Algebra

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Definition:

study is the use of letters.

Use of letters will allow us to write rules and formulas in a general way.

letters stand for numbers, operations can be performed on them as on numbers.

This leads to the study of algebraic expressions and their properties

Variable:

The word ‘variable’ means something that can vary, i.e. change.

The value of a variable is not fixed. It can take different values.

One may use any letter as m, l, p, x, y, z etc. to show a variable.

Applications : In Geometry : In calculating Area, perimeters etc, In Arithmetic: In Properties of Addition, Subtraction, Multiplication

An expression like (4x + 5), which contains the variable x, cannot be evaluated. Only if x is given some value, an expression like (4x + 5) can be evaluated.

Equation

is a condition on a variable.

It is satisfied only for a definite value of the variable.

An equation has an equal sign (=) between its two sides.

The equation says that the value of the left hand side (LHS) is equal to the value of the right hand side (RHS).

If the LHS is not equal to the RHS, we do not get an equation.

An equation remains the same, when the expressions on the left and on the right are interchanged.

Thus if we add the same number to both sides of an equality, it still holds.

if we subtract the same number from both sides of an equality, it still holds.

if we multiply or divide both sides of the equality by the same non-zero number, it still holds.

Changing side is called transposing. While transposing a number, we change its sign.

Transposing a number (i.e., changing the side of the number) is the same as adding or subtracting the number from both sides.

The statement 2n is greater than 10, i.e. 2n > 10 is not an equation.

The value of the variable in an equation which satisfies the equation is called a solution to the equation.

Thus, n = 5 is a solution to the equation 2 n = 10.

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Algebraic Expressions

Combination of Variables and Constants with Variables having changing value and constants having fixed values

we use the operations of addition, subtraction, multiplication and division.

Terms are added to form expressions.

Factors of a term

expression (4x2 – 3xy) consists of two terms 4x2 and –3xy.

The term 4x2 is a product of 4, x and x; we say that 4, x and x are the factors of the term 4x2.

A term is a product of its factors.

The term –3xy is a product of the factors –3, x and y.

Note, in the tree diagram, we have used dotted lines for factors and continuous lines for terms. This is to avoid mixing them.

Coefficients

The numerical factor is said to be the numerical coefficient or simply the coefficient of the term.

In the term 10xyz, 10 is the coefficient of xyz, in the term –7x2y2, –7 is the coefficient of x2y2.

Like and Unlike Terms

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Following simple steps will help you to decide whether the given terms are like or unlike terms:

(i) Ignore the numerical coefficients. Concentrate on the algebraic part of the terms.

(ii) Check the variables in the terms. They must be the same.

(iii) Next, check the powers of each variable in the terms. They must be the same.

Note that in deciding like terms, two things do not matter (1) the numerical coefficients of the terms and (2) the order in which the variables are multiplied in the terms.

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Monomials

An expression with only one term is called a monomial; for example, 7xy, – 5m, 3z2, 4 etc.

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Binomials

An expression which contains two unlike terms is called a binomial;

for example, x + y, m – 5, mn + 4m, a2 – b2 are binomials.

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Trinomials

An expression which contains three terms is called a trinomial;

for example, the expressions x + y + 7, ab + a +b, 3x2 – 5x + 2, m + n + 10 are trinomials.

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Polynomials

In general, an expression with one or more terms is called a polynomial.

Thus a monomial, a binomial and a trinomial are all polynomials.

The constant polynomial 0 is called the zero polynomial.

A polynomial can have any (finite) number of terms.

Highest power of the variable in a polynomial as the degree of the polynomial. So, the degree of the polynomial 3x7 – 4x6 + x + 9 is 7 and the degree of the polynomial 5y6 – 4y2 – 6 is 6. The degree of a non-zero constant polynomial is zero.

A polynomial of degree one is called a linear polynomial.

A polynomial of degree two is called a quadratic polynomial.

Polynomial of degree three a cubic polynomial.

zero of the polynomial p(x) is the value of the polynomial being 0 at some value of x

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.

zero of the linear polynomial ax + b is (Constant term) = -b/ a = - (constant term) / coefficient of x

Geometrical Meaning of the Zeroes of a Polynomial

Remainder Theorem

Dividend = (Divisor × Quotient) + Remainder

In general, if p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that:p(x) = g(x)q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x). Here we say that p(x) divided by g(x), gives q(x) as quotient and r(x) as remainder.

Find the remainder obtained on dividing p(x) = x3 + 1 by x + 1.

Division by Long Division Method

Remainder Theorem : Let p(x) be any polynomial of degree greater than or

equal to one and let a be any real number. If p(x) is divided by the linear

polynomial x – a, then the remainder is p(a).

Proof : Let p(x) be any polynomial with degree greater than or equal to 1. Suppose

that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x), i.e.,

p(x) = (x – a) q(x) + r(x)

Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a,

the degree of r(x) = 0. This means that r(x) is a constant, say r.

So, for every value of x, r(x) = r.

Therefore, p(x) = (x – a) q(x) + r

In particular, if x = a, this equation gives us

p(a) = (a – a) q(a) + r

= r,

which proves the theorem.

Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Solution : Here, p(x) = x4 + x3 – 2x2 + x + 1, and the zero of x – 1 is 1.

So, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1

= 2

So, by the Remainder Theorem, 2 is the remainder when x4 + x3 – 2x2 + x + 1 is

divided by x – 1.

Relationship between Zeroes and Coefficients of a Polynomial

zero of a linear polynomial ax + b is -b/a

So, the value of p(x) = 2x2 – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., whenx = 1 or x = 3. So, the zeroes of 2x2 – 8x + 6 are 1 and 3. Observe that :

Sum of its zeroes = alpha + beta= 1 +3 = 4 = -(-8)/ 2 = -Coefficient of x) / Coefficient of x2

Poduct of its Zeroes = alpha*beta = 1* 3 = 3 = 6/2 = constant term/ coeffiient of x2

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Addition & Subtraction of Algebric Expressions

The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.

Similarly, the difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.

Note, unlike terms cannot be added or subtracted the way like terms are added or subtracted.

Note, subtracting a term is the same as adding its inverse. Subtracting –10b is the same as adding +10b;

1. 3x+ 4x

Here the problem has two terms

3x and 4x

both are like terms of variable x

Hence add the co-efficients

7x is the answer

2. if we add the unlike terms 3xy and 7, the sum is 3xy + 7.

3. Add 3x + 11 and 7x – 5

3x +11 +7x -5

10x + 6

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Finding the Value of an expression

We know that the value of an algebraic expression depends on the values of the variables forming the expression.

There are a number of situations in which we need to find the value of an expression, such as when we wish to check whether a particular value of a variable satisfies a given equation or not.

Eg: 5n – 2 where n=2

5(2) - 2 = 10 -2 =8

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Using Algebric Expressions: Formulas & Rules

Perimeter formulas

the perimeter of the equilateral triangle = 3l

the perimeter of a square = 4l

Perimeter of a regular pentagon = 5l

Area formulas

1. If we denote the length of a square by l, then the area of the square = l22. If we denote the length of a rectangle by l and its breadth by b, then the area of the rectangle = l × b = lb.

3. Similarly, if b stands for the base and h for the height of a triangle, then the area of the triangle = (b × h)/ 2 = bh/ 2

2 2

.

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Multiplication of Algebraic Expressions:

Multiplying a Monomial by a Monomial

5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy

Multiplying three or more monomials

2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz

Multiplying a monomial by a binomial

(5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x

Multiplying a monomial by a trinomialConsider 3p × (4p2 + 5p + 7). As in the earlier case, we use distributive law;

3p × (4p2 + 5p + 7) = (3p × 4p2) + (3p × 5p) + (3p × 7)

= 12p3 + 15p2 + 21p

Multiplying a binomial by a binomial

(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)

= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)

= 6a2 + 9ab + 8ba + 12b2

= 6a2 + 17ab + 12b2

In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.

Multiplying a binomial by a trinomial

(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)= 2a2 – 3ab + ac + 2ab – 3b2 + bc

= 2a2 – ab – 3b2 + bc + ac

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Identity

We shall find that for any value of a, LHS = RHS. Such an equality, true for every value of the variable in it, is called an identity

Thus, (a + 1) (a + 2) = a2 + 3a + 2 is an identity.

An equation is true for only certain values of the variable in it. It is not true for all values of the variable.

For example, consider the equation a2 + 3a + 2 = 132 . It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc. Try it: Show that a2 + 3a + 2 = 132 is not true for a = –5 and for a = 0.

Standard Identities

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

(a + b) (a – b) = a2 – b2

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Factorisation

When we factorise an algebraic expression, we write it as a product of factors.

These factors may be numbers, algebraic variables or algebraic expressions

Method of common factors

We begin with a simple example: Factorise 2x + 4.

We shall write each term as a product of irreducible factors;

2x = 2 × x

4 = 2 × 2

Hence 2x + 4 = (2 × x) + (2 × 2)

Factorisation by regrouping terms

2xy + 2y = (2 × x × y) + (2 × y)= (2 × y × x) + (2 × y × 1)

= (2y × x) + (2y × 1) = 2y (x + 1)

Similarly, 3x + 3 = (3 × x) + (3 × 1)

= 3 × (x + 1) = 3 ( x + 1)

Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)

Observe, now we have a common factor (x + 1) in both the terms on the right hand

side. Combining the two terms,

2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)

The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its

factors are (x + 1) and (2y + 3). Note, these factors are irreducible.

Factorisation using identities

Factorise x2 + 8x + 16

a2 + 2ab + b2 = x2 + 2 (x) (4) + 42

= x2 + 8x + 16

Since a2 + 2ab + b2 = (a + b)2,

by comparison x2 + 8x + 16 = ( x + 4)2 (the required factorisation)

Factors of the form ( x + a) ( x + b)

(x + a) (x + b) = x2 + (a + b) x + ab

If a b = 6, it means that a and b are factors of 6. Let us try a = 6, b = 1. For these

values a + b = 7, and not 5, So this choice is not right.

Let us try a = 2, b = 3. For this a + b = 5 exactly as required.

The factorised form of this given expression is then (x +2) (x + 3).

Factorisation of Polynomials

Factor Theorem : If p(x) is a polynomial of degree n > 1 and a is any real number,

then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).

Proof: By the Remainder Theorem, p(x)=(x – a) q(x) + p(a).

(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).

(ii) Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x).

In this case, p(a) = (a – a) g(a) = 0.

Factorise y2 – 5y + 6 by using the Factor Theorem.

Solution : Let p(y) = y2 – 5y + 6. Now, if p(y) = (y – a) (y – b), you know that the

constant term will be ab. So, ab = 6. So, to look for the factors of p(y), we look at the

factors of 6.

The factors of 6 are 1, 2 and 3.

Now, p(2) = 22 – (5 × 2) + 6 = 0

So, y – 2 is a factor of p(y).

Also, p(3) = 32 – (5 × 3) + 6 = 0

So, y – 3 is also a factor of y2 – 5y + 6.

Therefore, y2 – 5y + 6 = (y – 2)(y – 3)

Note that y2 – 5y + 6 can also be factorised by splitting the middle term –5y.

Now, let us consider factorising cubic polynomials. Here, the splitting method will not

be appropriate to start with. We need to find at least one factor first, as you will see in

the following example.

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Division of Algebraic Expressions

Division of a monomial by another monomial

6x3 ÷ 2x = = 2 *3*X*X*X/ 2 *X = 3 *X*X

Division of a polynomial by a monomial

Let us consider the division of the trinomial 4y3 + 5y2 + 6y by the monomial 2y

4y3 + 5y2 + 6y =2 × y × (2 × y × y) + 2 × y ×5

2

⎛⎜⎝ × y⎞⎟⎠ + 2 × y × 3

= 2y (2y2) + 2y

5

2

⎛ y⎞

⎜⎝ ⎟⎠ + 2y (3)

= 2 2 2 5 3

2

y ⎛⎜⎝ y + y + ⎞⎟⎠ (The common factor 2y is shown separately.

Division of Algebraic Expressions (Polynomial ÷ Polynomial)

Consider (7x2 + 14x) ÷ (x + 2)

We shall factorise (7x2 + 14x) first to check and match factors with the denominator:

7x2 + 14x = (7 × x × x) + (2 × 7 × x)

= 7 × x × (x + 2) = 7x(x + 2)

Division Algorithm for Polynomials

Divide 2x2 + 3x + 1 by x + 2.

If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find

polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x).

This result is known as the Division Algorithm for polynomials.

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Linear Equations in One Variable

equations with linear expressions in one variable only. Such equations are known as linear equations in one variable.

Eg: 2x – 3 = 7

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Linear Equation with two Variables

a linear equation in two variables has infinitely many solutions.

an equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y.

So, every solution of the equation is a point on the line representing it.

Graph of a Linear Equation in Two Variables

The graph of the equation of the form y = kx is a line which always passes through the origin.

Grph to be plotted

Graphical Method of Solution of a Pair of Linear Equations(Intersecting, coinciding and parallel lines)

Algebraic Methods of Solving a Pair of Linear Equations

Substitution Method

Elimination Method

Cross - Multiplication Method

Equations Reducible to a Pair of Linear Equations in Two Variables

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Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0.

For example, 2x2 + x – 300 = 0 is a quadratic equation.

That is, ax2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

Solution of a Quadratic Equation by Factorisation

In general, a real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a α2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same.

Solution of a Quadratic Equation by Completing the Square

Consider the quadratic equation ax2 + bx + c = 0 (a ≠ 0). Dividing throughout bya, we get x2+ b/a)x +c/a)= 0

(x+b/2a)2 - (b/2a)2 +c/a =0

(x+b/2a)2 - (b2-4ac)/ 2a =0

Thus, if b2 – 4ac ≥ 0, then the roots of the quadratic equation

ax2 + bx + c = 0 are given by

– ± 2 – 4

2

b b ac

a

This formula for finding the roots of a quadratic equation is known as the

quadratic formula.

Nature of Roots

If b2 – 4ac > 0, we get two distinct real roots

If b2 – 4ac = 0, then x

Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 hasreal roots or not, b2 – 4ac is called the discriminant of this quadratic equation.

So, a quadratic equation ax2 + bx + c = 0 has

(i) two distinct real roots, if b2 – 4ac > 0,

(ii) two equal real roots, if b2 – 4ac = 0,

(iii) no real roots, if b2 – 4ac < 0.