Two terms only. Check whether both are perfect squares, or both perfect cubes.
Factor completely
50x² − 8
Take the GCF of 2 out first: 2(25x² − 4). Now 25x² and 4 are both perfect squares, so the bracket factors as (5x − 2)(5x + 2), giving 2(5x − 2)(5x + 2). B stops one step early — it is correct as far as it goes, but "completely" means going on. Neither 50 nor 8 is a perfect square, which is exactly why the GCF has to come out first.
x² + 49 has no factors over the real numbers. There is no "sum of squares"
pattern — only the difference works. This is the one exception worth memorising.
The third example is not finished: x² − 9 factors again into (x + 3)(x − 3), so
the complete answer is (x² + 9)(x + 3)(x − 3). Always re-check each bracket.
2
Recognising a perfect square.
Term
Square of
Term
Square of
x²
x
25x⁴
5x²
49
7
100m⁶
10m³
9y²
3y
x²y⁴
xy²
A term is a perfect square when its coefficient is a perfect square and
every exponent is even. Halve the exponents to get the root.
3
Sum and difference of cubes.
a³ + b³ = (a + b)(a² − ab + b²) · a³ − b³ = (a − b)(a² + ab + b²)
x³ − 27
a = x, b = 3
= (x − 3)(x² + 3x + 9)
same, opposite, always positive
The trinomial that comes out almost never factors again. If you find yourself trying
to factor x² + 3x + 9, stop — no pair of integers multiplies to 9 and adds to 3.
Note the middle term is ab, not 2ab. That is the difference between this and a squared binomial, and it is the detail most often slipped.
5
Which pattern, and in what order.
Step 1take out the GCFalways first
Two terms, both squares?a² − b² patterndifference only
Two terms, both cubes?SOAPsum or difference
Step 4re-check every bracketx⁴ − 81 needs two rounds
37² − 33²
mental arithmetic
= (37 + 33)(37 − 33)
the same pattern
= 70 × 4 = 280
no squaring needed
Remember the flow
GCF first · Both perfect squares? difference only · Both perfect cubes? SOAP · Re-check every bracket
A sum of squares never factors. A sum of cubes always does.
6
Two letters, and a GCF hiding the pattern.
If neither number is a perfect square or cube, look for a common factor first.
Factor 16a² − 81b².
16a² = (4a)², 81b² = (9b)²
both terms are squares
= (4a + 9b)(4a − 9b)
a² − b² with a = 4a, b = 9b
Each letter rides along with its number. Check one product: 4a · 4a = 16a² ✓.
Factor 12x² − 27 completely.
= 3(4x² − 9)
GCF 3 first
= 3(2x + 3)(2x − 3)
now both terms are squares
"Neither is a square, so it does not factor"
12 and 27 are not perfect squares, so it is tempting to call 12x² − 27 prime. But the common factor 3 is hiding a difference of squares. Always take the GCF out before deciding.
Factor 2x³ + 16 completely.
= 2(x³ + 8)
GCF 2 first; 8 = 2³
= 2(x + 2)(x² − 2x + 4)
sum of cubes: same, opposite, always +
7
Variations: a fraction, a bracket as a term, the number first.
The patterns accept any a and b — even a bracket.
x² − 19
19 = (13)²
a fraction can be a perfect square
= (x + 13)(x − 13)
difference of squares
Square-root the top and the bottom separately: √1 = 1 and √9 = 3.
(x + 1)² − 25
= [(x + 1) + 5][(x + 1) − 5]
a = x + 1, b = 5
= (x + 6)(x − 4)
tidy each bracket
Check by expanding both: x² + 2x − 24 either way ✓. Treating the bracket as one block saves the expansion.
1 − 125y³
1 = 1³, 125y³ = (5y)³
a = 1, b = 5y
= (1 − 5y)(1 + 5y + 25y²)
same, opposite, always +
The number can come first. Then a = 1, so a² = 1 and ab = 5y.
On the exam
Two-term factoring items test recognition: both perfect squares with a minus between them, or both perfect cubes. The trap choices are a "factored" sum of squares, an answer that stops after the GCF, and 2ab instead of ab in the cube trinomial. For a 10-second check, put x = 2 into the question and your answer.
Now you try
Practice problem
Factor completely
x⁴ − 16
First round: (x² + 4)(x² − 4). But x² − 4 is itself a difference of squares, so it factors again into (x + 2)(x − 2), giving (x² + 4)(x + 2)(x − 2). C stops after one round.D factors x² + 4 as well, which is impossible — a sum of squares does not factor.
Hint: after the first factoring, look at each bracket again. One of
them can go further; the other genuinely cannot.
