Factoring
Factoring is multiplication read backwards, so every identity below can be checked by expanding it again — and you should, every time. The skill is not memorising nine patterns; it is looking at an expression and knowing which one it is before you start writing. That recognition step is what the worked solutions lead with, and each example here names the tell before it names the move.
Factoring exercises with solutions
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1. Greatest common factor, always first. Take out everything every term shares: the numeric GCF and the lowest power of each common letter. A GCF pulled out first often turns what is left into a pattern you can already see. Skipping it is the single most common reason an answer is marked incomplete.
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2. Count the terms. Two terms: look for a difference of squares, or a sum or difference of cubes. Three terms: a perfect square trinomial or an ordinary trinomial. Four terms: grouping. The count tells you which shelf to reach for.
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3. Look for a substitution. If the exponents come in a 2n-and-n pattern (z4 and z2, m4 and m2) or the same binomial appears twice, set u equal to the repeating part. The problem collapses to one you have already solved.
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4. Factor completely. Look at every factor you produced and ask whether it factors again. x4−y4 is not finished at (x2+y2)(x2−y2), and (4z2−1)(z2−2) is not finished either.
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5. Check by expanding. Multiply your factors back out. This is not optional politeness — it catches every sign error, and it is exactly how the worked solutions verify themselves.
- Common factorab + ac + ad = a(b + c + d)
The distributive law, used backwards. Works for any number of terms.
- Difference of squaresa2 − b2 = (a + b)(a − b)
Two terms, both perfect squares, separated by a minus. The middle terms cancel when you expand, which is why nothing is left over.
- Sum of squaresa2 + b2 does not factor over the reals
There is no identity for it. Recognising it as unfactorable is itself the answer — unless a trick term can be added and subtracted (see below).
- Perfect square trinomiala2 + 2ab + b2 = (a + b)2
Three terms, first and last are perfect squares, middle is exactly twice the product of their roots.
- Perfect square trinomial, negativea2 − 2ab + b2 = (a − b)2
Same shape with a minus in the middle. The sign of the middle term is the sign inside the bracket.
- Sum of cubesa3 + b3 = (a + b)(a2 − ab + b2)
Signs read Same, Opposite, Always Positive. The trinomial never factors further.
- Difference of cubesa3 − b3 = (a − b)(a2 + ab + b2)
Same pattern, same sign rule. Only the binomial's sign changes.
- Groupingax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)
Four terms: split into two pairs, take a GCF from each, and the leftover brackets must match exactly. If they do not, try re-pairing.
- Simple trinomialx2 + (p + q)x + pq = (x + p)(x + q)
Leading coefficient 1: find two numbers that multiply to the constant and add to the middle coefficient.
- General trinomialax2 + bx + c → split b into two numbers with product ac
The ac method. Splitting the middle term turns any trinomial into a grouping problem.
- a2 + b2 = (a + b)2(a + b)2 = a2 + 2ab + b2
A sum of squares does not factor over the reals at all. Expanding (a+b)2 shows the missing 2ab: test at a = b = 1, where 2 ≠ 4.
- x2 − 6xy + 9y2 = (x − 3)2x2 − 6xy + 9y2 = (x − 3y)2
The y has to come along. (x−3)2 expands to x2−6x+9 — no y anywhere. Always expand the square you wrote and compare it term by term.
- a3 + b3 = (a + b)(a2 + ab + b2)a3 + b3 = (a + b)(a2 − ab + b2)
The middle sign of the trinomial is always opposite the binomial's. Same, Opposite, Always Positive — and the reason is the cancellation: only with −ab do the a2b and ab2 terms disappear on expanding.
- x4 − y4 = (x2 + y2)(x2 − y2)x4 − y4 = (x2 + y2)(x + y)(x − y)
Correct as far as it goes, but not factored completely. Re-inspect every factor you produce before calling it done.
- 3ax − ay − 3bx + by = a(3x − y) + b(−3x + y)3ax − ay − 3bx + by = a(3x − y) − b(3x − y)
Both are true, but only the second has matching brackets, which is the point of grouping. When the second bracket comes out backwards, factor out the negative instead.
- 6x2 + 3x = 3x(2x)6x2 + 3x = 3x(2x + 1)
Dividing a term by the GCF leaves 1, not nothing. Expanding 3x(2x) gives 6x2 and loses the 3x entirely.
Type an answer and it gets marked. What you get right is kept on this device.
- 1.2x3 − 18x
- 2.27a3 − 8b3
- 3.x2 + 10x + 25 − y2
- 4.10x2 + 11x − 6
- 5.x6 − 64
- 6.3(2a − b)2 − 11(2a − b) − 4