Reference Sheet · SAT Math · Algebra
Algebra & Exponents
Exponent rules, factoring patterns, and algebraic identities you should be able to recognize on sight. Each card includes a brief worked example.
Exponent rules
Multiplication
$$a^b \cdot a^c = a^{b+c}$$
Same base, multiply → add the exponents.
EXAMPLE
Simplify $x^5 \cdot x^3$.
$x^5 \cdot x^3 = x^{5+3} = $ $x^8$
Division
$$\frac{a^b}{a^c} = a^{b-c}$$
Same base, divide → subtract the exponents.
EXAMPLE
Simplify $\dfrac{x^7}{x^2}$.
$\dfrac{x^7}{x^2} = x^{7-2} = $ $x^5$
Power of a power
$$(a^b)^c = a^{bc}$$
Raising a power to a power → multiply the exponents.
EXAMPLE
Simplify $(x^2)^4$.
$(x^2)^4 = x^{2 \cdot 4} = $ $x^8$
Power of a product
$$(ab)^c = a^c b^c$$
The exponent distributes over multiplication. (Not over addition.)
EXAMPLE
Simplify $(2x)^3$.
$(2x)^3 = 2^3 \cdot x^3 = $ $8x^3$
Negative exponent
$$a^{-b} = \frac{1}{a^b}$$
A negative exponent flips the base to the denominator. The negative sign goes away after the flip.
EXAMPLE
Rewrite $3x^{-2}$ without negative exponents.
$3x^{-2} = $ $\dfrac{3}{x^2}$
Note: only the $x$ has the negative exponent, so only the $x$ moves down.
Fractional exponent
$$a^{b/c} = \sqrt[c]{a^b}$$
numerator = power
denominator = root
denominator = root
"Power over root" — the bottom of the fraction tells you what root to take.
EXAMPLE
Evaluate $8^{2/3}$.
$8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = $ $4$
Take the cube root first, then square — easier than cubing $8$ first.
Zero exponent
$$a^0 = 1 \quad (a \neq 0)$$
Anything (other than zero) raised to the zero power equals 1.
EXAMPLE
What is $(5x^2)^0$?
$1$ — the entire expression is being raised to the zero power.
Common mistakes
$(a + b)^2 \neq a^2 + b^2$
The exponent does not distribute over addition. See "perfect square trinomial" below for the correct expansion.
$a^b + a^c \neq a^{b+c}$
Adding powers is not the same as multiplying powers. The rules only combine exponents when bases are multiplied.
The exponent does not distribute over addition. See "perfect square trinomial" below for the correct expansion.
$a^b + a^c \neq a^{b+c}$
Adding powers is not the same as multiplying powers. The rules only combine exponents when bases are multiplied.
Imaginary numbers
Definition of $i$
$$i = \sqrt{-1} \quad\Rightarrow\quad i^2 = -1$$
$i$ is defined as the square root of $-1$. Squaring it gives $-1$.
EXAMPLE
Simplify $i^2 \cdot i^2$.
$i^2 \cdot i^2 = (-1)(-1) = $ $1$
Powers of $i$ cycle every four
$$i^1 = i,\ \ i^2 = -1,\ \ i^3 = -i,\ \ i^4 = 1$$
To compute $i^n$ for any $n$: divide $n$ by 4 and use the remainder.
EXAMPLE
Find $i^{15}$.
$15 \div 4 = 3$ remainder $3$, so $i^{15} = i^3 = $ $-i$
Algebraic identities
Difference of squares
$$a^2 - b^2 = (a + b)(a - b)$$
Two perfect squares with a minus sign between them. Always factorable.
EXAMPLE
Factor $x^2 - 25$.
$x^2 - 5^2 = $ $(x + 5)(x - 5)$
Perfect square trinomial
$$(a \pm b)^2 = a^2 \pm 2ab + b^2$$
The middle term is twice the product of $a$ and $b$. Spotting this saves a lot of work on factoring problems.
EXAMPLE
Factor $x^2 + 10x + 25$.
Half of $10$ is $5$, and $5^2 = 25$ ✓
$(x + 5)^2$
Sum of cubes
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
Less common on the SAT but worth knowing. The second factor is not a perfect square — note the $-ab$ middle term.
EXAMPLE
Factor $x^3 + 8$.
$8 = 2^3$, so $x^3 + 2^3 = $ $(x + 2)(x^2 - 2x + 4)$
Difference of cubes
$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
Memory aid: SOAP — Same sign, Opposite sign, Always Plus (signs in the factored form, in order).
EXAMPLE
Factor $x^3 - 27$.
$27 = 3^3$, so $x^3 - 3^3 = $ $(x - 3)(x^2 + 3x + 9)$
Completing the square
The procedure
$$x^2 + bx \ \ \longrightarrow \ \ \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2$$
Used for finding the vertex of a parabola, finding the center of a circle from expanded form, and solving quadratics that don't factor cleanly. The trick is that you can always force any $x^2 + bx$ expression into the shape of a perfect square — at the cost of one leftover constant.
- Identify $b$. It's the coefficient of the $x$ term (the middle one). Watch the sign — if you see $x^2 - 6x$, then $b = -6$, not $6$.
- Take half of $b$. Call the result $h$. So $h = \dfrac{b}{2}$. This number is going to be the new constant inside the squared parenthesis.
- Square it. Compute $h^2 = \left(\dfrac{b}{2}\right)^2$. This is the magic number you'll add and immediately subtract.
- Add and subtract $h^2$. Inside the expression, write $+ h^2 - h^2$. Net change is zero, so the value stays the same — but now the first three terms form a perfect square.
- Factor the perfect square. The first three terms factor as $(x + h)^2$. The leftover $-h^2$ stays outside, combined with any original constant.
WORKED EXAMPLE
Convert $x^2 + 6x + 5$ to vertex form.
Step 1. $b = 6$.
Step 2. $h = \dfrac{b}{2} = \dfrac{6}{2} = 3$.
Step 3. $h^2 = 3^2 = 9$.
Step 4. Add and subtract $9$:
$x^2 + 6x + 5 = \underbrace{x^2 + 6x + 9}_{\text{perfect square}} - 9 + 5$
Step 5. Factor the perfect square and combine constants:
$= (x + 3)^2 - 4$
The vertex is at $(-3, -4)$ — read directly off vertex form.
If the leading coefficient isn't 1
Factor the leading coefficient out of the first two terms first, complete the square inside the parenthesis, then distribute back at the end.
EXAMPLE
Convert $2x^2 + 12x + 5$ to vertex form.
Factor $2$ out of the first two terms: $2(x^2 + 6x) + 5$
Inside the parenthesis: half of $6$ is $3$, squared is $9$.
Add and subtract $9$ inside: $2(x^2 + 6x + 9 - 9) + 5$
Factor and separate: $2[(x + 3)^2 - 9] + 5$
Distribute the $2$: $2(x + 3)^2 - 18 + 5 = $ $2(x + 3)^2 - 13$