Simplify without losing restrictions

Name the structure before calculating

$3x + 5x$ becomes $8x$ because both terms are multiples of the same variable. In $3(x + 5)$ , distribute $3$ to both terms: $3x + 15$ .

For a square, $(x + 3)^2$ means $(x + 3)(x + 3)$ . Its expansion is $x^2 + 6x + 9$ . The middle term comes from two cross products.

For integer powers of the same nonzero base, multiply by adding exponents, divide by subtracting exponents, and raise a power to another integer power by multiplying exponents. These are three different operations.

An even root uses the nonnegative result. Thus $\sqrt{(x - 2)^2} = \lvert x - 2 \rvert$ . At $x = 1$ , the root is $1$ , while $x - 2 = -1$ . In a ratio, $\frac{\sqrt{(x - 2)^2}}{(x - 2)^2} = \frac{1}{\lvert x - 2 \rvert}$ only for $x \ne 2$ . Keep both the absolute value and the excluded point.

x^2 \cdot x^3 = x^5

\frac{x^5}{x^2} = x^3

(x^2)^3 = x^6

Identify the operation before choosing the exponent rule.

Cancel factors and keep the missing point

Simplify $\frac{x^2 - 9}{x - 3}$ . First record $x \ne 3$ , because the original denominator cannot be zero.

Factor $x^2 - 9$ as $(x - 3)(x + 3)$ . Cancel the shared factor $x - 3$ only for allowed $x$ . The simplified formula is $x + 3$ , still with $x \ne 3$ .

At $x = 3$ , the simplified formula gives $6$ but the original fraction is $\frac{0}{0}$ and undefined. An equivalent expression must retain that restriction. Cancellation does not fill the missing point.

For $\frac{x + 2}{x - 3} - \frac{x - 2}{x + 3}$ , exclude $3$ and $-3$ , then use the common denominator $x^2 - 9$ . The numerator is $(x + 2)(x + 3) - (x - 2)(x - 3)$ . Expanding and subtracting the entire second product cancels the square and constant terms, leaving $10x$ . The result is $\frac{10x}{x^2 - 9}$ , with both exclusions retained.

\frac{x^2 - 9}{x - 3}

x + 3;\quad x \ne 3

The formula becomes shorter; its original restriction remains.

Use a relationship directly

Suppose $x > 0$ and $x + \frac{1}{x} = 5$ . Find $x^2 + \frac{1}{x^2}$ without solving for $x$ .

Square the given sum. The middle term is $2 \cdot x \cdot \frac{1}{x} = 2$ . Remove that cross term to leave the quantity being asked for.

x + \frac{1}{x} = 5

x^2 + 2 + \frac{1}{x^2} = 25

The cross term has a fixed value even though x is unknown.

Try it

If $x > 0$ and $x + \frac{1}{x} = 5$ , what is $x^2 + \frac{1}{x^2}$ ?

$23$
$25$

Show the answer

$23$

Squaring gives $x^2 + 2 + \frac{1}{x^2} = 25$ , so subtract $2$ to obtain $23$ .

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Practice Algebraic expressions

Simplify without losing restrictions

Name the structure before calculating

Lesson help

About this lesson

Name the structure before calculating

Cancel factors and keep the missing point

Use a relationship directly

Try it

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