| Content On This Page | ||
|---|---|---|
| Addition of Algebraic Expressions | Subtraction of Algebraic Expressions | Multiplication of Algebraic Expressions |
| Division of Algebraic Expressions (excluding Polynomial division) | ||
Operations on Algebraic Expressions
Addition of Algebraic Expressions
Adding Algebraic Expressions
Adding algebraic expressions involves combining quantities represented by variables and constants. The key principle here is that you can only directly combine terms that are "alike". Like terms are defined as terms that have the exact same variable part, meaning they must have the same variables raised to the exact same powers. Constant terms (terms without any variables) are also considered like terms among themselves.
Understanding like terms is fundamental because addition and subtraction in algebra are performed only on like terms.
Examples of like terms:
- $2x$ and $5x$: Both have the variable $x$ raised to the power of $1$. We can add them: $2x + 5x = (2+5)x = 7x$.
- $3y^2$ and $-7y^2$: Both have the variable $y$ raised to the power of $2$. We can add them: $3y^2 + (-7y^2) = (3-7)y^2 = -4y^2$.
- $ab$ and $4ab$: Both have the variable part $ab$ ($a^1 b^1$). We can add them: $ab + 4ab = 1ab + 4ab = (1+4)ab = 5ab$.
- $5$ and $-10$: Both are constant terms. We can add them: $5 + (-10) = 5 - 10 = -5$.
- $x^2 y$ and $6x^2 y$: Both have the variable part $x^2 y$ ($x^2 y^1$). We can add them: $x^2 y + 6x^2 y = 1x^2 y + 6x^2 y = (1+6)x^2 y = 7x^2 y$.
Examples of unlike terms:
- $2x$ and $5y$: Different variables ($x$ vs $y$). They cannot be combined into a single term through addition or subtraction; the sum remains $2x + 5y$.
- $3x^2$ and $3x$: Same variable ($x$), but different powers ($x^2$ vs $x^1$). They cannot be combined; the sum remains $3x^2 + 3x$.
- $ab$ and $a$: Different variable parts ($ab$ vs $a$). They cannot be combined; the sum remains $ab + a$.
- $xy^2$ and $x^2 y$: Same variables ($x, y$), but the powers on each variable are different ($x^1 y^2$ vs $x^2 y^1$). They cannot be combined; the sum remains $xy^2 + x^2 y$.
The process of adding like terms by adding their coefficients is a direct application of the distributive property. For example, $2x + 5x$ can be thought of as having $x$ as a common factor, so by the distributive property, $2x + 5x = (2+5) \times x = 7x$. Similarly, for terms with multiple variables like $-2ab + 5ab$, the common factor is $ab$, so $-2ab + 5ab = (-2+5) \times ab = 3ab$.
Unlike terms cannot be combined into a single term through addition or subtraction because they represent fundamentally different quantities. Adding $2$ metres and $5$ kilograms does not yield a single combined unit; similarly, $2x + 5y$ remains as it is.
Methods for Adding Algebraic Expressions
There are two primary methods commonly used for adding algebraic expressions:
Method 1: Horizontal Method
This method involves writing the expressions in a line and then rearranging and combining like terms.
Steps:
- Write all the given algebraic expressions in a single line, using plus signs to connect them. If the expressions are enclosed in parentheses, you can usually remove the parentheses, especially if they are preceded by a plus sign, as it does not change the sign of the terms inside.
- Identify and group the like terms together. It can be helpful to rearrange the terms so that like terms are adjacent to each other. Be careful to keep the correct sign with each term.
- Add the coefficients of each group of like terms. Write the resulting term (coefficient plus the common variable part).
Example 1. Add the expressions $3x + 5y - 2$ and $2x - 3y + 7$.
Answer:
Step 1: Write the expressions in a line, connected by a plus sign. Since the expressions are being added, we can remove the parentheses without changing signs:
$$ (3x + 5y - 2) + (2x - 3y + 7) = 3x + 5y - 2 + 2x - 3y + 7 $$Step 2: Group the like terms together:
$$ (3x + 2x) + (5y - 3y) + (-2 + 7) $$Step 3: Add the coefficients of the like terms:
$$ (3+2)x + (5-3)y + (-2+7) $$ $$ 5x + 2y + 5 $$The sum of the two expressions is $\textbf{5x + 2y + 5}$.
Example 2. Add the expressions $4a^2 - 2ab + 3b^2$ and $-a^2 + 5ab - 2b^2 + 1$.
Answer:
Step 1: Write the expressions in a line:
$$ (4a^2 - 2ab + 3b^2) + (-a^2 + 5ab - 2b^2 + 1) $$Remove parentheses:
$$ 4a^2 - 2ab + 3b^2 - a^2 + 5ab - 2b^2 + 1 $$Step 2: Group like terms. Remember that $-a^2$ has a coefficient of $-1$, $3b^2$ is $+3b^2$, and $1$ is a constant term.
$$ (4a^2 - a^2) + (-2ab + 5ab) + (3b^2 - 2b^2) + 1 $$Step 3: Add the coefficients of like terms:
$$ (4-1)a^2 + (-2+5)ab + (3-2)b^2 + 1 $$ $$ 3a^2 + 3ab + 1b^2 + 1 $$Simplify $1b^2$ to $b^2$:
$$ 3a^2 + 3ab + b^2 + 1 $$The sum of the two expressions is $\textbf{3a\textsuperscript{2} + 3ab + b\textsuperscript{2} + 1}$.
Method 2: Vertical (Column) Method
This method is similar to the column method used for adding numbers. It helps in organizing terms, especially when dealing with multiple expressions or expressions with many terms.
Steps:
- Arrange the given expressions in rows, ensuring that like terms are placed directly one below the other in columns. If an expression is missing a term that is present in another expression, you can write a $0$ for that term's coefficient in its column, or simply leave the space blank, treating it as having a $0$ coefficient.
- Add the coefficients of the terms in each column, just as you would add numbers column by column.
Example 3. Add $3x + 5y - 2$ and $2x - 3y + 7$ using the vertical method.
Answer:
Arrange the expressions in columns, aligning the like terms ($x$ terms, $y$ terms, and constant terms):
$$ \begin{array}{cccc} & 3x & + 5y & - 2 \\ + & 2x & - 3y & + 7 \\ \hline \end{array} $$Now, add the coefficients in each column:
- For the $x$ column: $3 + 2 = 5$. The term is $5x$.
- For the $y$ column: $5 + (-3) = 5 - 3 = 2$. The term is $+2y$.
- For the constant column: $-2 + 7 = 5$. The term is $+5$.
Combining the results from each column:
$$ \begin{array}{cccc} & 3x & + 5y & - 2 \\ + & 2x & - 3y & + 7 \\ \hline & 5x & + 2y & + 5 \\ \hline \end{array} $$The sum is $\textbf{5x + 2y + 5}$.
Example 4. Add $4a^2 - 2ab + 3b^2$ and $-a^2 + 5ab - 2b^2 + 1$ using the vertical method.
Answer:
Arrange the expressions in columns, aligning the like terms ($a^2$ terms, $ab$ terms, $b^2$ terms, and constant terms). Note that the first expression has no constant term (implicitly $0$), and the second expression has no $a^2b^2$ or similar terms.
$$ \begin{array}{ccccc} & 4a^2 & - 2ab & + 3b^2 & \phantom{+0} \\ + & -a^2 & + 5ab & - 2b^2 & + 1 \\ \hline \end{array} $$Add the coefficients in each column:
- For the $a^2$ column: $4 + (-1) = 4 - 1 = 3$. The term is $3a^2$.
- For the $ab$ column: $-2 + 5 = 3$. The term is $+3ab$.
- For the $b^2$ column: $3 + (-2) = 3 - 2 = 1$. The term is $+1b^2$ or $+b^2$.
- For the constant column: $0 + 1 = 1$. The term is $+1$.
Combining the results from each column:
$$ \begin{array}{ccccc} & 4a^2 & - 2ab & + 3b^2 & \phantom{+0} \\ + & -a^2 & + 5ab & - 2b^2 & + 1 \\ \hline & 3a^2 & + 3ab & + b^2 & + 1 \\ \hline \end{array} $$The sum is $3a^2 + 3ab + b^2 + 1$.
Both the horizontal and vertical methods are effective for adding algebraic expressions. Choose the method that you find easier to manage, especially when dealing with multiple expressions or complex terms.
Subtraction of Algebraic Expressions
Subtracting Algebraic Expressions
Subtracting algebraic expressions is closely related to addition, as it relies on the concept of combining like terms. The fundamental idea behind subtracting an expression is to add its additive inverse (or opposite). The additive inverse of an algebraic expression is obtained by changing the sign of each and every term within that expression.
For example:
- The additive inverse of the term $5x$ is $-5x$.
- The additive inverse of the term $-3y^2$ is $+3y^2$.
- The additive inverse of the expression $3x + 5$ is $-(3x + 5)$, which, by distributing the minus sign, becomes $-3x - 5$. Notice that the sign of both terms ($+3x$ and $+5$) is changed.
- The additive inverse of the expression $a^2 - 2ab + 7$ is $-(a^2 - 2ab + 7) = -a^2 + 2ab - 7$. The signs of all terms ($+a^2$, $-2ab$, $+7$) are changed.
Subtracting an expression $A$ from an expression $B$ (written as $B - A$) is mathematically equivalent to adding the additive inverse of $A$ to $B$ (written as $B + (-A)$). Therefore, the process of subtraction transforms into a process of changing signs and then performing addition of like terms.
To subtract an algebraic expression, you essentially change the sign of every term in the expression being subtracted and then combine the resulting terms with the terms of the other expression using the rules of addition of algebraic expressions (combining like terms).
Methods for Subtracting Algebraic Expressions
Similar to addition, there are two common methods for subtracting algebraic expressions:
Method 1: Horizontal Method
This method involves writing the expressions in a line and systematically changing the signs of the terms being subtracted before combining like terms.
Steps:
- Write the expression from which you are subtracting (the minuend) first.
- Place a minus sign after the first expression.
- Write the expression being subtracted (the subtrahend) immediately after the minus sign, enclosing it within parentheses. This is crucial because the minus sign applies to the entire expression, not just the first term.
- Remove the parentheses by multiplying each term inside the parentheses by $-1$. This means changing the sign of every term within the parentheses.
- Once the parentheses are removed and signs are changed, the problem becomes an addition problem. Group the like terms together.
- Add the coefficients of the like terms to get the final simplified expression.
Example 1. Subtract $2x - 3y + 7$ from $3x + 5y - 2$.
Answer:
We are subtracting $(2x - 3y + 7)$ from $(3x + 5y - 2)$. Write this horizontally:
$$ (3x + 5y - 2) - (2x - 3y + 7) $$Remove the parentheses preceding the second expression by changing the sign of each term inside it:
$$ 3x + 5y - 2 \underbrace{- (2x)}_{(-1)(2x) = -2x} \underbrace{- (-3y)}_{(-1)(-3y) = +3y} \underbrace{- (+7)}_{(-1)(+7) = -7} $$The expression becomes:
$$ 3x + 5y - 2 - 2x + 3y - 7 $$Now, group the like terms:
$$ (3x - 2x) + (5y + 3y) + (-2 - 7) $$Add the coefficients of the like terms:
$$ (3-2)x + (5+3)y + (-2-7) $$ $$ 1x + 8y + (-9) $$Simplify $1x$ to $x$ and $+(-9)$ to $-9$:
$$ x + 8y - 9 $$The difference is $\textbf{x + 8y - 9}$.
Example 2. Subtract $p^2 + 5pq - 2q^2 + 1$ from $4p^2 - 2pq + 3q^2$.
Answer:
We are subtracting $(p^2 + 5pq - 2q^2 + 1)$ from $(4p^2 - 2pq + 3q^2)$. Write this horizontally:
$$ (4p^2 - 2pq + 3q^2) - (p^2 + 5pq - 2q^2 + 1) $$Remove the parentheses preceding the second expression by changing the sign of each term inside it:
$$ 4p^2 - 2pq + 3q^2 \underbrace{- (p^2)}_{(-1)(p^2) = -p^2} \underbrace{- (5pq)}_{(-1)(5pq) = -5pq} \underbrace{- (-2q^2)}_{(-1)(-2q^2) = +2q^2} \underbrace{- (+1)}_{(-1)(+1) = -1} $$The expression becomes:
$$ 4p^2 - 2pq + 3q^2 - p^2 - 5pq + 2q^2 - 1 $$Now, group the like terms:
$$ (4p^2 - p^2) + (-2pq - 5pq) + (3q^2 + 2q^2) - 1 $$Add the coefficients of like terms:
$$ (4-1)p^2 + (-2-5)pq + (3+2)q^2 - 1 $$ $$ 3p^2 + (-7)pq + 5q^2 - 1 $$Simplify $+(-7)pq$ to $-7pq$:
$$ 3p^2 - 7pq + 5q^2 - 1 $$The difference is $\textbf{3p\textsuperscript{2} - 7pq + 5q\textsuperscript{2} - 1}$.
Method 2: Vertical (Column) Method
This method arranges the expressions vertically, aligning like terms, which can be helpful for keeping track of terms when expressions are long or when subtracting multiple expressions (though subtraction of multiple expressions sequentially is less common than adding them).
Steps:
- Write the expression from which you are subtracting (minuend) in the first row.
- Write the expression being subtracted (subtrahend) below it in the second row, ensuring that like terms are placed directly one below the other in columns. If a term is missing in either expression, you can add it with a coefficient of $0$ or leave the space blank for clarity in alignment.
- To perform the subtraction using addition, conceptually (or by writing a third row), change the sign of each term in the subtrahend (the second row).
- Now, treat this as an addition problem. Add the coefficients of the terms in each column, combining the terms from the first row and the "changed-sign" second row.
Example 3. Subtract $2x - 3y + 7$ from $3x + 5y - 2$ using the vertical method.
Answer:
Arrange the expressions vertically, aligning like terms:
$$ \begin{array}{cccc} & 3x & + 5y & - 2 \\ - & (2x & - 3y & + 7) \\ \hline \end{array} $$Change the signs of the terms in the expression being subtracted ($2x - 3y + 7$) and add:
$$ \begin{array}{cccc} & 3x & + 5y & - 2 \\ + & -2x & + 3y & - 7 \\ % Signs of the second expression are changed \hline & (3-2)x & (5+3)y & (-2-7) \\ % Add the coefficients column-wise \hline & 1x & + 8y & - 9 \\ \hline & x & + 8y & - 9 % Simplify the result \end{array} $$The difference is $\textbf{x + 8y - 9}$.
Example 4. Subtract $p^2 + 5pq - 2q^2 + 1$ from $4p^2 - 2pq + 3q^2$ using the vertical method.
Answer:
Arrange the expressions vertically, aligning like terms. The first expression has no constant term (implicitly $+0$).
$$ \begin{array}{ccccc} % 5 columns for p^2, pq, q^2, constant, and the signs/operators & 4p^2 & - 2pq & + 3q^2 & \phantom{+0} \\ % Added phantom space for constant alignment - & (p^2 & + 5pq & - 2q^2 & + 1) \\ \hline \end{array} $$Change the signs of the terms in the expression being subtracted ($p^2 + 5pq - 2q^2 + 1$) and add:
$$ \begin{array}{ccccc} & 4p^2 & - 2pq & + 3q^2 & \phantom{+0} \\ + & -p^2 & - 5pq & + 2q^2 & - 1 \\ % Signs of the second expression are changed \hline & (4-1)p^2 & (-2-5)pq & (3+2)q^2 & (0-1) \\ % Add the coefficients column-wise \hline & 3p^2 & - 7pq & + 5q^2 & - 1 \\ \hline \end{array} $$The difference is $\textbf{3p\textsuperscript{2} - 7pq + 5q\textsuperscript{2} - 1}$.
Both methods are effective for subtracting algebraic expressions. The key step is always to correctly change the sign of every term in the expression that is being subtracted, thereby converting the subtraction problem into an addition problem.
Multiplication of Algebraic Expressions
Multiplying Algebraic Expressions
Multiplication is a fundamental operation in algebra used to find the product of two or more algebraic expressions. Unlike addition and subtraction which only combine like terms, multiplication can be performed between any terms or expressions. The process relies heavily on the distributive property and the laws of exponents.
The general principle of multiplying algebraic expressions is to multiply every term in one expression by every term in the other expression and then combine any resulting like terms to simplify the product.
Rules for Multiplication of Terms
When multiplying individual terms, follow these rules:
Multiply the Coefficients:
Multiply the numerical coefficients of the terms together, paying careful attention to the signs (positive or negative). Remember the rules for multiplying signs:- Positive number $\times$ Positive number = Positive number
- Negative number $\times$ Negative number = Positive number
- Positive number $\times$ Negative number = Negative number
- Negative number $\times$ Positive number = Negative number
Multiply the Variables:
Multiply the variable parts together. If the same variable appears in both terms, use the law of exponents which states that when multiplying powers with the same base, you add their exponents: $a^m \times a^n = a^{m+n}$. If variables are different, just write them next to each other to indicate multiplication (e.g., $x \times y = xy$).
Applying these rules allows us to multiply any two terms. When multiplying expressions with multiple terms, we use the distributive property.
Multiplying Different Types of Expressions
1. Multiplying a Monomial by a Monomial
To multiply two monomials, multiply their coefficients and then multiply their variable parts using the laws of exponents.
Example 1. Multiply $3x$ by $5y$.
Answer:
The two monomials are $3x$ and $5y$.
Multiply the coefficients: $3 \times 5 = 15$.
Multiply the variable parts: $x \times y = xy$.
Combine the results:
$$ (3x) \times (5y) = (3 \times 5) \times (x \times y) = 15xy $$The product is $\textbf{15xy}$.
Example 2. Multiply $-2a^2$ by $4a^3$.
Answer:
The two monomials are $-2a^2$ and $4a^3$.
Multiply the coefficients: $-2 \times 4 = -8$.
Multiply the variable parts: $a^2 \times a^3$. Using the law $a^m \times a^n = a^{m+n}$, we get $a^{2+3} = a^5$.
Combine the results:
$$ (-2a^2) \times (4a^3) = (-2 \times 4) \times (a^2 \times a^3) = -8a^5 $$The product is $\textbf{-8a\textsuperscript{5}}$.
2. Multiplying a Monomial by a Polynomial
To multiply a monomial by a polynomial (which could be a binomial, trinomial, or a polynomial with more terms), we use the distributive property. This property states that $a(b + c) = ab + ac$. We multiply the monomial outside the parentheses by each term inside the parentheses and then add the products.
Formula: $a(b + c + d + \ldots) = ab + ac + ad + \ldots$
Example 3. Multiply $3x$ by $(2x + 5)$.
Answer:
We multiply the monomial $3x$ by each term inside the binomial $(2x+5)$.
$$ 3x(2x + 5) = (3x \times 2x) + (3x \times 5) $$Now, perform the monomial multiplications:
- $3x \times 2x = (3 \times 2) \times (x \times x) = 6 \times x^{1+1} = 6x^2$.
- $3x \times 5 = (3 \times 5) \times x = 15x$.
Add the results:
$$ 6x^2 + 15x $$The product is $\textbf{6x\textsuperscript{2} + 15x}$.
Example 4. Multiply $-2ab$ by $(a^2 - 3b + 1)$.
Answer:
We multiply the monomial $-2ab$ by each term inside the trinomial $(a^2 - 3b + 1)$.
$$ -2ab(a^2 - 3b + 1) = (-2ab \times a^2) + (-2ab \times -3b) + (-2ab \times 1) $$Now, perform the monomial multiplications:
- $-2ab \times a^2 = (-2 \times 1) \times (a^1 \times a^2 \times b) = -2 \times a^{1+2} \times b = -2a^3 b$.
- $-2ab \times -3b = (-2 \times -3) \times (a \times b^1 \times b^1) = +6 \times a \times b^{1+1} = 6ab^2$.
- $-2ab \times 1 = (-2 \times 1) \times ab = -2ab$.
Add the results:
$$ -2a^3 b + 6ab^2 - 2ab $$These resulting terms are unlike terms (different variable parts), so they cannot be combined further.
The product is $\textbf{-2a\textsuperscript{3}b + 6ab\textsuperscript{2} - 2ab}$.
3. Multiplying a Polynomial by a Polynomial
To multiply a polynomial by another polynomial, we extend the distributive property. We must multiply each term of the first polynomial by every term of the second polynomial and then combine any like terms in the resulting expression.
General approach:
If you have a polynomial with terms $(t_1 + t_2 + \ldots)$ and another with terms $(u_1 + u_2 + \ldots)$, the product is found by calculating $(t_1 \times u_1) + (t_1 \times u_2) + \ldots + (t_2 \times u_1) + (t_2 \times u_2) + \ldots$ and then combining like terms.
For multiplying a Binomial by a Binomial:
Let the binomials be $(a+b)$ and $(c+d)$.
Using the distributive property, distribute the first term ($a$) of the first binomial to the second binomial, and then distribute the second term ($b$) of the first binomial to the second binomial:
$$ (a+b)(c+d) = a(c+d) + b(c+d) $$Now, apply the monomial-by-polynomial rule to each part:
$$ a(c+d) = ac + ad $$ $$ b(c+d) = bc + bd $$So, the product is:
$$ (a+b)(c+d) = ac + ad + bc + bd $$The acronym FOIL is a mnemonic often used for binomial by binomial multiplication, representing the pairs of terms multiplied:
- First terms: $a \times c = ac$
- Outer terms: $a \times d = ad$
- Inner terms: $b \times c = bc$
- Last terms: $b \times d = bd$
Adding these four products gives $ac + ad + bc + bd$. FOIL is just a specific way to remember the distribution process for two binomials.
Example 5. Multiply $(x + 3)$ by $(x - 5)$.
Answer:
We multiply each term of $(x+3)$ by each term of $(x-5)$.
Distribute $x$ from the first binomial to $(x-5)$, and distribute $3$ from the first binomial to $(x-5)$:
$$ (x + 3)(x - 5) = x(x - 5) + 3(x - 5) $$Now, apply the distributive property (monomial by polynomial) to each part:
$$ x(x - 5) = (x \times x) + (x \times -5) = x^2 - 5x $$ $$ 3(x - 5) = (3 \times x) + (3 \times -5) = 3x - 15 $$Add the results:
$$ (x^2 - 5x) + (3x - 15) = x^2 - 5x + 3x - 15 $$Combine the like terms ($-5x$ and $+3x$):
$$ x^2 + (-5 + 3)x - 15 $$ $$ x^2 - 2x - 15 $$The product is $\textbf{x\textsuperscript{2} - 2x - 15}$.
Example 6. Multiply $(2y + 1)$ by $(y^2 - 3y + 4)$.
Answer:
We multiply each term of the binomial $(2y+1)$ by each term of the trinomial $(y^2 - 3y + 4)$.
Distribute $2y$ from the binomial to the trinomial, and distribute $1$ from the binomial to the trinomial:
$$ (2y + 1)(y^2 - 3y + 4) = 2y(y^2 - 3y + 4) + 1(y^2 - 3y + 4) $$Now, apply the distributive property (monomial by polynomial) to each part:
$$ 2y(y^2 - 3y + 4) = (2y \times y^2) + (2y \times -3y) + (2y \times 4) $$ $$ = (2 \times 1 \times y^1 \times y^2) + (2 \times -3 \times y^1 \times y^1) + (2 \times 4 \times y) $$ $$ = 2y^{1+2} - 6y^{1+1} + 8y = 2y^3 - 6y^2 + 8y $$ $$ 1(y^2 - 3y + 4) = (1 \times y^2) + (1 \times -3y) + (1 \times 4) $$ $$ = y^2 - 3y + 4 $$Add the results from the two distributions:
$$ (2y^3 - 6y^2 + 8y) + (y^2 - 3y + 4) $$ $$ = 2y^3 - 6y^2 + 8y + y^2 - 3y + 4 $$Combine the like terms ($-6y^2$ with $y^2$, and $8y$ with $-3y$):
$$ 2y^3 + (-6y^2 + y^2) + (8y - 3y) + 4 $$ $$ 2y^3 + (-6+1)y^2 + (8-3)y + 4 $$ $$ 2y^3 - 5y^2 + 5y + 4 $$The product is $\textbf{2y\textsuperscript{3} - 5y\textsuperscript{2} + 5y + 4}$.
Multiplication of algebraic expressions is a core skill that is used extensively in simplifying expressions, solving equations, factorising, and working with polynomials and functions.
Division of Algebraic Expressions (excluding Polynomial division)
Division of Algebraic Expressions
Division in algebra is essentially the inverse operation of multiplication. When we divide an algebraic expression by another, we are looking for an expression that, when multiplied by the divisor, gives the original expression (the dividend). This section focuses on simpler cases of algebraic division, specifically dividing a monomial by a monomial, and dividing a polynomial by a monomial. More complex cases, such as dividing a polynomial by another polynomial (where the divisor has two or more terms), are typically covered under the topic of Polynomial Long Division.
The process of dividing algebraic expressions often involves simplifying fractions where the numerator is the dividend and the denominator is the divisor. The rules for division are derived from the rules of arithmetic for fractions and the laws of exponents for variables.
Rules for Division of Terms
When dividing algebraic terms (or simplifying a fraction of terms), we apply the following rules:
Divide the Coefficients:
Divide the numerical coefficient of the numerator term by the numerical coefficient of the denominator term. Pay attention to the signs:- Positive $\div$ Positive = Positive
- Negative $\div$ Negative = Positive
- Positive $\div$ Negative = Negative
- Negative $\div$ Positive = Negative
Divide the Variables:
Divide the variable parts. For variables with the same base (the same letter), use the law of exponents for division: $\frac{a^m}{a^n} = a^{m-n}$, where $a \neq 0$. If a variable appears only in the numerator or only in the denominator, it remains there in the simplified result. If a variable appears with the same power in both the numerator and the denominator, it cancels out (since $\frac{x^n}{x^n} = x^{n-n} = x^0 = 1$, provided $x \neq 0$).Simplify the Result:
Combine the results from dividing the coefficients and the variable parts. Ensure the numerical part is in its simplest fractional form if it's not a whole number.
It is critically important to remember that division by zero is undefined. This means that the algebraic expression in the denominator cannot take any value(s) that would make its numerical value equal to zero. Any result obtained through division is valid only under the assumption that the divisor is non-zero.
Dividing Different Types of Expressions (Excluding Polynomial Long Division)
Let's look at the types of division covered in this section:
1. Dividing a Monomial by a Monomial
To divide a monomial by another monomial, we apply the rules mentioned above directly: divide the coefficients and then divide the variable parts.
Example 1. Divide $15xy$ by $5y$. (Assume $y \neq 0$).
Answer:
We write the division as a fraction:
$$ \frac{15xy}{5y} $$Divide the coefficients: $\frac{15}{5} = 3$.
Divide the variable parts: $\frac{xy}{y}$. We can write this as $x \times \frac{y}{y}$. Since $y \neq 0$, $\frac{y}{y} = 1$. So, $\frac{xy}{y} = x \times 1 = x$.
Alternatively, using the exponent rule $\frac{y^1}{y^1} = y^{1-1} = y^0 = 1$ (for $y \neq 0$).
Combine the numerical and variable results:
$$ \frac{15xy}{5y} = \left(\frac{15}{5}\right) \times \left(\frac{xy}{y}\right) = 3 \times x = 3x $$We can also use cancellation of common factors, both numerical and variable, in the fractional form:
$$ \frac{15xy}{5y} = \frac{\cancel{15}^{\;\;3} \times x \times \cancel{y}}{\cancel{5}^{\;\;1} \times \cancel{y}} = \frac{3 \times x \times 1}{1 \times 1} = \frac{3x}{1} = 3x $$The result of the division is $\textbf{3x}$, provided $y \neq 0$.
Example 2. Divide $-24a^5 b^3$ by $8a^2 b$. (Assume $a \neq 0, b \neq 0$).
Answer:
Write the division as a fraction:
$$ \frac{-24a^5 b^3}{8a^2 b} $$Divide the coefficients: $\frac{-24}{8} = -3$. (Negative divided by positive is negative).
Divide the variable parts:
- For variable $a$: $\frac{a^5}{a^2} = a^{5-2} = a^3$. (Using the rule $\frac{x^m}{x^n} = x^{m-n}$).
- For variable $b$: $\frac{b^3}{b}$. Here, $b$ means $b^1$. So, $\frac{b^3}{b^1} = b^{3-1} = b^2$.
Combine the numerical and variable results:
$$ \frac{-24a^5 b^3}{8a^2 b} = \left(\frac{-24}{8}\right) \times \left(\frac{a^5}{a^2}\right) \times \left(\frac{b^3}{b}\right) = -3 \times a^3 \times b^2 = -3a^3 b^2 $$Alternatively, using cancellation:
$$ \frac{-24a^5 b^3}{8a^2 b} = \frac{\cancel{-24}^{\;\;-3} \times a^{\cancel{5}^{\;\;3}} \times b^{\cancel{3}^{\;\;2}}}{\cancel{8}^{\;\;1} \times \cancel{a^2} \times \cancel{b}} = \frac{-3 \times a^3 \times b^2}{1} = -3a^3 b^2 $$The result of the division is $\textbf{-3a\textsuperscript{3}b\textsuperscript{2}}$, provided $a \neq 0$ and $b \neq 0$.
2. Dividing a Polynomial by a Monomial
To divide a polynomial by a monomial, we use the distributive property of division over addition and subtraction. This property states that when a sum or difference is divided by a single term, each term in the sum or difference is divided separately by that term.
Formula: $\frac{a+b+c}{d} = \frac{a}{d} + \frac{b}{d} + \frac{c}{d}$, where $d \neq 0$.
So, to divide a polynomial by a monomial, you divide each term of the polynomial (numerator) by the monomial (denominator) and then add or subtract the resulting monomials.
Example 3. Divide $(6x^2 + 9x)$ by $3x$. (Assume $x \neq 0$).
Answer:
Write the division as a fraction:
$$ \frac{6x^2 + 9x}{3x} $$Using the distributive property of division, divide each term in the numerator ($6x^2$ and $9x$) by the monomial denominator ($3x$).
$$ = \frac{6x^2}{3x} + \frac{9x}{3x} $$Now, perform each monomial division separately:
- First term: $\frac{6x^2}{3x} = \left(\frac{6}{3}\right) \times \left(\frac{x^2}{x}\right) = 2 \times x^{2-1} = 2x$.
- Second term: $\frac{9x}{3x} = \left(\frac{9}{3}\right) \times \left(\frac{x}{x}\right) = 3 \times x^{1-1} = 3 \times x^0 = 3 \times 1 = 3$ (since $x \neq 0$).
Add the results of the individual divisions:
$$ = 2x + 3 $$The result of the division is $\textbf{2x + 3}$, provided $x \neq 0$.
Example 4. Divide $(10a^3 b^2 - 15a^2 b^3 + 5ab)$ by $-5ab$. (Assume $a \neq 0, b \neq 0$).
Answer:
Write the division as a fraction:
$$ \frac{10a^3 b^2 - 15a^2 b^3 + 5ab}{-5ab} $$Divide each term in the numerator ($10a^3 b^2$, $-15a^2 b^3$, and $+5ab$) by the monomial denominator ($-5ab$).
$$ = \frac{10a^3 b^2}{-5ab} + \frac{-15a^2 b^3}{-5ab} + \frac{5ab}{-5ab} $$Perform each monomial division separately, paying close attention to the signs:
- First term: $\frac{10a^3 b^2}{-5ab} = \left(\frac{10}{-5}\right) \times \left(\frac{a^3}{a^1}\right) \times \left(\frac{b^2}{b^1}\right) = -2 \times a^{3-1} \times b^{2-1} = -2a^2 b$. (Positive divided by negative is negative).
- Second term: $\frac{-15a^2 b^3}{-5ab} = \left(\frac{-15}{-5}\right) \times \left(\frac{a^2}{a^1}\right) \times \left(\frac{b^3}{b^1}\right) = +3 \times a^{2-1} \times b^{3-1} = +3ab^2$. (Negative divided by negative is positive).
- Third term: $\frac{5ab}{-5ab} = \left(\frac{5}{-5}\right) \times \left(\frac{ab}{ab}\right) = -1 \times 1 = -1$. (Positive divided by negative is negative, and any non-zero term divided by itself is 1).
Combine the results of the individual divisions:
$$ = -2a^2 b + 3ab^2 - 1 $$The result of the division is $\textbf{-2a\textsuperscript{2}b + 3ab\textsuperscript{2} - 1}$, provided $a \neq 0$ and $b \neq 0$.
This method simplifies division when the divisor is a single term. When the divisor is a polynomial with two or more terms, a different technique called polynomial long division (or synthetic division in specific cases) is required, which allows us to find both a quotient and potentially a remainder.