1
Add and subtract: like terms only, and the minus hits everything.
Like terms share the same letter and the same exponent. x² and x never combine.
combine
3x² − 2x² = x²
−4x + 6x = 2x
do not
3x² + 2x
already simplified
(3x² − 4x + 1) − 2(x² − 3x + 5)
the opening problem
−2 · x² = −2x², −2 · (−3x) = +6x, −2 · 5 = −10
every term, sign included
3x² − 4x + 1 − 2x² + 6x − 10
now it is one long sum
= x² + 2x − 9
collect the three families
The exponent never moves
3x² − 2x² = x², not x⁰ and not 1x⁴. You are counting how many x² you have; the
power stays exactly where it is.
2
Multiply: every term meets every term.
FOIL is only the two-by-two case. Two terms times three terms means six products.
(2x − 3)(x² + 4x − 5)
2x · x² = 2x³, 2x · 4x = 8x², 2x · (−5) = −10x
the 2x meets all three
−3 · x² = −3x², −3 · 4x = −12x, −3 · (−5) = +15
then the −3 does the same
2x³ + (8x² − 3x²) + (−10x − 12x) + 15
gather like terms
= 2x³ + 5x² − 22x + 15
standard form, highest power first
A square is a multiplication too: (a + b)² = a² + 2ab + b².
(2x + 5)² = (2x + 5)(2x + 5)
write it twice if it helps
(2x)² = 4x², 2(2x)(5) = 20x, 5² = 25
square, double the cross, square
= 4x² + 20x + 25
three terms, never two
(2x + 5)² is not 4x² + 25
The middle term 20x is the whole point, and the SAT always offers the answer that
leaves it out. Squaring a sum is not squaring each piece.
3
The difference of squares, read in both directions.
a² − b² = (a + b)(a − b). Two squares with a minus between them, and the middle terms always cancel.
(x + 7)(x − 7) = x² − 7x + 7x − 49
expanding: the x terms kill each other
= x² − 49
two terms out of four
9x² − 16 = (3x)² − 4²
factoring: name the two squares
= (3x + 4)(3x − 4)
same pair, opposite signs
| Expression | Factors as | Note |
|---|
| x² − 25 | (x + 5)(x − 5) | the plain case |
|---|
| 49x² − 36y² | (7x + 6y)(7x − 6y) | two letters, same idea |
|---|
| x⁴ − 16 | (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2) | it can happen twice |
|---|
| x² + 25 | does not factor | a sum of squares is prime |
|---|
This is also the fastest expansion on the test. If a stem shows (2x + 9)(2x − 9),
do not FOIL it — write 4x² − 81 and move on.
4
Leading coefficient 1: find the pair that multiplies to c and adds to b.
Read the signs before you test anything — they tell you which pairs are even possible.
x² + 7x − 18
a = 1
b = 7
c = −18
| Pair | Product | Sum | Works? |
|---|
| −1 and 18 | −18 | 17 | no |
| −2 and 9 | −18 | 7 | ✓ yes |
| −3 and 6 | −18 | 3 | no |
| 3 and −6 | −18 | −3 | no |
x² + 7x − 18 = (x − 2)(x + 9)
What the signs tell you
c is negative, so the two numbers have opposite signs. b is positive, so the bigger
of the two is the positive one. That leaves only a handful of pairs to test — and
on the SAT the factored form is often the answer choice itself.
5
Leading coefficient not 1: multiply the ends, split, group.
When a ≠ 1 the pair you need lives in a × c, because that a gets shared between the brackets.
8x² − 2x − 15
a × c = 8 × (−15) = −120
multiply the two ends
−12 and 10: product −120, sum −2
the pair that does both jobs
8x² − 12x + 10x − 15
split the middle term
4x(2x − 3) + 5(2x − 3)
GCF out of each pair
= (2x − 3)(4x + 5)
the matching bracket comes out
4x out of 8x² − 12x
8x² ÷ 4x = 2x
−12x ÷ 4x = −3
5 out of 10x − 15
10x ÷ 5 = 2x
−15 ÷ 5 = −3
If the two brackets do not match, the GCF was wrong
Sometimes you must pull out a negative to make them agree — from −9x + 3 take −3,
not 3, and the bracket becomes (3x − 1). Matching brackets are the signal that the
split was right.
6
Two terms that are both cubes: SOAP.
a³ − b³ = (a − b)(a² + ab + b²) · a³ + b³ = (a + b)(a² − ab + b²)
S — Same
the binomial keeps the original sign
O — Opposite
the middle term flips it
AP — Always Positive
the last term never changes
1
8
27
64
125
216
343
512
x³ − 64, so a = x and b = 4
4³ = 64
= (x − 4)(x² + 4x + 16)
same, opposite, always positive
27x³ + 8, so a = 3x and b = 2
27x³ = (3x)³, 8 = 2³
= (3x + 2)(9x² − 6x + 4)
a² = 9x², ab = 6x, b² = 4
Square the whole thing, coefficient included
With a = 3x, a² is 9x², not 3x², and ab is 3x · 2 = 6x, not 2x. Unlike a sum of
squares, a sum of cubes does factor — and the second bracket never factors
again.
7
The SAT's signature item: expand everything, then match a form.
Three pieces, three separate expansions, one collection at the end. Do not try to do it in your head.
(2x + 5)² − (x − 2) + 2(x + 3)
(2x + 5)² = 4x² + 20x + 25
square, double the cross, square
−(x − 2) = −x + 2
the minus flips both signs
2(x + 3) = 2x + 6
plain distribution
x terms: 20x − x + 2x = 21x
one family at a time
constants: 25 + 2 + 6 = 33
and only one x² term
= 4x² + 21x + 33
the equivalent expression
| Form offered | Where it comes from | Match? |
|---|
| 4x² + x + 33 | squared (2x + 5) term by term, losing the 20x | no |
|---|
| 4x² + 21x + 29 | read −(x − 2) as −x − 2 | no |
|---|
| 4x² + 21x + 33 | every sign handled | ✓ yes |
|---|
| 4x² + 23x + 33 | added the x instead of subtracting it | no |
|---|
One substitution settles it
At x = 1 the original is 49 + 1 + 8 = 58. Only 4 + 21 + 33 = 58 among the four
forms, so the other three are gone without any more algebra. Type both into Desmos
if the numbers are uglier — you have the calculator all section.
✓
Check: these are identities, so any number works.
Run x = 2 through the question and through the answer. Two different values means a slip.
the opener at x = 2
(12 − 8 + 1) − 2(4 − 6 + 5) = 5 − 6 = −1
the original
our answer
4 + 4 − 9 = −1
x² + 2x − 9 ✓
the six-product expansion
(1)(7) = 7 vs 16 + 20 − 44 + 15 = 7
2x³ + 5x² − 22x + 15 ✓
the ac factoring
32 − 4 − 15 = 13 vs (1)(13) = 13
(2x − 3)(4x + 5) ✓
the cube
8 − 64 = −56 vs (−2)(4 + 8 + 16) = −56
(x − 4)(x² + 4x + 16) ✓
Remember the flow
Distribute · Collect by family · Factor on sight · Test one number
Every minus sign in front of a bracket reaches every term inside it.
