How to solve SAT equivalent expressions questions

Equivalent expressions questions ask you to rewrite an expression in a different form without changing its value for any $x$. On the Digital SAT that includes expanding and factoring polynomials, applying exponent and radical rules, simplifying rational expressions, and matching coefficients in an identity. Some questions also ask you to rearrange a formula to isolate one variable. Advanced Math is about 35% of the Math section, or 13 to 15 of the 44 questions, and has only three skills, so expect several questions on this one.

What the questions look like

Pick the equivalent form. You get an expression like $(3x - 4)(2x + 5) - (x - 2)^2$ and choose the expanded or factored version from four choices.

Exponents and radicals. You rewrite something like $\sqrt[3]{x^2}$ as $x^{2/3}$, or simplify $\dfrac{(x^3y^{-2})^2}{x^{1/2}}$.

Match coefficients. An equation such as $(ax + 3)(2x - b) = 10x^2 - 4x - 6$ holds for all $x$, and you find $a$, $b$, or a combination of them.

Rational expressions. You add two fractions with variable denominators, or rewrite $\dfrac{x^2 + 5x + 7}{x + 2}$ as a polynomial plus a remainder over $x + 2$.

Step-by-step method

1. Look at the choices first. If they are expanded, expand the given expression. If they are factored, factor it. This tells you which direction to work.
2. Expand carefully. Multiply every term in one factor by every term in the other. For a squared binomial, use $(a - b)^2 = a^2 - 2ab + b^2$.
3. Use the exponent rules.
   • $x^m \cdot x^n = x^{m+n}$ and $\dfrac{x^m}{x^n} = x^{m-n}$
   • $(x^m)^n = x^{mn}$ and $(xy)^n = x^n y^n$
   • $x^{-n} = \dfrac{1}{x^n}$ and $x^{m/n} = \sqrt[n]{x^m}$
4. Match coefficients for identities. Expand the side with unknowns, then set the coefficient of $x^2$ equal to the coefficient of $x^2$, the $x$ term to the $x$ term, and the constant to the constant.
5. Check with a test value. Substitute a number like $x = 2$ or $x = 3$ into the original and into your answer. Equivalent expressions give the same output. Avoid 0 and 1, since they make different expressions look equal too often.

Worked example

Example 1. Which expression is equivalent to $(3x - 4)(2x + 5) - (x - 2)^2$?

Expand the product:

$$(3x - 4)(2x + 5) = 6x^2 + 15x - 8x - 20 = 6x^2 + 7x - 20$$

Expand the square:

$$(x - 2)^2 = x^2 - 4x + 4$$

Subtract the whole square, distributing the negative to all three terms:

$$6x^2 + 7x - 20 - x^2 + 4x - 4 = 5x^2 + 11x - 24$$

Test with $x = 2$. The original gives $(2)(9) - 0^2 = 18$. The answer gives $5(4) + 11(2) - 24 = 18$. They match.

Example 2. The equation $(ax + 3)(2x - b) = 10x^2 - 4x - 6$ is true for all values of $x$, where $a$ and $b$ are constants. What is the value of $a + b$?

Expand the left side:

$$(ax + 3)(2x - b) = 2ax^2 - abx + 6x - 3b = 2ax^2 + (6 - ab)x - 3b$$

Match coefficients with $10x^2 - 4x - 6$:

• $x^2$ terms: $2a = 10$, so $a = 5$.
• Constants: $-3b = -6$, so $b = 2$.
• $x$ terms, as a check: $6 - ab = 6 - 10 = -4$. It matches.

So $a + b = 7$. Always use the third coefficient to confirm, since a mistake in $a$ or $b$ will show up there.

Common traps

• Squaring a binomial term by term. $(x - 2)^2$ is $x^2 - 4x + 4$, not $x^2 + 4$ or $x^2 - 4$. The middle term is the one students drop.
• Subtracting only the first term. In Example 1, $-(x^2 - 4x + 4)$ becomes $-x^2 + 4x - 4$. Leaving the $+4x$ or $-4$ unchanged produces a wrong choice that the test writers expect.
• Mixing up exponent rules. $(x^3)^2 = x^6$, but $x^3 \cdot x^2 = x^5$. Powers of powers multiply, and products of the same base add exponents.
• Canceling terms instead of factors. $\dfrac{x + 6}{x + 2}$ does not simplify to $3$. You can only cancel a factor that multiplies the entire numerator and the entire denominator.
• Testing with 0 or 1. At $x = 1$, $x^2$, $x^3$, and $x^{1/2}$ all equal 1, so a wrong choice can pass the test. Use 2 or 3, and test a second value if two choices tie.

Desmos tip

Graph the original expression as $y = (3x - 4)(2x + 5) - (x - 2)^2$, then graph each answer choice. The equivalent choice draws exactly on top of the original curve. If two graphs look close, click the curves or add a table at a few $x$ values to compare outputs. For coefficient matching, you can also graph both sides with sliders for $a$ and $b$ and adjust until the curves coincide, then confirm the values by algebra.