Page 1.4: Algebraic Expressions and Operations
We will often deal with algebraic expressions in calculus. In order to understand algebraic expressions, we must first discuss variables.
Definition: A variable is a letter which denotes a real number.
In calculus, we will often use \(x\) and \(y\) as our variables but there will be instances when we may use \(t\) or \(s\). Note that a real number in front of a variable is called the coefficient.
Definition: If we start with variables and some real numbers and combine them via addition, subtraction, multiplication, division, powers, or roots, then we obtain an algebraic expression .
The following are examples of algebraic expressions:
\[ 141x^2 + 142x + 143, ~~ \sqrt{5x} + 12, ~~ \dfrac{12x^2}{\sqrt[4]{x+5} + 9} , 12xyz + 15s \nonumber\ \]
One common type of algebraic expression we will encounter is a polynomial.
Definition: A polynomial in the variable \(x\) is an expression of the form \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + \ a_0 \) where \( a_n, a_{n-1}, \ldots, a_1, a_0\) are real numbers.
Here are some examples of polynomials in the variable \(x\): \(3 x ^ { 2 }\), \(7 x + 5\), \(\frac { 3 } { 2 } x ^ { 3 } + 3 x ^ { 2 } - \frac { 1 } { 2 } x + 1\), \(6 x ^ { 2 } - 4 x + 7\).
The degree of a polynomial is defined to be the highest power of the variable that appears in the polynomial.
| Term | Degree |
|---|---|
| \(3 x ^ { 2 }\) | \(2\) |
| \(7 x + 5\) | \(1\) |
| \(\frac { 3 } { 2 } x ^ { 3 } + 3 x ^ { 2 } - \frac { 1 } { 2 } x + 1\) | \(3\) |
| \(6 x ^ { 2 } - 4 x + 7\) | \(2\) |
| \(x\) | \(1\) |
Typically, we arrange terms of polynomials in descending order based on the degree of each term. Such an arrangement is called standard form . The leading coefficient is the coefficient of the variable with the highest power, in this case, \(a_{n}\).
Write in standard form: \(3 x - 4 x ^ { 2 } + 5 x ^ { 3 } + 7 - 2 x ^ { 4 }\).
Answer :
\(- 2 x ^ { 4 } + 5 x ^ { 3 } - 4 x ^ { 2 } + 3 x + 7\)
We classify polynomials by the number of terms and the degree:
| Expression | Classification | Degree |
|---|---|---|
| \(5x^{7}\) | Monomial (one term) | \(7\) |
| \(8x^{6}-1\) | Binomial (two terms) | \(6\) |
| \(-3x^{2} +x-1\) | Trinomial (three terms) | \(2\) |
| \(5 x ^ { 3 } - 2 x ^ { 2 } + 3 x - 6\) | Polynomial (many terms) | \(3\) |
We can further classify polynomials with one variable by their degree:
In this text, we call any polynomial of degree \(n ≥ 4\) an \(n\)th-degree polynomial. In other words, if the degree is \(4\), we call the polynomial a fourth-degree polynomial. If the degree is \(5\), we call it a fifth-degree polynomial, and so on.
State whether the following polynomial is linear or quadratic and give the leading coefficient: \(25 + 4 x - x ^ { 2 }\).
Solution
The highest power is \(2\); therefore, it is a quadratic polynomial. Rewriting in standard form we have
\(- x ^ { 2 } + 4 x + 25\)
Here \(- x ^ { 2 } = - 1 x ^ { 2 }\) and thus the leading coefficient is \(−1\).
Answer :
Quadratic; leading coefficient: \(−1\)
Adding and Subtracting Algebraic Expressions
We begin by simplifying algebraic expressions that look like \(+ (a + b)\) or \(− (a + b)\). Here, the coefficients are actually implied to be \(+1\) and \(−1\) respectively and therefore the distributive property applies. Multiply each term within the parentheses by these factors as follows:
\(\begin{array} { l } { + ( a + b ) = + 1 ( a + b ) = ( + 1 ) a + ( + 1 ) b = a + b } \\ { - ( a + b ) = - 1 ( a + b ) = ( - 1 ) a + ( - 1 ) b = - a - b } \end{array}\)
Use this idea as a means to eliminate parentheses when adding and subtracting polynomials.
Add: \(9 x ^ { 2 } + \left( x ^ { 2 } - 5 \right)\).
Solution
The property \(+ (a + b) = a + b\) allows us to eliminate the parentheses, after which we can then combine like terms.
\(\begin{aligned} 9 x ^ { 2 } + \left( x ^ { 2 } - 5 \right) & = 9 x ^ { 2 } + x ^ { 2 } - 5 \\ & = 10 x ^ { 2 } - 5 \end{aligned}\)
Answer :
\(10x^{2} − 5\)
Add: \(\left( 3 x ^ { 2 } y ^ { 2 } - 4 x y + 9 \right) + \left( 2 x ^ { 2 } y ^ { 2 } - 6 x y - 7 \right)\).
Solution
Remember that the variable parts have to be exactly the same before we can add the coefficients.
\(\begin{array} { l } { \left( 3 x ^ { 2 } y ^ { 2 } - 4 x y + 9 \right) + \left( 2 x ^ { 2 } y ^ { 2 } - 6 x y - 7 \right) } \\ { = \color{Cerulean}{\underline{ 3 x ^ { 2 } y ^ { 2 }}} \color{Black}{-} \color{OliveGreen}{\underline{\underline {4 x y}}} \color{Black}{+ \underline{\underline{\underline{9}}}} + \color{Cerulean}{\underline{2 x ^ { 2 } y ^ { 2 }}} \color{Black}{-} \color{OliveGreen}{\underline{\underline {6 x y}}} \color{Black} {- \underline{\underline{\underline{7}}}}} \\ { = 5 x ^ { 2 } y ^ { 2 } - 10 x y + 2 } \end{array}\)
Answer :
\(5 x ^ { 2 } y ^ { 2 } - 10 x y + 2\)
When subtracting algebraic expressions, the parentheses become very important.
Subtract: \(4 x ^ { 2 } - \left( 3 x ^ { 2 } + 5 x \right)\).
Solution
The property \(− (a + b) = −a − b\) allows us to remove the parentheses after subtracting each term.
\(\begin{aligned} 4 x ^ { 2 } - \left( 3 x ^ { 2 } + 5 x \right) & = 4 x ^ { 2 } - 3 x ^ { 2 } - 5 x \\ & = x ^ { 2 } - 5 x \end{aligned}\)
Answer :
\(x^{2} − 5x\)
Subtracting a quantity is equivalent to multiplying it by \(−1\).
Subtract: \(\left( 3 x ^ { 2 } - 2 x y + y ^ { 2 } \right) - \left( 2 x ^ { 2 } - x y + 3 y ^ { 2 } \right)\).
Solution
Distribute the \(−1\), remove the parentheses, and then combine like terms. Multiplying the terms of a polynomial by \(−1\) changes all the signs.
\(\begin{array} { l } { = 3 x ^ { 2 } - 2 x y + y ^ { 2 } - 2 x ^ { 2 } + x y - 3 y ^ { 2 } } \\ { = x ^ { 2 } - x y - 2 y ^ { 2 } } \end{array}\)
Answer :
\(x^{2} − xy − 2y^{2}\)
Subtract: \(\left( 7 a ^ { 2 } - 2 a b + b ^ { 2 } \right) - \left( a ^ { 2 } - 2 a b + 5 b ^ { 2 } \right)\).
- Answer
-
\(6 a ^ { 2 } - 4 b ^ { 2 }\)
www.youtube.com/v/IDtREB_PQ3A
Multiplying Algebraic Expressions
Use the product rule for exponents, \(x ^ { m } \cdot x ^ { n } = x ^ { m + n }\), to multiply a monomial times a polynomial. In other words, when multiplying two expressions with the same base, add the exponents. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,
\(\begin{aligned} 7 x ^ { 4 } \cdot 8 x ^ { 3 } & = 7 \cdot 8 \cdot x ^ { 4 } \cdot x ^ { 3 } \color{Cerulean} { Commutative \:property } \\ & = 56 x ^ { 4 + 3 } \quad\quad \color {Cerulean} { Product \:rule \:for \:exponents } \\ & = 56 x ^ { 7 } \end{aligned}\)
To multiply a polynomial by a monomial, apply the distributive property, and then simplify each term.
Multiply: \(5 x y ^ { 2 } \left( 2 x ^ { 2 } y ^ { 2 } - x y + 1 \right)\).
Solution
Apply the distributive property and then simplify.
\(\begin{array} { l } { = \color{Cerulean}{5 x y ^ { 2 }} \color{Black}{\cdot} 2 x ^ { 2 } y ^ { 2 } - \color{Cerulean}{5 x y ^ { 2 }}\color{Black}{ \cdot} x y + \color{Cerulean}{5 x y ^ { 2 }} \color{Black}{ \cdot 1 }} \\ { = 10 x ^ { 3 } y ^ { 4 } - 5 x ^ { 2 } y ^ { 3 } + 5 x y ^ { 2 } } \end{array}\)
Answer :
\(10 x ^ { 3 } y ^ { 4 } - 5 x ^ { 2 } y ^ { 3 } + 5 x y ^ { 2 }\)
To summarize, multiplying a polynomial by a monomial involves the distributive property and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add exponents of variables where the bases are the same.
In the same manner that we used the distributive property to distribute a monomial, we use it to distribute a binomial.
\(\begin{aligned} \color{Cerulean}{( a + b )}\color{Black}{ ( c + d )} & = \color{Cerulean}{( a + b )}\color{Black}{ \cdot} c + \color{Cerulean}{( a + b )}\color{Black}{ \cdot} d \\ & = a c + b c + a d + b d \\ & = a c + a d + b c + b d \end{aligned}\)
Here we apply the distributive property multiple times to produce the final result. This same result is obtained in one step if we apply the distributive property to \(a\) and \(b\) separately as follows:
This is often called the FOIL method. Multiply the first, outer, inner, and then last terms.
Multiply: \(( 6 x - 1 ) ( 3 x - 5 )\).
Solution
Distribute \(6x\) and \(−1\) and then combine like terms.
\(\begin{aligned} ( 6 x - 1 ) ( 3 x - 5 ) & = \color{Cerulean}{6 x}\color{Black}{ \cdot} 3 x - \color{Cerulean}{6 x}\color{Black}{ \cdot} 5 + ( \color{OliveGreen}{- 1}\color{Black}{ )} \cdot 3 x - ( \color{OliveGreen}{- 1}\color{Black}{ )} \cdot 5 \\ & = 18 x ^ { 2 } - 30 x - 3 x + 5 \\ & = 18 x ^ { 2 } - 33 x + 5 \end{aligned}\)
Answer :
\(18 x ^ { 2 } - 33 x + 5\)
Consider the following two calculations:
| \(\begin{aligned} ( a + b ) ^ { 2 } & = ( a + b ) ( a + b ) \\ & = a ^ { 2 } + a b + b a + b ^ { 2 } \\ & = a ^ { 2 } + a b + a b + b ^ { 2 } \\ & = a ^ { 2 } + 2 a b + b ^ { 2 } \end{aligned}\) | \(\begin{aligned} ( a - b ) ^ { 2 } & = ( a - b ) ( a - b ) \\ & = a ^ { 2 } - a b - b a + b ^ { 2 } \\ & = a ^ { 2 } - a b - a b + b ^ { 2 } \\ & = a ^ { 2 } - 2 a b + b ^ { 2 } \end{aligned}\) |
This leads us to two formulas that describe perfect square trinomials :
\(\begin{array} { l } { ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 } } \\ { ( a - b ) ^ { 2 } = a ^ { 2 } - 2 a b + b ^ { 2 } } \end{array}\)
We can use these formulas to quickly square a binomial.
Multiply: \((3x+5)^{2}\)
Solution
Here \(a=3x\) and \(b=5\). Apply the formula:
Answer :
\(9 x ^ { 2 } + 30 x + 25\)
This process should become routine enough to be performed mentally. Our third special product follows:
\(\begin{aligned} ( a + b ) ( a - b ) & = a ^ { 2 } - a b + b a - b ^ { 2 } \\ & = a ^ { 2 } \color{red}{- a b + a b}\color{Black}{ -} b ^ { 2 } \\ & = a ^ { 2 } - b ^ { 2 } \end{aligned}\)
This product is called difference of squares :
\(( a + b ) ( a - b ) = a ^ { 2 } - b ^ { 2 }\)
The binomials \((a + b)\) and \((a − b)\) are called conjugate binomials . When multiplying conjugate binomials the middle terms are opposites and their sum is zero; the product is itself a binomial.
Multiply: \((3xy + 1) (3xy − 1)\).
Solution
\(\begin{aligned} ( 3 x y + 1 ) ( 3 x y - 1 ) & = ( 3 x y ) ^ { 2 } - 3 x y + 3 x y - 1 ^ { 2 } \\ & = 9 x ^ { 2 } y ^ { 2 } - 1 \end{aligned}\)
Answer :
\(9x^{2}y^{2} − 1\)
Multiply: \(\left( x ^ { 2 } + 5 y ^ { 2 } \right) \left( x ^ { 2 } - 5 y ^ { 2 } \right)\).
- Answer
-
\(\left( x ^ { 4 } - 25 y ^ { 4 } \right)\)
www.youtube.com/v/p7R3FdPp6_s
Multiply: \((5x − 2)^{3}\).
Solution
Here we perform one product at a time.
Answer :
\(125x^{2} − 150x^{2} + 60x − 8\)
Dividing Polynomials
Use the quotient rule for exponents, \(\frac { x ^ { m } } { x ^ { n } } = x ^ { m - n }\), to divide a polynomial by a monomial. In other words, when dividing two expressions with the same base, subtract the exponents. In this section, we will assume that all variables in the denominator are nonzero.
Divide: \(\frac { 24 x ^ { 7 } y ^ { 5 } } { 8 x ^ { 3 } y ^ { 2 } }\).
Solution
Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.
\(\begin{aligned} \frac { 24 x ^ { 7 } y ^ { 5 } } { 8 x ^ { 3 } y ^ { 2 } } & = \frac { 24 } { 8 } x ^ { 7 - 3 } y ^ { 5 - 2 } \\ & = 3 x ^ { 4 } y ^ { 3 } \end{aligned}\)
Answer :
\(3 x ^ { 4 } y ^ { 3 }\)
When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and break up the fraction using the following property:
\(\frac { a + b } { c } = \frac { a } { c } + \frac { b } { c }\)
Applying this property will result in terms that can be treated as quotients of monomials.
Divide: \(\frac { - 5 x ^ { 4 } + 25 x ^ { 3 } - 15 x ^ { 2 } } { 5 x ^ { 2 } }\).
Solution
Break up the fraction by dividing each term in the numerator by the monomial in the denominator, and then simplify each term.
\(\begin{aligned} \frac { - 5 x ^ { 4 } + 25 x ^ { 3 } - 15 x ^ { 2 } } { 5 x ^ { 2 } } & = - \frac { 5 x ^ { 4 } } { 5 x ^ { 2 } } + \frac { 25 x ^ { 3 } } { 5 x ^ { 2 } } - \frac { 15 x ^ { 2 } } { 5 x ^ { 2 } } \\ & = - \frac { 5 } { 5 } x ^ { 4 - 2 } + \frac { 25 } { 5 } x ^ { 3 - 2 } - \frac { 15 } { 5 } x ^ { 2 - 2 } \\ & = - 1 x ^ { 2 } + 5 x ^ { 1 } - 3 x ^ { 0 } \\ & = - x ^ { 2 } + 5 x - 3 \cdot 1 \end{aligned}\)
Answer :
\(- x ^ { 2 } + 5 x - 3\)
We can check our division by multiplying our answer, the quotient, by the monomial in the denominator, the divisor, to see if we obtain the original numerator, the dividend.
| \({\frac{Dividend}{Divisor}=Quotient}\) | \(\frac { - 5 x ^ { 4 } + 25 x ^ { 3 } - 15 x ^ { 2 } } { 5 x ^ { 2 } } = - x ^ { 2 } + 5 x-3\) |
|---|---|
| or | or |
| \(Dividend = Divisor\cdot Quotient\) | \(- 5 x ^ { 4 } + 25 x ^ { 3 } - 15 x ^ { 2 } = 5 x ^ { 2 } ( - x ^ { 2 } + 5x-3)\) |
The same technique outlined for dividing by a monomial does not work for polynomials with two or more terms in the denominator. In this section, we will outline a process called polynomial long division , which is based on the division algorithm for real numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.
Divide \(\frac { x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4 } { x - 2 }\):
Solution
Here \(x−2\) is the divisor and \(x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4\) is the dividend. To determine the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor.
Multiply the first term of the quotient by the divisor, remembering to distribute, and line up like terms with the dividend.
Subtract the resulting quantity from the dividend. Take care to subtract both terms.
Bring down the remaining terms and repeat the process.
Notice that the leading term is eliminated and that the result has a degree that is one less. The complete process is illustrated below:
Polynomial long division ends when the degree of the remainder is less than the degree of the divisor. Here, the remainder is \(0\). Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown above the division bar.
\(\frac { x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4 } { x - 2 } = x ^ { 2 } + 5 x + 2\)
To check the answer, multiply the divisor by the quotient to see if you obtain the dividend as illustrated below:
\(x ^ { 3 } + 3 x ^ { 2 } - 8 x - 4 = ( x - 2 ) \left( x ^ { 2 } + 5 x + 2 \right)\)
This is left to the reader as an exercise.
Answer :
\(x ^ { 2 } + 5 x + 2\)
Next, we demonstrate the case where there is a nonzero remainder.
Just as with real numbers, the final answer adds to the quotient the fraction where the remainder is the numerator and the divisor is the denominator. In general, when dividing we have:
\(\frac{Dividend}{Divisor}=\color{Cerulean}{Quotient}\color{Black}{+}\frac{\color{OliveGreen}{Remainder}}{\color{Black}{Divisor}}\)
If we multiply both sides by the divisor we obtain,
\(Dividend=\color{Cerulean}{Quotient}\color{Black}{\times}Divisor +\color{OliveGreen}{Remainder}\)
Divide: \(\frac { 6 x ^ { 2 } - 5 x + 3 } { 2 x - 1 }\).
Solution
Since the denominator is a binomial, begin by setting up polynomial long division.
To start, determine what monomial times \(2x−1\) results in a leading term \(6x^{2}\).This is the quotient of the given leading terms: \((6x^{2})÷(2x)=3x\). Multiply \(3x\) times the divisor \(2x−1\), and line up the result with like terms of the dividend.
Subtract the result from the dividend and bring down the constant term \(+3\).
Subtracting eliminates the leading term. Multiply \(2x−1\) by \(−1\) and line up the result.
Subtract again and notice that we are left with a remainder.
The constant term \(2\) has degree \(0\) and thus the division ends. Therefore,
\(\frac { 6 x ^ { 2 } - 5 x + 3 } { 2 x - 1 } = \color{Cerulean}{3 x - 1}\color{Black}{ +} \frac { \color{OliveGreen}{2} } {\color{Black}{ 2 x - 1} }\)
To check that this result is correct, we multiply as follows:
\(\begin{aligned} \color{Cerulean}{ {quotient }}\color{Black}{ \times} divisor + \color{OliveGreen} {remainder} & \color{Black}{=} \color{Cerulean}{( 3 x - 1 )}\color{Black}{ (} 2 x - 1 ) + \color{OliveGreen}{2} \\ & = 6 x ^ { 2 } - 3 x - 2 x + 1 + 2 \\ & = 6 x ^ { 2 } - 5 x + 2 = dividend\:\: \color{Cerulean}{✓} \end{aligned} \)
Answer :
\(3 x - 1 + \frac { 2 } { 2 x - 1 }\)
Occasionally, some of the powers of the variables appear to be missing within a polynomial. This can lead to errors when lining up like terms. Therefore, when first learning how to divide polynomials using long division, fill in the missing terms with zero coefficients, called placeholders .
Divide: \(\frac { 27 x ^ { 3 } + 64 } { 3 x + 4 }\).
Solution
Notice that the binomial in the numerator does not have terms with degree \(2\) or \(1\). The division is simplified if we rewrite the expression with placeholders:
\(27 x ^ { 3 } + 64 = 27 x ^ { 3 } + \color{OliveGreen}{0 x ^ { 2 }}\color{Black}{ +}\color{OliveGreen}{ 0 x}\color{Black}{ +} 64\)
Set up polynomial long division:
We begin with 27x3÷3x=9x2 and work the rest of the division algorithm.
Answer :
\(9 x ^ { 2 } - 12 x + 16\)
Divide: \(\frac { 3 x ^ { 4 } - 2 x ^ { 3 } + 6 x ^ { 2 } + 23 x - 7 } { x ^ { 2 } - 2 x + 5 }\).
Solution
Begin the process by dividing the leading terms to determine the leading term of the quotient \(3x^{4}÷x^{2}=\color{Cerulean}{3x^{2}}\). Take care to distribute and line up the like terms. Continue the process until the remainder has a degree less than \(2\).
The remainder is \(x−2\). Write the answer with the remainder:
\(\frac { 3 x ^ { 4 } - 2 x ^ { 3 } + 6 x ^ { 2 } + 23 x - 7 } { x ^ { 2 } - 2 x + 5 } = 3 x ^ { 2 } + 4 x - 1 + \frac { x - 2 } { x ^ { 2 } - 2 x + 5 }\)
Answer :
\(3 x ^ { 2 } + 4 x - 1 + \frac { x - 2 } { x ^ { 2 } - 2 x + 5 }\)
Polynomial long division takes time and practice to master. Work lots of problems and remember that you may check your answers by multiplying the quotient by the divisor (and adding the remainder if present) to obtain the dividend.
Divide: \(\frac { 6 x ^ { 4 } - 13 x ^ { 3 } + 9 x ^ { 2 } - 14 x + 6 } { 3 x - 2 }\).
- Answer
-
\(2 x ^ { 3 } - 3 x ^ { 2 } + x - 4 - \frac { 2 } { 3 x - 2 }\)
www.youtube.com/v/K9NRVMreKAQ