.

3. xⁿ + aⁿ is never divisble by x – a.
4. xⁿ – aⁿ is exactly divisible by (x + a) when n is even, e.g., x¹⁰ – a¹⁰ is exactly divisible by x + a.
5. xⁿ-aⁿ is exactly divisible by x – a irrespective of n being odd or even.

Example : Find the last two digits of the expansion of 4⁶ⁿ – 6⁴ⁿ when n is any positive integer.

Sol.: 4⁶ⁿ – 6⁴ⁿ = (4³)²ⁿ – (6²)²ⁿ = (64)²ⁿ – (36)²ⁿ = x²ⁿ – a²ⁿ (Let)

Because 2n is an even integer, it must be divisible by (x + a) i.e. (64 + 36 = 100). Hence, 100 is a factor of above expression. Therefore, the last two digits of its expansion (i.e. Unit’s and Ten’s place) must be 00.

Factorizing a Polynomial
List of Important Formulae :

• (a + b)² = a² + 2ab + b²
• (a – b)² = a² – 2ab + b²
• (a + b)² = (a – b)² + 4ab
• (a – b)² = (a + b)² – 4ab
• a² – b² = (a + b) (a – b) .
• a³ + b³ = (a + b) (a² – ab + b²)
• a³ – b³ = (a – b) (a² + ab + b²)
• (a + b)³ = a³ + b³ + 3ab (a + b)
• (a – b)³ = a³ – b³ – 3ab (a – b)
• a³ + b³ = (a + b)³ – 3ab (a + b)
• a³ -b³ = (a-b)³ + 3ab (a-b)

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – be – ac) = (a + b + c) — (2a² + 2b² + 2c² – 2ab – 2bc – 2ac)

• If a + b + c = 0, then a³ + b³ + c³ = 3abc
• (a + b + c)³ = a³ + b³ + c³ + 3 (b + c) (c + a) (a + b)
• a² + b² = (a + b)² – 2ab
• a² + b² = (a – b)² + 2ab
• (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc)

Method of Splitting the Middle Term

Factors of the polynomial of the form x² + px + q. Examine the following examples carefully (x + 2) (x + 5) = x² + (2 + 5) x + 2 x 5 = x² + 7x + 10 Note that, in the product x² + 7x + 10, 10 is obtained by multiplying together 2 and 5 and 7 is got by adding together 2 and 5.

Hence, in order to find out factor of x² + 7x x 10, we must find two numbers whose product is 10 and sum is 7. Since (x + a) (x + b)
= x² + (a + b) x + ab
.-. x² + (a + b)x + ab = (x + a) (x + b)
Example: x² + 14x + 45
Sol.: x² + 14x + 45
14 = 9 + 5 and 45 = 9 x 5
.-. x² + 14x + 45 = x² + 9x + 5x + 45
= x (x + 9) + 5 (x + 9)

Linear Equations in one Variable
An equation involving only linear polynomials, is called a linear equation, e.g.,
(i) 5x + 2= 12                       (ii) 12x + 4 = 4x + 28

Solution of Linear Equation
A value of the variable or unknown, that satisfies the equation, is known as the solution of the equation. Consider the linear equation :
5x-2= 18

If we substitute x = 4, we get L.H.S. = 5x-2 = 5×4-2=18 R.H.S. = 18 .-. L.H.S. = R.H.S.
Thus, 4 is a solution of 5x – 2 = 18
Hence, a value of the variable which, when substituted for the variable in the equation, makes the two sides of the equation equal is the solution of equation.

Solving an equation : Solving a linear equation means finding a value of the variable which satisfies the equation.
An equation remains unaffected by any of the following operations :

• Adding the same quantity on both sides of the equal­ity.
• Subtracting the same quantity from both sides of the equality.
• Multiplying both sides of an equality by the same non-zero number.
• Dividing both sides of an equality by the same non­ zero number.

Examples: 1. 2 + x = 2x + 4 or, 2x + 4 = 2 + x or, 2x – x = 2 – 4 or, x = -2

2. 5 (x + 1) + 9 = 2 (3x + 1) or, 5x + 5 + 9 = 6x + 2 or, 5x – 6x = 2 – 5 – 9 = -12 or, – x = – 12 or, x = 12

Some problems involving linear equations.

The application of linear equations in the solution of problems is of great use. It is usual to take the following steps in the solutions of problems :

1. Read the problems carefully and denote the unknown quantity by the letter x.
2. Express the condition of the problem in symbolical lan­guage and form an equation.
3. Solve the equation. Some steps can be left for quick calculation.

Linear Equations in Two Variables

Definition : An equation of the form ax + by + c = 0 or ax + by = c where a, b, c are real numbers, a + 0, b + 0 and x, y are variables is called a linear equation in two variables.
Example: (i) 2 + 2y = 3
(ii) 3x + 4y = 6 etc.
Simultaneous linear equations in two variables. Definition : A pairs of linear equations, in two variables is said to form a system of simultaneous linear equations.
Solution : A pair of values of x and y satisfying each one of the equations in a given system of two simultaneous equations in x and y is called a solution of the system.
Algebraic method of solving simultaneous linear equations in two variables.
Method of elimination by equating the co-efficients.
Algorithm:
Step I: Obtain the two equations.
Step II : Multiply the equations so as to make the coefficient of the variable to be eliminated equal.
Step III : Add or subtract the equations obtained in Step II according as the terms having the same co-efficients are of opposite or of the same sign.
Step IV : Solve the equation in one variable obtained in Step III.
Step V : Substitute the value found in step IV in any one of the given equations and find the value of the other variable.
The values of the variables in Steps IV and V constitute the solution of the given system of equations.

Quadratic Equations And Inequation

Quadratic Equation : A quadratic equation is an equation of the form ax² + bx + c = 0 where a, b, c are given numbers (constant) and a # 0 and x is an unknown quan­tity (variable).              .•
The numbers a, b and c are called the coefficients of the quadratic equation; a is the coefficient of x², b is the coef­ficient of x and c is the constant term.

Ex. 3x² + 2x – 1= 0 ,is a quadratic equation.
Here, a=3, b = 2, c = -l
Roots or zeros of the quadratic equation : The values of the variable (x) satisfying the given equation are called its roots.
For example, x = 1 is a root of the equation x² – 3.x + 2 = 0 because l²-3 x 1 +2 = 0