| Classwise Concept with Examples | ||||||
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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
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| Algebraic Expression and Polynomial | Addition and Subtraction of Algebraic Expressions | Multiplication of Algebraic Expressions |
| Algebraic Identity | Division of Algebraic Expressions | |
Chapter 9 Algebraic Expressions and Identities (Concepts)
Welcome to this crucial chapter that significantly advances our journey into the world of algebra, building directly upon the foundational work with algebraic expressions undertaken in Class 7. We will revisit key concepts but quickly pivot to more complex operations, focusing particularly on the multiplication of polynomials and introducing a set of indispensable tools known as standard algebraic identities. Mastering these techniques is paramount for simplifying complex expressions, solving higher-level equations, and laying the groundwork for future algebraic manipulations like factorization.
Let's begin with a brief recap of essential terminology. Recall that an algebraic expression is a combination of constants and variables linked by mathematical operations. The parts separated by '+' or '-' signs are terms. Within each term, the numerical part is the coefficient, and the variable parts are crucial. We differentiate between like terms (identical variable parts with identical exponents, e.g., $3x^2y$ and $-5x^2y$) and unlike terms (differing variable parts or exponents, e.g., $2ab^2$ and $2a^2b$). Remember, only like terms can be combined through addition or subtraction. Our previous work involved adding and subtracting expressions by carefully grouping and combining these like terms.
The core new skill introduced here is the multiplication of algebraic expressions. This process is approached systematically, increasing in complexity:
- Multiplication of two Monomials: Multiply their coefficients and multiply their variable parts. When multiplying variables with the same base, we add their exponents according to the law of exponents ($x^m \times x^n = x^{m+n}$). For example, $(4x^2y) \times (3xy^3) \ $$ = (4 \times 3) \ $$ \times (x^2 \times x^1) \ $$ \times (y^1 \times y^3) \ $$ = 12x^{2+1}y^{1+3} = \ $$ 12x^3y^4$.
- Multiplication of a Monomial by a Polynomial: Apply the distributive property. Multiply the monomial by each term of the polynomial separately. For example, $2a(3a^2 - 5b) = (2a \times 3a^2) - (2a \times 5b) = 6a^3 - 10ab$.
- Multiplication of two Polynomials: (e.g., Binomial by Binomial, Binomial by Trinomial). The fundamental principle is to multiply each term of the first polynomial by each term of the second polynomial. After performing all these individual multiplications, combine any resulting like terms. For instance, multiplying $(x+2)$ by $(y+3)$: $x(y+3) + 2(y+3) = xy + 3x + 2y + 6$.
A major focus of this chapter is the introduction and application of standard Algebraic Identities. An identity is a special type of equality that holds true for all possible numerical values substituted for the variables involved, unlike an equation which is true only for specific values. We will focus on four fundamental identities:
- $(a + b)^2 = a^2 + 2ab + b^2$
- $(a - b)^2 = a^2 - 2ab + b^2$
- $(a + b)(a - b) = a^2 - b^2$
- $(x + a)(x + b) = x^2 + (a + b)x + ab$
We will see how these identities can be derived through direct multiplication (e.g., $(a+b)^2 = (a+b)(a+b)$). More importantly, we will learn to recognize the patterns associated with these identities and apply them strategically. They serve as powerful shortcuts for:
- Simplifying numerical calculations: For example, calculating $102^2$ can be done easily by seeing it as $(100 + 2)^2 = 100^2 + 2(100)(2) + 2^2 = 10000 + 400 + 4 = 10404$. Similarly, $98 \times 102$ fits the pattern $(100 - 2)(100 + 2) = 100^2 - 2^2 = 10000 - 4 = 9996$.
- Simplifying complex algebraic expressions: Recognizing expressions that match the structure of an identity allows for immediate simplification without performing lengthy step-by-step multiplication.
These identities are not mere formulas to memorize; they are fundamental structural relationships in algebra that form the bedrock for techniques like factorization in subsequent studies. Careful application, especially regarding signs, is crucial. Extensive practice is provided to build fluency in both direct multiplication and the strategic use of these vital identities.
Algebraic Expression and Polynomial
In earlier classes, you were introduced to the basics of algebra, which involves using letters (variables) to represent unknown quantities and combining them with numbers (constants) using mathematical operations. This chapter will delve deeper into algebraic expressions and introduce the concept of polynomials, along with operations on them and special identities.
Algebraic Expression
An algebraic expression is a combination of constants and variables connected by the fundamental mathematical operations of addition (+), subtraction (-), multiplication ($\times$), and division ($\div$).
- Variables: These are symbols, usually represented by letters from the alphabet (like $x, y, z, a, b, p, m, n$), that can take on different numerical values. Their value is not fixed.
- Constants: These are symbols, usually numbers (like 2, -5, 7, $\frac{1}{2}, \sqrt{3}, \pi$), that have a fixed numerical value.
Examples of algebraic expressions:
- $x+3$ (Variable $x$, constant 3, operation +)
- $2y-5$ (Variable $y$, constants 2, 5, operations $\times$, -)
- $3x^2$ (Variable $x$, constant 3, operation $\times$, power)
- $\frac{p}{4}$ (Variable $p$, constant 4, operation $\div$)
- $x^2 + 2xy + y^2$ (Variables $x, y$, constant 2, operations +, $\times$, powers)
- $ab - 5$ (Variables $a, b$, constant 5, operations $\times$, -)
Sometimes expressions might involve roots or variables in the denominator, like $\sqrt{x} + 1$ or $\frac{1}{x}$. While these are algebraic expressions, they might not fit into the category of polynomials, which we will define shortly.
Terms of an Expression
The different parts of an algebraic expression that are separated by the addition (+) or subtraction (-) signs are called the terms of the expression.
Example: In the expression $5x + 3y - 7$:
- The terms are $5x$, $3y$, and $-7$. Note that the sign before a term is part of the term.
A term itself is a product of factors. These factors can be numerical or variable.
Example: In the term $5x$, the factors are 5 and $x$. In the term $3y$, the factors are 3 and $y$. In the term $-7$, the only factor is -7.
Example: In the term $2x^2y$, the factors are 2, $x$, $x$, and $y$.
Coefficient:
In a term, any factor or group of factors is called the coefficient of the remaining part of the term.
The numerical factor in a term is called the numerical coefficient or simply the coefficient (when referring to the numerical part). If there is no numerical factor explicitly written, the coefficient is 1 (e.g., $x = 1 \times x$) or -1 (e.g., $-y = -1 \times y$).
Examples:
- In the term $5x$: The numerical coefficient is 5. The coefficient of $x$ is 5. The coefficient of 5 is $x$.
- In the term $-7xy$: The numerical coefficient is -7. The coefficient of $xy$ is -7. The coefficient of $x$ is $-7y$. The coefficient of $y$ is $-7x$. The coefficient of $-7x$ is $y$.
- In the term $m^2$: The numerical coefficient is 1 (since $m^2 = 1 \times m^2$). The coefficient of $m^2$ is 1.
- In the term $-p$: The numerical coefficient is -1 (since $-p = -1 \times p$). The coefficient of $p$ is -1. The coefficient of -1 is $p$.
- In the term $\frac{1}{2}ab$: The numerical coefficient is $\frac{1}{2}$. The coefficient of $ab$ is $\frac{1}{2}$. The coefficient of $a$ is $\frac{1}{2}b$.
Polynomial: A Detailed Definition
A polynomial is a specific type of algebraic expression that consists of variables, coefficients, and the operations of addition, subtraction, and multiplication. The key characteristic that defines a polynomial is that the exponents of all its variables must be non-negative integers (i.e., whole numbers like 0, 1, 2, 3, ...).
code CodeWhat is a Polynomial?
Let's break down the conditions for an expression to be a polynomial:
- It consists of one or more terms separated by '+' or '-' signs.
- Each term is a product of a constant (called a coefficient) and one or more variables.
- The power (or exponent) of each variable in every term must be a whole number (0, 1, 2, ...). A constant term like '5' can be written as $5x^0$, so its exponent is 0, which is a whole number.
Examples of Polynomials:
- $7x^2 - 4x + 2$ (Exponents are 2, 1, and 0)
- $5a^3b^2 - 3ab$ (All exponents are positive integers)
- $9y + 1$ (Exponents are 1 and 0)
- $15$ (A constant is a polynomial of degree 0, as it can be written as $15x^0$)
What is NOT a Polynomial?
An algebraic expression is not a polynomial if it contains any of the following:
- Negative Exponents: This occurs when a variable is in the denominator.
- Example: $3x^2 + \frac{2}{x} + 5$ is not a polynomial because $\frac{2}{x}$ is equivalent to $2x^{-1}$, and the exponent -1 is negative.
- Fractional Exponents: This occurs when there is a root of a variable.
- Example: $5\sqrt{y} - 3 = 5y^{1/2} - 3$ is not a polynomial because the exponent $\frac{1}{2}$ is not an integer.
- Variables in the Exponent:
- Example: $3^x + 2$ is not a polynomial because the variable is in the exponent.
Classification of Polynomials
Polynomials can be classified based on different criteria: the number of terms they contain, the number of variables they involve, and their degree.
1. Classification Based on the Number of Terms
This classification focuses on how many terms are present in the polynomial.
| Name | Number of Terms | Examples |
|---|---|---|
| Monomial | One term | $5x$, $-3y^2$, $10$, $7xyz$ |
| Binomial | Two terms | $x+y$, $a^2 - 4$, $2p+3q$ |
| Trinomial | Three terms | $x+y+z$, $a^2 - ab + b^2$, $2x^2 - 3x + 5$ |
| Polynomial | Four or more terms | $a+b+c+d$, $x^3 - 2x^2 + 5x - 1$ |
2. Classification Based on the Number of Variables
This classification depends on how many different variables appear in the polynomial.
- Polynomial in one variable: An expression containing only one variable. Example: $P(x) = 3x^3 - 4x^2 + 7x - 9$.
- Polynomial in two variables: An expression containing two different variables. Example: $P(x, y) = 5x^2y - 3xy^2 + 2x - y$.
- Polynomial in three variables: An expression containing three different variables. Example: $P(a, b, c) = 2ab + 3bc - 4ca$.
3. Classification Based on Degree
The degree is a fundamental property of a polynomial. To find it, we first need to understand the degree of a term.
- Degree of a term: The sum of the exponents of the variables in that term.
- The degree of $5x^3$ is 3.
- The degree of $7x^2y^4$ is $2+4=6$.
- The degree of a constant term like 8 is 0 (since $8 = 8x^0$).
- Degree of a polynomial: The highest degree among all its terms.
- The degree of $P(x) = 3x^4 - 5x^2 + 2$ is 4 (the highest power of x).
- The degree of $P(x, y) = 6x^3y^2 + 2xy^5 - 7$ is 7, because the degree of the first term is $3+2=5$, and the degree of the second term is $1+5=6$. The highest degree is 6. Wait, correction: The degree of the second term is $1+5=6$. No, $2xy^5$ means $2x^1y^5$, so the degree is $1+5=6$. Let's recheck. Ah, the example I wrote in my thought process was better. Let's use a clear one. The degree of $P(x, y) = 8x^2y^3 - 3xy^2 + 9$ is 5, because the degree of the first term is $2+3=5$, the degree of the second term is $1+2=3$, and the degree of the third term is 0. The highest is 5.
Based on their degree, polynomials (usually in one variable) are given special names:
| Degree | Name | Standard Form (in one variable 'x') | Example |
|---|---|---|---|
| 0 | Constant Polynomial | $P(x) = c$ | $7$ |
| 1 | Linear Polynomial | $P(x) = ax + b$ | $2x - 5$ |
| 2 | Quadratic Polynomial | $P(x) = ax^2 + bx + c$ | $x^2 + 3x - 2$ |
| 3 | Cubic Polynomial | $P(x) = ax^3 + bx^2 + cx + d$ | $4x^3 - x^2 + 8$ |
| 4 | Biquadratic (or Quartic) Polynomial | $P(x) = ax^4 + bx^3 + cx^2 + dx + e$ | $x^4 - 2x^3 + 5x - 1$ |
| Not defined | Zero Polynomial | $P(x) = 0$ | $0$ |
Note: The degree of the zero polynomial (the number 0) is not defined because it can be written as $0 = 0x^1 = 0x^2 = 0x^3$, etc., so there is no unique highest power.
Like Terms and Unlike Terms
Understanding like and unlike terms is crucial for performing operations like addition and subtraction on algebraic expressions.
Like terms are terms that have the same variables raised to the same powers. The numerical coefficients of like terms can be different.
Examples of like terms:
- $2x$ and $-5x$ (Same variable $x$, same power 1)
- $3xy$ and $7xy$ (Same variables $x, y$, same powers 1 for $x$ and 1 for $y$)
- $a^2b$ and $9a^2b$ (Same variables $a, b$, same powers 2 for $a$ and 1 for $b$)
- 5 and 12 (Both are constant terms, which can be thought of as terms with variable raised to power 0, e.g., $5x^0 = 5$)
- $y^2$ and $-8y^2$ (Same variable $y$, same power 2)
Unlike terms are terms that do not have the same variables or have the same variables but raised to different powers.
Examples of unlike terms:
- $2x$ and $3y$ (Different variables)
- $3xy$ and $7x$ (Same variable $x$, but $3xy$ also has $y$)
- $a^2b$ and $ab^2$ (Same variables $a, b$, but powers are different: $a^2$ vs $a^1$, $b^1$ vs $b^2$)
- $2x$ and $2x^2$ (Same variable $x$, but different powers 1 vs 2)
- 5 (constant term) and $5x$ (variable term)
The concept of like terms is important because we can only add or subtract like terms. Unlike terms cannot be combined through addition or subtraction in a simple manner (e.g., $2x + 3y$ remains $2x+3y$).
Addition and Subtraction of Algebraic Expressions
Performing arithmetic operations on algebraic expressions involves combining like terms. Just like you can add apples to apples but not directly apples to oranges, you can add or subtract terms that have the same variable parts. Unlike terms cannot be combined through addition or subtraction; they remain as separate terms in the expression.
Addition of Algebraic Expressions
To add two or more algebraic expressions, we follow these steps:
- Write down all the expressions being added. If using the horizontal method, place parentheses around each expression, separated by plus signs.
- Remove the parentheses. Remember that a plus sign before a parenthesis does not change the sign of the terms inside.
- Identify and group the like terms together.
- Add the coefficients of the like terms.
- Write the sum of the like terms and the unlike terms (if any) to get the final result.
There are two common methods for adding expressions: the horizontal method and the column method.
Example 1. Add the algebraic expressions $3x + 5y$ and $2x - 2y + 7$.
Answer:
Method 1: Horizontal Method
Write the expressions horizontally, separated by a plus sign:
$(3x + 5y) + (2x - 2y + 7)$
Remove the parentheses. Since there is a '+' sign before the second parenthesis, the signs of the terms inside remain unchanged.
$= 3x + 5y + 2x - 2y + 7$
Group the like terms together. Terms with $x$ are $3x$ and $2x$. Terms with $y$ are $5y$ and $-2y$. The constant term is 7.
$= (3x + 2x) + (5y - 2y) + 7$
Add the coefficients of the like terms: For the $x$ terms, $3+2=5$. For the $y$ terms, $5-2=3$.
$= (3+2)x + (5-2)y + 7$
$= 5x + 3y + 7$
The sum of the given expressions is $5x + 3y + 7$.
Method 2: Column Method
Write the expressions vertically, arranging the like terms in columns. If a term is missing in an expression, you can leave a blank space or write '+ 0' for that term's column.
$\begin{array}{c@{}c@{\,}c@{}c@{\,}c} & 3x & {}+{} & 5y & {} \\ {} + {} & 2x & {}-{} & 2y & {}+{} & 7 \\ \hline & \quad 5x & {}+{} & 3y & {}+{} & 7 \end{array}$Add the coefficients column by column:
- For the $x$ column: $3 + 2 = 5$. Write $5x$.
- For the $y$ column: $5 + (-2) = 5 - 2 = 3$. Write $+3y$.
- For the constant column: $0 + 7 = 7$. Write $+7$. (Implicitly, the first expression has a constant term of 0).
The sum is $5x + 3y + 7$. Both methods yield the same result.
Subtraction of Algebraic Expressions
To subtract an algebraic expression from another, we use the concept of adding the additive inverse. The additive inverse of an expression is obtained by changing the sign of each term in the expression.
To subtract Expression B from Expression A (A - B):
- Write down Expression A.
- Change the sign of each term in Expression B to get its additive inverse (-B).
- Add Expression A and the additive inverse of Expression B (A + (-B)).
- Combine the like terms as in addition.
Similar to addition, subtraction can also be done using horizontal or column methods.
Example 2. Subtract $4a - 7b + 3c - 2$ from $5a - 2b - 6c + 8$.
Answer:
We want to calculate $(5a - 2b - 6c + 8) - (4a - 7b + 3c - 2)$.
Step 1 & 2: Find the additive inverse of the expression being subtracted ($4a - 7b + 3c - 2$). Change the sign of each term: $-(4a - 7b + 3c - 2) = -4a + 7b - 3c + 2$.
Step 3: Add the first expression and the additive inverse of the second expression.
Method 1: Horizontal Method
$(5a - 2b - 6c + 8) + (-4a + 7b - 3c + 2)$
Remove parentheses and group like terms:
$= (5a - 4a) + (-2b + 7b) + (-6c - 3c) + (8 + 2)$
Combine like terms by adding their coefficients:
$= (5-4)a + (-2+7)b + (-6-3)c + (8+2)$
$= 1a + 5b + (-9)c + 10$
$= a + 5b - 9c + 10$
The difference is $a + 5b - 9c + 10$.
Method 2: Column Method
Write the expressions vertically, aligning like terms. Write the expression to be subtracted below the first one. For subtraction in columns, it is helpful to write the expression to be subtracted with its signs changed below the first expression, and then add.
Original setup:
$\begin{array}{c@{}c@{\,}c@{}c@{\,}c@{}c@{\,}c} & 5a & {}-{} & 2b & {}-{} & 6c & {}+{} & 8 \\ {} - {} & 4a & {}-{} & 7b & {}+{} & 3c & {}-{} & 2 \\ \hline \end{array}$Change the signs of the terms in the second expression and add:
$\begin{array}{c@{}c@{\,}c@{}c@{\,}c@{}c@{\,}c} & 5a & {}-{} & 2b & {}-{} & 6c & {}+{} & 8 \\ {} + {} & -4a & {}+{} & 7b & {}-{} & 3c & {}+{} & 2 \\ \hline & \quad & \quad & \quad & \quad & \quad & \quad & \quad \\ & (5-4)a & {}+{} & (-2+7)b & {}+{} & (-6-3)c & {}+{} & (8+2) \\ \hline & 1a & {}+{} & 5b & {}+{} & (-9)c & {}+{} & 10 \\ \hline & a & {}+{} & 5b & {}-{} & 9c & {}+{} & 10 \end{array}$Add the coefficients column by column based on the changed signs:
- For the $a$ column: $5 + (-4) = 1$. Write $a$.
- For the $b$ column: $-2 + 7 = 5$. Write $+5b$.
- For the $c$ column: $-6 + (-3) = -9$. Write $-9c$.
- For the constant column: $8 + 2 = 10$. Write $+10$.
The difference is $a + 5b - 9c + 10$. Both methods yield the same result.
Multiplication of Algebraic Expressions
Multiplying algebraic expressions is a fundamental operation. It involves multiplying the numerical coefficients and combining the variable parts according to the rules of exponents. The distributive property plays a crucial role in multiplying polynomials with more than one term.
Multiplication of a Monomial by a Monomial
A monomial is an expression with a single term. To multiply two monomials, we multiply their numerical coefficients together and multiply their variable parts together. When multiplying variable parts with the same base, we add their exponents using the rule $x^m \times x^n = x^{m+n}$.
Example 1. Multiply the following pairs of monomials:
(i) $3x$ by $5y$
(ii) $-2a^2$ by $4a^3$
(iii) $5xy$ by $-3x^2y$
Answer:
(i) Multiply $3x$ by $5y$:
$(3x) \times (5y)$
Group numerical coefficients and variable parts:
$= (3 \times 5) \times (x \times y)$
$= 15xy$
The product is $15xy$.
(ii) Multiply $-2a^2$ by $4a^3$:
$(-2a^2) \times (4a^3)$
Group numerical coefficients and variable parts. Use $a^m \times a^n = a^{m+n}$.
$= (-2 \times 4) \times (a^2 \times a^3)$
$= -8 \times a^{2+3} = -8a^5$
The product is $-8a^5$.
(iii) Multiply $5xy$ by $-3x^2y$:
$(5xy) \times (-3x^2y)$
Group numerical coefficients and variable parts. Use $x^m \times x^n = x^{m+n}$ and $y^m \times y^n = y^{m+n}$. Remember $x = x^1$ and $y = y^1$.
$= (5 \times -3) \times (x \times x^2) \times (y \times y)$
$= -15 \times x^{1+2} \times y^{1+1} = -15x^3y^2$
The product is $-15x^3y^2$.
Multiplication of a Monomial by a Polynomial (Binomial or Trinomial)
To multiply a monomial by a polynomial (an expression with more than one term), we use the distributive property of multiplication over addition or subtraction. This property states that to multiply a sum (or difference) by a number, you can multiply each addend (or the minuend and subtrahend) by the number and then add (or subtract) the products.
In algebraic terms:
$a(b+c) = a \times b + a \times c = ab + ac$
... (i)
$a(b-c) = a \times b - a \times c = ab - ac$
... (ii)
We apply this property by multiplying the monomial by each term of the polynomial separately and then adding the results.
Example 2. Multiply $2x$ by $(3x + 4y)$.
Answer:
We are multiplying the monomial $2x$ by the binomial $(3x + 4y)$.
Using the distributive property $a(b+c) = ab+ac$, with $a=2x$, $b=3x$, and $c=4y$:
$2x \times (3x + 4y) = (2x \times 3x) + (2x \times 4y)$
Now, perform the multiplication of monomials (as learned above) for each term:
$2x \times 3x = (2 \times 3) \times (x \times x) = 6x^2$
$2x \times 4y = (2 \times 4) \times (x \times y) = 8xy$
Add the results:
$= 6x^2 + 8xy$
The product is $6x^2 + 8xy$.
Example 3. Find the product of $-3m$ and $(m^2 - 2m + 5)$.
Answer:
We are multiplying the monomial $-3m$ by the trinomial $(m^2 - 2m + 5)$.
Apply the distributive property: multiply $-3m$ by each term inside the parenthesis ($m^2$, $-2m$, and $+5$).
$-3m \times (m^2 - 2m + 5) = (-3m \times m^2) + (-3m \times -2m) + (-3m \times 5)$
Now, perform the multiplication of monomials for each term:
$-3m \times m^2 = (-3 \times 1) \times (m^1 \times m^2) = -3m^{1+2} = -3m^3$
$-3m \times -2m = (-3 \times -2) \times (m^1 \times m^1) = 6m^{1+1} = 6m^2$
$-3m \times 5 = (-3 \times 5) \times m = -15m$
Add the results:
$= -3m^3 + 6m^2 - 15m$
The product is $-3m^3 + 6m^2 - 15m$.
Multiplication of a Polynomial by a Polynomial (Binomial by Binomial)
To multiply a polynomial by another polynomial, we extend the distributive property. If the first polynomial has $p$ terms and the second has $q$ terms, the product will initially have $p \times q$ terms before combining like terms.
To multiply a binomial $(a+b)$ by another binomial $(c+d)$, we multiply each term of the first binomial by the entire second binomial and then distribute again, or multiply each term of the first binomial by each term of the second binomial and add the results.
$(a+b)(c+d) = a(c+d) + b(c+d)$
Applying distributive property again:
$= (a \times c) + (a \times d) + (b \times c) + (b \times d)$
$= ac + ad + bc + bd$
... (iii)
This means we multiply the first term of the first binomial by both terms of the second binomial, and then multiply the second term of the first binomial by both terms of the second binomial, and finally add all the resulting terms and combine like terms if any.
Example 4. Multiply $(x+3)$ by $(x+5)$.
Answer:
We are multiplying the binomial $(x+3)$ by the binomial $(x+5)$.
Using the distributive method $(a+b)(c+d) = a(c+d) + b(c+d)$, with $a=x$, $b=3$, $c=x$, $d=5$:
$(x+3)(x+5) = x(x+5) + 3(x+5)$
Apply the distributive property again for each term:
$= (x \times x) + (x \times 5) + (3 \times x) + (3 \times 5)$
Perform the monomial multiplications:
$= x^2 + 5x + 3x + 15$
Identify and combine like terms ($5x$ and $3x$):
$= x^2 + (5+3)x + 15$
$= x^2 + 8x + 15$
The product is $x^2 + 8x + 15$.
Example 5. Multiply $(2a - 1)$ by $(3a + 4)$.
Answer:
We are multiplying the binomial $(2a - 1)$ by the binomial $(3a + 4)$.
Using the distributive method $(a-b)(c+d) = a(c+d) - b(c+d)$, with $a=2a$, $b=1$, $c=3a$, $d=4$:
$(2a - 1)(3a + 4) = 2a(3a + 4) - 1(3a + 4)$
Apply the distributive property again for each term:
$= (2a \times 3a) + (2a \times 4) + (-1 \times 3a) + (-1 \times 4)$
Perform the monomial multiplications:
$= 6a^2 + 8a - 3a - 4$
Identify and combine like terms ($8a$ and $-3a$):
$= 6a^2 + (8-3)a - 4$
$= 6a^2 + 5a - 4$
The product is $6a^2 + 5a - 4$.
Multiplication of a Binomial by a Trinomial
The distributive property extends to multiplying polynomials with any number of terms. To multiply a binomial by a trinomial, multiply each term of the binomial by every term of the trinomial and add the results. Then, combine any like terms.
To multiply $(a+b)$ by $(c+d+e)$:
$(a+b)(c+d+e) = a(c+d+e) + b(c+d+e)$
Apply distributive property again:
$= (a \times c) + (a \times d) + (a \times e) + (b \times c) + (b \times d) + (b \times e)$
$= ac + ad + ae + bc + bd + be$
... (iv)
Example 6. Multiply $(x+y)$ by $(x^2 - xy + y^2)$.
Answer:
We are multiplying the binomial $(x+y)$ by the trinomial $(x^2 - xy + y^2)$.
Multiply each term of the first expression by each term of the second expression:
$(x+y)(x^2 - xy + y^2) = x(x^2 - xy + y^2) + y(x^2 - xy + y^2)$
Apply distributive property for each term:
$= (x \times x^2) + (x \times -xy) + (x \times y^2) + (y \times x^2) + (y \times -xy) + (y \times y^2)$
Perform the monomial multiplications:
$= x^{1+2} - x^{1+1}y + xy^2 + yx^2 - xy^{1+1} + y^{1+2}$
$= x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$
(Rewriting $yx^2$ as $x^2y$ for easier grouping)
Identify and combine like terms: $-x^2y$ and $+x^2y$ are like terms, $xy^2$ and $-xy^2$ are like terms.
$= x^3 + (-x^2y + x^2y) + (xy^2 - xy^2) + y^3$
$= x^3 + 0 + 0 + y^3$
$= x^3 + y^3$
The product is $x^3 + y^3$. This is a notable result, as $(x+y)(x^2 - xy + y^2)$ is the factorisation of the sum of cubes. Similarly, $(x-y)(x^2 + xy + y^2) = x^3 - y^3$.
Algebraic Identity
In algebra, we often work with equalities. An equality can be either an equation or an identity. An equation is an equality that holds true for only specific values of the variable(s) involved. For example, $x+3=10$ is true only when $x=7$.
An identity, on the other hand, is an equality that holds true for all possible values of the variable(s) involved. It is like a rule that is always true in algebra.
Example: $x+x = 2x$. No matter what value you substitute for $x$, this equality is always true. So, $x+x=2x$ is an identity.
Example: $(a+b)^2 = a^2 + 2ab + b^2$. No matter what real numbers you choose for $a$ and $b$, the square of their sum is always equal to the sum of their squares plus twice their product. This is an identity.
Algebraic identities are powerful tools that help in simplifying expressions, performing multiplications quickly, and in factorisation.
Standard Algebraic Identities
There are a few fundamental algebraic identities that are used very frequently. You should learn these identities and understand how they are derived.
Identity I: Square of a Sum
$(a+b)^2 = a^2 + 2ab + b^2$
... (i)
Derivation:
$(a+b)^2$ means $(a+b)$ multiplied by itself.
$(a+b)^2 = (a+b) \times (a+b)$
Using the method for multiplying binomials (multiply each term of the first binomial by each term of the second):
$= a(a+b) + b(a+b)$
Apply the distributive property:
$= (a \times a) + (a \times b) + (b \times a) + (b \times b)$
$= a^2 + ab + ba + b^2$
Since multiplication is commutative, $ab = ba$. Combine the like terms $ab$ and $ba$:
$= a^2 + ab + ab + b^2$
$= a^2 + 2ab + b^2$
Thus, $(a+b)^2 = a^2 + 2ab + b^2$ is an identity.
Example 1. Expand $(x+4)^2$ using identity I.
Answer:
We use the identity $(a+b)^2 = a^2 + 2ab + b^2$.
In the expression $(x+4)^2$, we can see that it is in the form $(a+b)^2$ where $a=x$ and $b=4$.
Substitute $a=x$ and $b=4$ into the identity:
$(x+4)^2 = (x)^2 + 2(x)(4) + (4)^2$
Simplify the terms:
$= x^2 + (2 \times 4 \times x) + (4 \times 4)$
$= x^2 + 8x + 16$
So, the expansion of $(x+4)^2$ is $x^2 + 8x + 16$.
Identity II: Square of a Difference
$(a-b)^2 = a^2 - 2ab + b^2$
... (ii)
Derivation:
$(a-b)^2 = (a-b) \times (a-b)$
Using binomial multiplication:
$= a(a-b) - b(a-b)$
Apply the distributive property:
$= (a \times a) + (a \times -b) + (-b \times a) + (-b \times -b)$
$= a^2 - ab - ba + b^2$
Since $ab = ba$. Combine the like terms $-ab$ and $-ba$:
$= a^2 - ab - ab + b^2$
$= a^2 - 2ab + b^2$
Thus, $(a-b)^2 = a^2 - 2ab + b^2$ is an identity.
Example 2. Expand $(3p-5)^2$ using identity II.
Answer:
We use the identity $(a-b)^2 = a^2 - 2ab + b^2$.
In the expression $(3p-5)^2$, it is in the form $(a-b)^2$ where $a=3p$ and $b=5$.
Substitute $a=3p$ and $b=5$ into the identity:
$(3p-5)^2 = (3p)^2 - 2(3p)(5) + (5)^2$
Simplify the terms:
$= (3^2 \times p^2) - (2 \times 3 \times 5 \times p) + (5 \times 5)$
$= 9p^2 - 30p + 25$
So, the expansion of $(3p-5)^2$ is $9p^2 - 30p + 25$.
Identity III: Product of a Sum and a Difference
$(a+b)(a-b) = a^2 - b^2$
... (iii)
Derivation:
Multiply the binomial $(a+b)$ by the binomial $(a-b)$:
$(a+b)(a-b) = a(a-b) + b(a-b)$
Apply the distributive property:
$= (a \times a) + (a \times -b) + (b \times a) + (b \times -b)$
$= a^2 - ab + ba - b^2$
Since $ab = ba$, the terms $-ab$ and $+ba$ are like terms with opposite signs, so they cancel each other out (their sum is 0):
$= a^2 + (-ab + ab) - b^2$
$= a^2 + 0 - b^2$
$= a^2 - b^2$
Thus, $(a+b)(a-b) = a^2 - b^2$ is an identity. This is also known as the difference of squares formula.
Example 3. Expand $(2x+3y)(2x-3y)$ using identity III.
Answer:
We use the identity $(a+b)(a-b) = a^2 - b^2$.
In the expression $(2x+3y)(2x-3y)$, it is in the form $(a+b)(a-b)$ where $a=2x$ and $b=3y$.
Substitute $a=2x$ and $b=3y$ into the identity:
$(2x+3y)(2x-3y) = (2x)^2 - (3y)^2$
Simplify the terms:
$= (2^2 \times x^2) - (3^2 \times y^2)$
$= 4x^2 - 9y^2$
So, the expansion of $(2x+3y)(2x-3y)$ is $4x^2 - 9y^2$.
Example 4. Evaluate $103 \times 97$ using a suitable identity.
Answer:
We can express 103 and 97 in terms of a common number, which is 100.
$103 = 100 + 3$
$97 = 100 - 3$
So, the product $103 \times 97$ can be written as $(100 + 3)(100 - 3)$.
$103 \times 97 = (100 + 3)(100 - 3)$
This expression is in the form $(a+b)(a-b)$, where $a=100$ and $b=3$. We can use Identity III:
$(a+b)(a-b) = a^2 - b^2$
Substitute $a=100$ and $b=3$ into the identity:
$(100 + 3)(100 - 3) = (100)^2 - (3)^2$
Calculate the squares:
$= 10000 - 9$
Perform the subtraction:
$= 9991$
So, $103 \times 97 = 9991$. Using the identity allows for a quicker calculation than direct multiplication.
Identity IV: Product of Two Binomials with a Common Term
$(x+a)(x+b) = x^2 + (a+b)x + ab$
... (iv)
Derivation:
Multiply the binomial $(x+a)$ by the binomial $(x+b)$:
$(x+a)(x+b) = x(x+b) + a(x+b)$
Apply the distributive property:
$= (x \times x) + (x \times b) + (a \times x) + (a \times b)$
$= x^2 + bx + ax + ab$
Identify and combine like terms ($bx$ and $ax$). Factor out the common variable $x$ from $bx+ax$:
$= x^2 + (b+a)x + ab$
Since $b+a = a+b$ (commutative property of addition):
$= x^2 + (a+b)x + ab$
Thus, $(x+a)(x+b) = x^2 + (a+b)x + ab$ is an identity.
Example 5. Expand $(y+2)(y+5)$ using identity IV.
Answer:
We use the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
In the expression $(y+2)(y+5)$, it is in the form $(x+a)(x+b)$ where the common term is $y$ (so $x=y$) and $a=2$, $b=5$.
Substitute $x=y$, $a=2$, and $b=5$ into the identity:
$(y+2)(y+5) = y^2 + (2+5)y + (2)(5)$
Simplify the terms:
$= y^2 + 7y + 10$
So, the expansion of $(y+2)(y+5)$ is $y^2 + 7y + 10$.
Example 6. Expand $(p-3)(p+8)$ using identity IV.
Answer:
We use the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.
In the expression $(p-3)(p+8)$, it is in the form $(x+a)(x+b)$ where the common term is $p$ (so $x=p$) and $a=-3$, $b=8$. Remember that $p-3$ is $p+(-3)$.
Substitute $x=p$, $a=-3$, and $b=8$ into the identity:
$(p-3)(p+8) = p^2 + (-3+8)p + (-3)(8)$
Simplify the terms:
$= p^2 + (5)p + (-24)$
$= p^2 + 5p - 24$
So, the expansion of $(p-3)(p+8)$ is $p^2 + 5p - 24$.
Division of Algebraic Expressions
Division is the inverse operation of multiplication. In algebra, dividing expressions follows similar principles as dividing numbers and involves the rules of exponents. Just like division by zero is undefined for numbers, it is also undefined for algebraic expressions – we cannot divide by an expression that evaluates to zero.
Division of a Monomial by a Monomial
To divide one monomial by another non-zero monomial, we divide their numerical coefficients and divide their variable parts separately. When dividing powers with the same base, we subtract the exponent of the denominator from the exponent of the numerator using the rule $x^m \div x^n = x^{m-n}$ (assuming $x \neq 0$ and $m \ge n$). If $m < n$, the result will involve a variable in the denominator or a negative exponent, which might lead to an expression that is not a polynomial.
Example 1. Divide $12x^3$ by $4x$.
Answer:
We need to calculate $\frac{12x^3}{4x}$. Assume $x \neq 0$.
Separate the numerical and variable parts:
$\frac{12x^3}{4x} = (\frac{12}{4}) \times (\frac{x^3}{x^1})$
Perform the division for the numerical part and the variable part using the exponent rule $x^m \div x^n = x^{m-n}$ ($m=3, n=1$):
$= 3 \times x^{3-1}$
$= 3x^2$
The quotient is $3x^2$ (provided $x \neq 0$).
Example 2. Divide $-18a^5b^3$ by $6a^2b$.
Answer:
We need to calculate $\frac{-18a^5b^3}{6a^2b}$. Assume $a \neq 0, b \neq 0$. Remember $b = b^1$.
Separate the numerical and variable parts:
$\frac{-18a^5b^3}{6a^2b} = (\frac{-18}{6}) \times (\frac{a^5}{a^2}) \times (\frac{b^3}{b^1})$
Perform the division for each part using the exponent rule:
$= -3 \times a^{5-2} \times b^{3-1}$
$= -3a^3b^2$
The quotient is $-3a^3b^2$ (provided $a \neq 0, b \neq 0$).
Division of a Polynomial by a Monomial
To divide a polynomial (an expression with one or more terms) by a non-zero monomial, we divide each term of the polynomial by the monomial separately and then combine the results using addition or subtraction as in the original polynomial. This is based on the distributive property of division over addition/subtraction, which states $\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$ and $\frac{a-b}{c} = \frac{a}{c} - \frac{b}{c}$, provided $c \neq 0$.
Example 3. Divide $(6x^2 + 4x)$ by $2x$.
Answer:
We need to calculate $\frac{6x^2 + 4x}{2x}$. Assume $2x \neq 0$, which means $x \neq 0$.
Divide each term of the polynomial $(6x^2 + 4x)$ by the monomial $2x$:
$\frac{6x^2 + 4x}{2x} = \frac{6x^2}{2x} + \frac{4x}{2x}$
Now, perform the division of monomials for each term (as learned above):
$\frac{6x^2}{2x} = (\frac{6}{2}) \times (\frac{x^2}{x^1}) = 3 \times x^{2-1} = 3x^1 = 3x$
$\frac{4x}{2x} = (\frac{4}{2}) \times (\frac{x^1}{x^1}) = 2 \times x^{1-1} = 2 \times x^0$
Remember that any non-zero number raised to the power of 0 is 1, so $x^0 = 1$ (since we assumed $x \neq 0$).
$2 \times x^0 = 2 \times 1 = 2$
Combine the results:
$= 3x + 2$
The quotient is $3x + 2$ (provided $x \neq 0$).
Example 4. Divide $(9a^3b^2 - 12a^2b^3 + 3ab)$ by $-3ab$.
Answer:
We need to calculate $\frac{9a^3b^2 - 12a^2b^3 + 3ab}{-3ab}$. Assume $-3ab \neq 0$, which means $a \neq 0$ and $b \neq 0$.
Divide each term of the polynomial by the monomial $-3ab$. Be careful with the signs.
$\frac{9a^3b^2 - 12a^2b^3 + 3ab}{-3ab} = \frac{9a^3b^2}{-3ab} + \frac{-12a^2b^3}{-3ab} + \frac{3ab}{-3ab}$
Now, perform the division of monomials for each term:
Term 1:
$\frac{9a^3b^2}{-3ab} = (\frac{9}{-3}) \times (\frac{a^3}{a^1}) \times (\frac{b^2}{b^1}) = -3 \times a^{3-1} \times b^{2-1} = -3a^2b^1 = -3a^2b$
Term 2:
$\frac{-12a^2b^3}{-3ab} = (\frac{-12}{-3}) \times (\frac{a^2}{a^1}) \times (\frac{b^3}{b^1}) = 4 \times a^{2-1} \times b^{3-1} = 4a^1b^2 = 4ab^2$
Term 3:
$\frac{3ab}{-3ab} = (\frac{3}{-3}) \times (\frac{a^1}{a^1}) \times (\frac{b^1}{b^1}) = -1 \times a^{1-1} \times b^{1-1} = -1 \times a^0 \times b^0 = -1 \times 1 \times 1 = -1$
Combine the results:
$= -3a^2b + 4ab^2 - 1$
The quotient is $-3a^2b + 4ab^2 - 1$ (provided $a \neq 0, b \neq 0$).
Division of a Polynomial by a Polynomial
When the divisor is a polynomial with more than one term (like a binomial or trinomial), the division can be performed by factoring the dividend and cancelling common factors (if possible), or by using the method of polynomial long division.
Method 1: Factoring and Cancelling
If the dividend can be factored such that one of the factors is identical to the divisor, the division becomes a simple cancellation. This method relies on your ability to factor the dividend correctly, often using identities or by splitting terms.
If the polynomial $P(x)$ can be factored as $P(x) = D(x) \times Q(x)$, where $D(x)$ is the divisor, then $\frac{P(x)}{D(x)} = \frac{D(x) \times Q(x)}{D(x)} = Q(x)$, provided $D(x) \neq 0$.
Example 5. Divide $(x^2 - 9)$ by $(x - 3)$.
Answer:
We need to calculate $\frac{x^2 - 9}{x - 3}$. Assume $x - 3 \neq 0$, which means $x \neq 3$.
Recognise the numerator $x^2 - 9$ as a difference of squares, $x^2 - 3^2$. We know the identity $a^2 - b^2 = (a+b)(a-b)$.
Factor the numerator using the identity, with $a=x$ and $b=3$:
$x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)$
Now substitute the factored form into the division:
$\frac{x^2 - 9}{x - 3} = \frac{(x+3)(x-3)}{x - 3}$
Since we assumed $x \neq 3$, $x-3 \neq 0$. We can cancel out the common factor $(x-3)$ from the numerator and the denominator:
$= \frac{(x+3)\cancel{(x-3)}^{\normalsize 1}}{\cancel{(x-3)}_{\normalsize 1}}$
$= x+3$
The quotient is $x+3$ (provided $x \neq 3$).
Example 6. Divide $(y^2 + 7y + 10)$ by $(y + 2)$.
Answer:
We need to calculate $\frac{y^2 + 7y + 10}{y + 2}$. Assume $y + 2 \neq 0$, which means $y \neq -2$.
Let's try to factor the numerator $y^2 + 7y + 10$. This is a quadratic trinomial of the form $x^2 + (a+b)x + ab$. We need to find two numbers that multiply to 10 and add up to 7. These numbers are 2 and 5 ($2 \times 5 = 10$ and $2 + 5 = 7$).
So, factor the numerator as $(y+2)(y+5)$.
$y^2 + 7y + 10 = (y+2)(y+5)$
Now substitute the factored form into the division:
$\frac{y^2 + 7y + 10}{y + 2} = \frac{(y+2)(y+5)}{y + 2}$
Since we assumed $y \neq -2$, $y+2 \neq 0$. We can cancel out the common factor $(y+2)$ from the numerator and the denominator:
$= \frac{\cancel{(y+2)}^{\normalsize 1}(y+5)}{\cancel{(y+2)}_{\normalsize 1}}$
$= y+5$
The quotient is $y+5$ (provided $y \neq -2$).
Method 2: Polynomial Long Division
Polynomial long division is a method similar to the long division of numbers. It can be used to divide any polynomial by another polynomial of the same or lower degree (provided the divisor is not zero). This method is particularly useful when factoring is not obvious or possible.
Steps for Polynomial Long Division (Dividing $P(x)$ by $D(x)$):
- Arrange both the dividend $P(x)$ and the divisor $D(x)$ in descending powers of the variable. If any power is missing in the dividend, you can write it with a coefficient of zero (e.g., $x^2+1 = x^2+0x+1$).
- Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.
- Multiply the divisor by this first term of the quotient.
- Subtract the result from the dividend.
- Bring down the next term(s) from the original dividend to form a new dividend.
- Repeat the process (divide the first term of the new dividend by the first term of the divisor) until the degree of the remainder is less than the degree of the divisor.
Let's apply this to Example 6:
Example 7. Divide $(y^2 + 7y + 10)$ by $(y + 2)$ using long division.
Answer:
We want to divide $y^2 + 7y + 10$ (dividend) by $y + 2$ (divisor). Both are already in descending powers of $y$.
Divide the first term of the dividend ($y^2$) by the first term of the divisor ($y$): $\frac{y^2}{y} = y$. This is the first term of the quotient. Write it above the dividend.
Multiply the divisor $(y+2)$ by the first term of the quotient ($y$): $y \times (y+2) = y^2 + 2y$. Write this below the dividend.
Subtract $(y^2 + 2y)$ from the dividend $(y^2 + 7y + 10)$. $(y^2 + 7y) - (y^2 + 2y) = y^2 - y^2 + 7y - 2y = 5y$. Bring down the next term (+10) to get the new dividend $5y + 10$.
Now, repeat the process with the new dividend $5y + 10$.
Divide the first term of the new dividend ($5y$) by the first term of the divisor ($y$): $\frac{5y}{y} = 5$. This is the next term of the quotient. Write $+5$ above the dividend.
Multiply the divisor $(y+2)$ by this term of the quotient (5): $5 \times (y+2) = 5y + 10$. Write this below the new dividend.
Subtract $(5y + 10)$ from $(5y + 10)$: $(5y + 10) - (5y + 10) = 0$. The remainder is 0.
The long division calculation is shown as:
$\begin{array}{r} y+5\phantom{....)} \\ y+2{\overline{\smash{\big)}\,y^2+7y+10}} \\ \underline{-~\phantom{(}(y^2+2y)} \\ 0+5y+10 \\ \underline{-~\phantom{()}(5y+10)} \\ 0+0 \end{array}$The quotient is $y+5$, and the remainder is 0.
Therefore, $\frac{y^2 + 7y + 10}{y + 2} = y+5$ (provided $y \neq -2$). This matches the result obtained by factoring.