Interpret the parts of an expression (terms, factors, coefficients) in context, and rewrite expressions using structure, including factoring and the properties of exponents, to reveal meaning such as a zero, a rate, or a percent change.

What this topic is asking

The Regents Algebra I exam (standards in the Seeing Structure in Expressions, A-SSE, cluster) wants you to read an algebraic expression, naming its terms, factors and coefficients and saying what they mean in context, and to rewrite an expression so a hidden feature becomes visible: a zero from factored form, a maximum from vertex form, or a percent change from an exponential base. The skill is not just manipulation, it is manipulation with a purpose.

Terms, factors, and coefficients

An expression is a combination of numbers, variables, and operations with no equals sign. Its building blocks have names you are expected to use:

• A term is a part separated by $+$ or $-$. In $5x^2 - 3x + 7$ the terms are $5x^2$, $-3x$, and $7$.
• A factor is something being multiplied. In the term $5x^2$ the factors are $5$, $x$, and $x$.
• A coefficient is the numerical factor of a term: the coefficient of $5x^2$ is $5$.

In a contextual problem you must connect these to meaning. If a phone plan costs $C = 30 + 0.10m$ dollars for $m$ minutes, the term $30$ is the fixed monthly fee, the coefficient $0.10$ is the cost per minute, and $m$ is the number of minutes.

Rewriting to reveal a zero or a maximum

The same quadratic can be written three ways, each revealing a different feature.

To expose the zeros, factor. To expose the turning point (the maximum height of a thrown ball, the minimum cost), complete the square into vertex form. Choosing the right rewrite for the question asked is exactly what the A-SSE standard rewards.

Rewriting to reveal a rate or percent change

Exponential expressions hide a growth or decay rate inside the base. Writing the base as $1 + r$ (growth) or $1 - r$ (decay) exposes the percent change.

For $V = 25000(0.85)^t$ modeling a car's value, the base $0.85 = 1 - 0.15$, so the car loses 15% per year. For $A = 500(1.03)^t$, the base $1.03 = 1 + 0.03$, so the account grows 3% per year. You can also rewrite to change the time unit: $(1.03)^t = \big((1.03)^{1/12}\big)^{12t}$ converts an annual rate into a monthly factor, a structure-rewrite the exam sometimes asks for.

A second idea worth stressing is that an equivalent expression must give the same value for every input, which is your built-in check. After factoring $x^2 - 5x + 6$ into $(x - 2)(x - 3)$, substituting $x = 0$ into both gives $6$, confirming the rewrite. Checking one easy value catches almost every sign or coefficient slip, and the Regents graders look for an answer that is genuinely equivalent, not merely similar.

Try this

Q1. Identify the coefficient and the constant term in $7 - 4x$. [1 credit]

• Cue. The coefficient of $x$ is $-4$; the constant term is $7$.

Q2. Rewrite $x^2 - 9$ in factored form and state its zeros. [2 credits]

• Cue. Difference of squares: $(x - 3)(x + 3)$, zeros $x = 3$ and $x = -3$.