Zero on one side, factor, then let each bracket take a turn.
Solve for x
x² + 7x + 12 = 0
Two numbers multiplying to 12 and adding to 7 are 3 and 4, so (x + 3)(x + 4) = 0 and x = −3 or −4. B forgets the sign flip — the factors carry +3 and +4, but the roots are their opposites.
If two things multiply to zero, one of them is zero.
so A = 0 or B = 0
tells you nothing
2 · 3, 1 · 6, 12 · ½ …
For (x + 1)(x + 2) = 6 you cannot set the brackets to 6. Expand, subtract 6, and factor again. The whole method depends on that zero.
Find two numbers that multiply to 12 and add to 7.
| Pair | Multiply | Add |
|---|---|---|
| 1 and 12 | 12 | 13 |
| 2 and 6 | 12 | 8 |
| 3 and 4 | 12 | 7 |
(x + 3)(x + 4) = 0
factored
x + 3 = 0 → x = −3
first bracket
x + 4 = 0 → x = −4
second bracket
The signs flip. Brackets with +3 and +4 give roots of −3 and −4.
Multiply a by c, then split the middle term.
2x² − 5x − 3
a = 2, c = −3
ac = −6, need sum −5
the pair is +1 and −6
2x² + x − 6x − 3
split the middle
x(2x + 1) − 3(2x + 1)
group in pairs
(2x + 1)(x − 3) = 0
factor out the bracket
2x + 1 = 0
x = −½
x − 3 = 0
x = 3
It looks harmless and it destroys a root every time.
x² = 5x
x = 5
one answer only
x² − 5x = 0
x(x − 5) = 0
x = 0 or x = 5
Dividing by x quietly assumes x ≠ 0 — and here x = 0 is a genuine solution. Move everything to one side instead; the common factor then hands you that root.
The same goes for a bracket. In x(x − 3) = 6(x − 3), dividing by (x − 3) loses x = 3. Instead: x(x − 3) − 6(x − 3) = 0, so (x − 3)(x − 6) = 0 and x = 3 or x = 6.
| Looks like | Factors as | Roots |
|---|---|---|
| x² − 16 = 0 | (x − 4)(x + 4) | ±4 |
| x² − 6x = 0 | x(x − 6) | 0 and 6 |
| x² + 6x + 9 = 0 | (x + 3)² | −3 twice |
| x² + 1 = 0 | will not factor | no real roots |
A perfect square gives one repeated root, not two. That is exactly when the discriminant is zero.
If no integer pair works after a minute, stop hunting and use the quadratic formula. It never fails and it is rarely slower.
Say what the letter stands for, and the story becomes a quadratic.
n(n + 2) = 143
n is the smaller odd integer
n² + 2n − 143 = 0
zero first
(n + 13)(n − 11) = 0
+13 and −11: multiply to −143, add to 2
n = 11 or n = −13
both are odd numbers
n = 11, so the integers are 11 and 13
the question said positive
Lengths, ages and counts cannot be negative either. A rectangle with w(w + 5) = 84 gives (w + 12)(w − 7) = 0, so the width is 7 m and −12 is thrown out.
A diagonal brings in Pythagoras. Sides w and w + 7 with diagonal 13: w² + (w + 7)² = 169, so 2w² + 14w − 120 = 0. Divide by 2: w² + 7w − 60 = 0, (w + 12)(w − 5) = 0, so w = 5 and the sides are 5 and 12.
If the question wants the larger integer, the answer is 13, not the 11 we solved for. Read the last line of the question again before choosing.
Three shapes that look harder than they are. Each one is tidied first, then factored as usual.
Negative leading term: −x² + x + 12 = 0
x² − x − 12 = 0
multiply every term by −1
(x − 4)(x + 3) = 0
−4 and +3: multiply to −12, add to −1
x = 4 or x = −3
each bracket = 0
Multiplying both sides of “= 0” by −1 leaves 0, so the roots do not change. Factoring with a = 1 is easier than with a = −1.
A number common to every term: 3x² − 12x − 36 = 0
x² − 4x − 12 = 0
divide every term by 3
(x − 6)(x + 2) = 0
−6 and +2: multiply to −12, add to −4
x = 6 or x = −2
each bracket = 0
Dividing by a number is safe; only dividing by x loses a root. Taking out the 3 turns an a ≠ 1 problem into an a = 1 problem.
A difference of squares with a ≠ 1: 9x² = 4
9x² − 4 = 0
zero first
(3x − 2)(3x + 2) = 0
9x² = (3x)², 4 = 2²
x = 23 or x = −23
3x = 2 and 3x = −2
The roots are fractions because the x-term carries a 3. No middle term means the two roots are opposites.
From (3x − 2)(3x + 2) = 0 it is tempting to read off x = ±2. Each bracket still has to be solved: 3x = 2 gives x = 23.
A root makes the equation true, so substitute it.
25 + 5k − 15 = 0
put x = 5 in
5k = −10, so k = −2
solve for k
x² − 2x − 15 = 0
the full equation
(x − 5)(x + 3) = 0
5 is a root, as promised
other root: x = −3
from the second bracket
Faster: with a = 1 the two roots multiply to c. So 5 × (other root) = −15 and the other root is −3 — no k needed.
The question asks for the other root. Stopping at k = −2 picks a choice that is there on purpose.
On UPCAT-type papers factoring rarely appears as a bare “solve”. It comes inside a story (consecutive integers, a rectangle’s area) or with a missing coefficient. The fastest check: put a choice back into the original equation — a true root gives exactly 0 — and make sure you are giving the quantity asked, not the letter you solved for.
Solve by factoring
3x² + 5x − 2 = 0
Split the middle with 6 and −1: 3x² + 6x − x − 2 = 3x(x + 2) − 1(x + 2) = (3x − 1)(x + 2). Setting each bracket to zero gives x = 13 or −2. C flips both signs.
