Algebraic Formulae and Operations
In this concept, we will discuss the formulae and operations required to solve algebraic expressions that contain numbers, variables and arithmetic operations. We will discuss binomial expansion, addition, subtraction, multiplication and division of algebraic expressions.
An algebraic expression is an
expression that contains one or more numbers, one or more variables (represented
by letters) and one or more arithmetic operations. For example, ‘3x +
4y’ is an algebraic expression. The parts of the expression that are
separated by a ‘+’ or ‘–’ sign are called terms of the expression. In the
above example, ‘3x’ and ‘4y’ are two terms joined with an
arithmetic operator ‘+’. A term is a product of a number and one or more
variables. In ‘3x’, x is the variable and 3 is the coefficient of
x. Similarly, in the expression ‘xy 
3x2’ the first term ‘xy’ is the product of the number 1
(coefficient) and two variables x and y.
An expression which consists of only one term is known as a monomial. For example, the expression ‘3xyz2’ is a monomial containing three variables x, y and z.
An expression which consists of two terms is known as a binomial. For example, the expression ‘9x + 3y’ is a binomial containing two variables x and y.
An expression which consists of three terms is known as a trinomial. For example, the expression ‘12x + y + z2’ is a trinomial containing three variables x, y and z.
An expression which consists of more than one term is known as multinomial or polynomial. For example, the expression ‘2xy + 3x + 12z + 2’ is a multinomial containing four terms and three variables x, y and z. (Binomial and trinomial are also examples of multinomial or polynomial.)
The degree of a term is the addition of the exponent of each variable present in that term. For example, the degree of the term ‘3xyz2’ is (1 + 1 + 2) = 4.
The degree of a polynomial is the degree of a term having the highest degree. For example, the degree of the polynomial ‘2xy + 3x + 12z + 2’ is the same as the degree of the first term i.e. (1 + 1) or 2.
The root(s) of an equation is/are value(s) of x at which a polynomial f(x) = 0. In other words, a will be a root of f(x) if f(a) = 0.
If the remainder when a polynomial f(x) is divided by binomial of the form (x – a) is R, then f (a) = R. This is also called the Polynomial Remainder Theorem. (Refer to the chapter on Number Theory)
Explanation:
Here, f(x) is the dividend, (x – a) is the divisor and R is the remainder. Also, let q(x) be the quotient. Then,
f(x) = (x – a)
q(x) + R
Substituting x = a, we get,
f (a) = 0
q(a) + R = R
Also, if (x – a) is a factor of f(x), then a will be a root of f(x). This is because, if (x – a) completely divides f(x) without leaving any remainder, then R = 0, and so f (a) = 0.
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Find the degree of the following algebraic expression: x + 2x2y
Solution: The given algebraic expression is a trinomial. The degree of the first, second and third terms is 1, 3 and 5 respectively. Hence, the degree of the given expression is the same as the degree of the term having the highest degree i.e. 5. |
We can add, subtract, multiply and divide algebraic expressions with rules similar to those we follow in arithmetic. The terms which differ only in numerical coefficient are known as like terms. For example, ‘3x’ and ‘4x’ are like terms. Like terms can be combined by addition or subtraction. Let us take few examples to understand the four basic algebraic operations.
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If A = 5x + 7y2,
B = 2x
Solution: D = A + B + C
= (5x +
2x) + (7 y2
= 7x –
10y2 + 13xy |
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Solve: 2x
Solution: 2x(3xy2 + 4z) = 6x2y2 + 8xz |
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Divide
(12x4y2z3) by (
Solution: Let A =
(12x4y2z3)
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(
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What is the remainder when 2x4
Solution: Using the long division method,
Hence, the remainder is 13x – 14. |
Binomial theorem is used to find the expansion of the powers of sums. If n is a natural number that is greater than or equal to 2, then according to the binomial theorem:
(x + a)n =
nC0 xn a0 +
nC1 xn
1 a1 + nC2 xn
2 a2 + nC3 xn
3 a3 + … + nCn
x0 an
For example,
(x + a)2 =
2C0 x2 – 0a0 +
2C1 x2
1a1 + 2C2x2 –
2a2
= x2 + 2xa + a2
(x + a)3 =
3C0 x3 a0 +
3C1 x3
1a1 + 3C2x3
2a2 +
3C3x0a3
= x3 + 3x2a + 3xa2 + a3
Hence, the (k + 1)th term in the expansion of (x + a)n is nCk xn – k ak
Explanation:
(x + a)n = (x + a)(x + a)(x + a)(x + a)… n times
(x + a)n = 
1xn
+ 2xn
1 + 3xn
2 + … + n
+ 1,
where 1,
2,
3,
…, n
+ 1 are the coefficients of the 1st, 2nd,
3rd, …, (n + 1)th term
xn is produced by
multiplying the x’s in all the n factors. So, its co-efficient,
i.e. 1,
will be 1.
2xn
1 is produced by multiplying the x’s in (n – 1) factors and
one of the a’s from the remaining factor. There will be several terms of
the form axn
1, and 2
will be the sum of the coefficients of all of these terms. Here, there will be
nC1 terms (
there are n number of a’s, and each term is formed by selecting
one a of these n a’s). Thus, 2
= nC1
a
3xn
– 2 is produced by multiplying the x’s in (n – 2)
factors and two a’s from the remaining factors. There will be several
terms of the form a2xn
2, and 3
will be the sum of the coefficients of all of these terms. Here, there will be
nC2 terms (
there are n number of a’s, and each term is formed by selecting
two a’s of these n a’s). Thus, 3
= nC2
a2
Similarly, 4=
nC3
a3
.
.
.
n+1
= nCn
an + 1
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Find the fourth term of the expansion (a + b)6.
Solution: The fourth term of (a + b)6
= 6C3 a6 |
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Calculate the value of (1.08)9, approximated up to the 4th decimal place.
Solution: (1.08)9 = (1 + 0.08)9
Using the Binomial Theorem, this becomes,
(1 + 0.08)9 = 9C0 19 (0.08)0 + 9C1 18 (0.08)1 + 9C2 17 (0.08)2 + 9C3 16 (0.08)3 + …
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1 + 9
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If the term that does not contain the variable a in the expansion of
is equal to 12285, then find the value of the constant p.
Solution:
Now, we need to find that term in the expansion for which the exponents of a in a1/2 and p/a2 become equal. That is,
Hence, the exponents will be equal for the 4th term. Hence,
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What is the number of distinct terms in the expansion of (a +b + c)20? [CAT 2008]
(1) 231(2) 253(3) 242 (4) 210(5) 228
Solution: Consider (a + b + c)20,
We have to divide 20 into three parts which be done by using the
distribution rule 
Where, n is number of things to be distributed. r is number of parts into which the things are to be distributed.
Alternately, This can be solved without using much knowledge of permutations and combinations as follows, (a + b + c)1 = a + b + c [3 terms = (1 + 2) terms] (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac [6 terms = (1 + 2 + 3) terms] (a + b + c)3 = a3 + b3 + c3 +6abc + 3ab2 + 3ac2 + 3a2b + 3bc2 + 3a2c + 3b2c [10 terms = (1 + 2 + 3 + 4) terms] Similarly, (a + b + c)n will have (1 + 2 + 3 + … + (n + 1)) terms
Hence, option 1. |
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The coefficient of x7 in the expansion of (1 – x2 + x3)(1 + x)10 is: [IIFT 2009]
(1) 75(2) 78 (3) 85(4) None of the above
Solution: Let P = (1 – x2 + x3)(1 + x)10 The terms containing x7, x5 and x4 in the expansion of (1 + x)10 give terms containing x7 in the expression of P. The rth term in the Binomial Expansion of (a + b)n is given by Tr = nCr – 1 a(n – r – 1)b(r – 1)
T8 = 10C7 x7 The term containing x5 is given by T6 = 10C5x5 The term containing x4 is given by T5 = 10C4x4
x7 = 10C7 – 10C5 + 10C4 = 120 – 252 + 210 = 78 Hence, option 2. |
Some very important formulae are given below.
a2
b2 = (a + b)(a
b)
(a + b)2 = a2 + 2ab + b2
(a
b)2 = a2
2ab + b2
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a
b)3 = a3
3a2b + 3ab2
b3
a3 + b3
= (a + b)(a2
ab + b2)
a3
b3 = (a
b)(a2 + ab + b2)
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Let N = 553 + 173 – 723. N is divisible by [CAT 2000]
(1) both 7 and 13(2) both 3 and 13 (3) both 17 and 7(4) both 3 and 17 Solution: N = 553 + 173 It is known that if a + b + c = 0 then a3 + b3 + c3 = 3abc. In the given expression, 55 + 17 + (
Hence, option 4. |
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If a2 – b2 = 158, a3 – b3 = 79 and ab = 33, then find the value of (a + b)
Solution:
Let (a + b) = x. Also, ab = 33. Hence, the above equation becomes a quadratic equation: 2x2 – x – 66 = 0
Hence, (a + b) = 6 or (a + b) = –5.5 |
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If x3 + y3 = 387 and x + y = 3, then find the values of x and y.
Solution:
Also,
Subtracting equation (i) from (iii), we get,
xy = –120/3 = –40
Also, subtracting 4xy from equation (iii), we get,
x2 + 2xy +
y2 – 4xy = 9 – 4
Solving equations (ii) and (iv) simultaneously, we get,
x = 8 and y = –5 or x = –5 and y = 8 |
13… (iv)
