Algebra · Paper 1
Factors
Higher Level · Tap NEXT to begin
Factoring isn't really doing anything — it's recognising a pattern. There are five families to know: grouping, difference of two squares (DOTS), sum and difference of cubes, trinomials, and combinations — problems that stack two of the above. Spot the pattern, apply the formula, you're done.
Section 1 of 5
Grouping
The simplest factoring move: take out what's common. With 2 terms, look for a shared factor across both. With 4 terms, group them in pairs — each pair gives up its own common factor, and then a shared bracket pops out as the final common factor.
Question 1 · Two terms — a single common factor. Factorise $ax + ay$.
What's in both terms? Both $ax$ and $ay$ contain $a$.
Pull $a$ out the front: $ax + ay = a(x + y)$
$a(x + y)$
Always check by expanding back: $a(x + y) = ax + ay$ ✓
Question 2 · Four terms — group in pairs, then factor twice. Factorise $ax + ay + bx + by$.
Group in pairs: $ax + ay + bx + by = (ax + ay) + (bx + by)$
Factor each pair: $= a(x + y) + b(x + y)$
The bracket on each side matches. $(x + y)$ is now a common factor of the whole expression — that's the test that the grouping was right. If the brackets disagree, try grouping differently.
Take out the matching bracket: $= (x + y)(a + b)$
$(x + y)(a + b)$
Question 3 · Four terms, order scrambled — rearrange first. Factorise $2px + 9 - 9x - 2p$.
Don't group as written — $2px$ and $9$ share nothing. Move terms around so each pair has a common factor: put the $p$-terms together and the constant-times-$x$ terms together.
Rearrange: $2px + 9 - 9x - 2p = 2px - 2p - 9x + 9$
Factor each pair: $= 2p(x - 1) - 9(x - 1)$
Watch the sign. Pulling $-9$ (not $+9$) out of $-9x + 9$ gives $-9(x - 1)$ — because $-9 \times (-1) = +9$. Pulling out the negative is what makes the second bracket match the first.
Take out the common bracket: $= (x - 1)(2p - 9)$
$(x - 1)(2p - 9)$
Section 2 of 5
Difference of Two Squares (DOTS)
Whenever you see something squared minus something else squared, you can factor it instantly. The two factors are (first $-$ second) and (first $+$ second).
The pattern — learn it once
$x^{2} - y^{2} = (x - y)(x + y)$
The $x$ and $y$ can be anything — a letter, a coefficient times a letter, a whole bracketed expression, even a number. As long as you've got square minus square, the formula applies. Only works for minus: $x^{2} + y^{2}$ does not factor over the reals.
Question 4 · Basic DOTS. Factorise $x^{2} - 25$.
Rewrite the constant as a square: $x^{2} - 25 = x^{2} - 5^{2}$
Apply $x^{2} - y^{2} = (x - y)(x + y)$: $= (x - 5)(x + 5)$
$(x - 5)(x + 5)$
Question 5 · The "$x$" can be a whole bracket. Factorise $(p + 2q)^{2} - 100$.
Rewrite as square $-$ square: $(p + 2q)^{2} - 100 = (p + 2q)^{2} - 10^{2}$
Treat $(p + 2q)$ as a single unit. DOTS doesn't care what's inside the bracket — it just needs a square minus a square.
Apply DOTS: $= (p + 2q - 10)(p + 2q + 10)$
$(p + 2q - 10)(p + 2q + 10)$
Question 6 · The bracket is being subtracted — mind the signs. Factorise $36 - (x - 2y)^{2}$.
Rewrite as square $-$ square: $36 - (x - 2y)^{2} = 6^{2} - (x - 2y)^{2}$
Apply DOTS: $= \big[\,6 - (x - 2y)\,\big]\big[\,6 + (x - 2y)\,\big]$
Drop the inner brackets carefully: $= (6 - x + 2y)(6 + x - 2y)$
The minus distributes. Removing $[\,6 - (x - 2y)\,]$ flips both signs inside: $-x$ stays $-x$, but $-(-2y) = +2y$. The other bracket is fine as written.
$(6 - x + 2y)(6 + x - 2y)$
Section 3 of 5
Sum & Difference of Two Cubes
Same idea as DOTS, but for cubes — and there are two formulas. Unlike $x^{2} + y^{2}$, the sum of cubes does factor. The key is recognising perfect cubes when you see them.
The two formulas — learn both
$x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})$
$x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2})$
"Same, Opposite, Plus." The first bracket's sign matches the original; the middle sign of the second bracket is opposite; the last sign is always plus.
Perfect cubes — memorise these
1³$1$
2³$8$
3³$27$
4³$64$
5³$125$
6³$216$
7³$343$
8³$512$
9³$729$
10³$1000$
Question 7 · Sum of cubes. Factorise $x^{3} + 8$.
Rewrite $8$ as a cube: $x^{3} + 8 = x^{3} + 2^{3}$
Apply $x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2})$ with $y = 2$: $= (x + 2)(x^{2} - 2x + 4)$
$(x + 2)(x^{2} - 2x + 4)$
The second bracket $(x^{2} - 2x + 4)$ doesn't factor further over the reals.
Question 8 · Difference of cubes with a coefficient. Factorise $27p^{3} - 125$.
Rewrite both terms as cubes: $27p^{3} - 125 = (3p)^{3} - 5^{3}$
Here $x = 3p$, $y = 5$: $x^{2} = 9p^{2}$, $xy = 15p$, $y^{2} = 25$
Apply $x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})$: $= (3p - 5)(9p^{2} + 15p + 25)$
$(3p - 5)(9p^{2} + 15p + 25)$
The careful bit is squaring the $3p$ — square both the $3$ and the $p$.
Section 4 of 5
Trinomials (Quadratics)
Three-term expressions of the form $ax^{2} + bx + c$. The reliable method is to split the middle term — then group as before.
Carl's method — the Guide Number
Guide number $\text{GN} = a \times c$
Find two numbers: product $= \text{GN}$, sum $= b$
"Add" or "Sub"? If the two numbers have the same sign, you add their magnitudes (Carl writes "Add"). Opposite signs, the magnitudes subtract to give $b$ (Carl writes "Sub"). A quick mental check on the pair you're hunting.
Question 9 · Leading coefficient other than 1 — the textbook example. Factorise $3x^{2} - 7x - 6$.
$\text{GN} = a \times c = (3)(-6) = -18$. Two numbers: product $-18$, sum $-7$.
Try $-9$ and $+2$: product $-18$ ✓, sum $-7$ ✓.
"Sub" pair. Opposite signs, so magnitudes subtract: $9 - 2 = 7$. The bigger magnitude ($9$, negative) wins, giving the net $-7$ we want.
Split the middle: $3x^{2} - 7x - 6 = 3x^{2} - 9x + 2x - 6$
Factor each pair: $= 3x(x - 3) + 2(x - 3)$
Common bracket: $= (x - 3)(3x + 2)$
$(x - 3)(3x + 2)$
Question 10 · Two-variable trinomial — same recipe, keep the $y$. Factorise $x^{2} - 7xy + 12y^{2}$.
The $y^{2}$ plays the role of the constant. $\text{GN} = (1)(12) = +12$, sum $= -7$.
Try $-3$ and $-4$: product $+12$ ✓, sum $-7$ ✓.
"Add" pair. Same sign, so magnitudes add: $3 + 4 = 7$. Both negative, so the sum is $-7$.
Split & group, keeping the $y$ on the middle terms: $x^{2} - 3xy - 4xy + 12y^{2}$
$= x(x - 3y) - 4y(x - 3y)$
$= (x - 3y)(x - 4y)$
$(x - 3y)(x - 4y)$
Section 5 of 5
Combinations
Expressions don't always come in one neat pattern. Often you do one technique, then another. The order matters: always pull out a common factor first. Then look at what's left and ask "is this DOTS? a perfect square? a cubic?" — apply whichever fits.
Question 11 · Common factor $\rightarrow$ DOTS. Factorise $2x^{2} - 50$.
Don't try DOTS straight away — $2x^{2}$ isn't a perfect square. Take out the common factor of $2$ first.
Take out the common factor: $2x^{2} - 50 = 2(x^{2} - 25)$
DOTS on the bracket: $= 2(x - 5)(x + 5)$
$2(x - 5)(x + 5)$
The $2$ stays out the front — it's part of the answer.
Question 12 · DOTS on a fourth power — apply DOTS twice. Factorise $x^{4} - y^{4}$.
Worth memorising: $x^{4} = (x^{2})^{2}$. Same for any even power: $x^{6} = (x^{3})^{2}$, $x^{8} = (x^{4})^{2}$. Even powers are always perfect squares.
Rewrite as square $-$ square: $x^{4} - y^{4} = (x^{2})^{2} - (y^{2})^{2}$
DOTS — first pass: $= (x^{2} - y^{2})(x^{2} + y^{2})$
DOTS again on the first bracket: $= (x - y)(x + y)(x^{2} + y^{2})$
$(x - y)(x + y)(x^{2} + y^{2})$
Don't try to factor $(x^{2} + y^{2})$ — sum of two squares does not factor over the reals. Three brackets is as far as it goes.
Question 13 · Hidden perfect square + DOTS. Factorise $a^{2} + 2ab + b^{2} - 121$.
⚠ Common mistake. Four terms tempts you to group: $a(a + b) + (b - 11)(b + 11)$. But $a(a + b) = a^{2} + ab$, not $a^{2} + 2ab$ — and the groups share no bracket. Grouping is the wrong tool here.
Spot the perfect square in the first three terms: $a^{2} + 2ab + b^{2} = (a + b)^{2}$
So the expression is: $(a + b)^{2} - 121 = (a + b)^{2} - 11^{2}$
Now it's DOTS: $= (a + b - 11)(a + b + 11)$
$(a + b - 11)(a + b + 11)$
Lesson: with four-term expressions, always check for a hidden perfect square in three of the terms before reaching for grouping.
Question 14 · The hard one — rearrange, then stack two patterns. Factorise $x^{3} - x^{2} - y^{3} + y^{2}$.
⚠ Common mistake. The natural pairing gives $x^{2}(x - 1) - y^{2}(y - 1)$. The brackets $(x - 1)$ and $(y - 1)$ are different — you can't pull a common factor from two different brackets. That "answer" is wrong.
Rearrange — $x$'s together, $y$'s together: $x^{3} - x^{2} - y^{3} + y^{2} = (x^{3} - y^{3}) - (x^{2} - y^{2})$
The insight. Group the cubes together and the squares together. Each is now a familiar pattern — difference of cubes, difference of squares.
Apply both formulas: $= (x - y)(x^{2} + xy + y^{2}) - (x - y)(x + y)$
Pull out the common $(x - y)$: $= (x - y)\big[(x^{2} + xy + y^{2}) - (x + y)\big]$
$(x - y)(x^{2} + xy + y^{2} - x - y)$
The second bracket doesn't simplify further.
Five patterns. Stack them when needed.
Grouping for 2 or 4 terms. DOTS when you see square $-$ square. Sum and difference of cubes — two formulas. Trinomials — split the middle term using $\text{GN} = a \times c$. Combinations — common factor first, then ask which of the four fits. The five tools never change — only the order you use them.
Stuck on a question, or want one looked at before the next test? Send it over (photo, screenshot, anything) and I'll come back with worked notes inside 24 hours. That's what these notes — and the grinds — are for. — Carl