Get zero on one side, factor, then set each bracket to zero.
Solve for x
x² = 5x
Move everything to one side: x² − 5x = 0, then factor: x(x − 5) = 0, giving x = 0 or 5. A is what you get by dividing both sides by x — and dividing by a variable that might be zero destroys a solution. Both answers check: 0² = 5(0) ✓ and 25 = 25 ✓. Never divide by the variable; always factor.
If a product is zero, at least one factor must be zero.
(x − 3)(x + 7) = 0
a product equal to zero
x − 3 = 0 or x + 7 = 0
one of them must be
x = 3 or x = −7
two solutions
(x − 3)(x + 7) = 12 tells you nothing about the individual brackets — there are endless ways to multiply to 12. Expand and rearrange to get zero on one side first.
Isolate the square, then take ± the root.
3x² − 48 = 0
no middle term
x² = 16
isolate
x = ±4
both roots
(x − 2)² = 25
a squared bracket
x − 2 = ±5
root both sides
x = 7 or x = −3
two branches
The ± is not decoration. Both 4 and −4 square to 16, so dropping the negative loses half the answer.
x² + 2x = 15
not zero yet
x² + 2x − 15 = 0
standard form first
(x + 5)(x − 3) = 0
product −15, sum 2
x = −5 or x = 3
check: 9 + 6 = 15 ✓
Standard form — ax² + bx + c = 0 — comes before everything. Trying to factor while a number sits on the right is where these go wrong.
2x² − 8x = 0
GCF is 2x
2x(x − 4) = 0
factor
x = 0 or x = 4
the 2 gives no solution
A constant factor like 2 can never be zero, so it contributes nothing. Only factors containing x produce solutions.
A rectangle is 3 m longer than wide
area 40 m²
w(w + 3) = 40
set it up
w² + 3w − 40 = 0
standard form
(w + 8)(w − 5) = 0
w = −8 or 5
Rejecting a root is not the same as making a mistake. The algebra gives both; the situation rules one out, and saying so explicitly is part of the answer.
Two consecutive positive integers multiply to 72. Find them.
n(n + 1) = 72
call them n and n + 1
n² + n − 72 = 0
standard form
(n + 9)(n − 8) = 0
n = −9 or 8
n = 8, so 8 and 9
reject −9: not positive; check 8 × 9 = 72 ✓
The same two methods, where the numbers stop being friendly.
A coefficient inside the square: (3x + 1)² = 16
3x + 1 = ±4
root both sides
3x = 3 or 3x = −5
subtract 1 on each branch
x = 1 or x = −53
check: 3(−53) + 1 = −4, and (−4)² = 16 ✓
Finish both branches; the negative one often ends in a fraction.
A square equal to a negative: 2x² + 18 = 0
2x² = −18
isolate the square
x² = −9
divide by 2
no real solution
no real number squares to a negative
Writing x = ±3 here is wrong: 3² = 9, not −9. "No real solution" is the answer.
Leading coefficient 2, no common factor: 2x² + x − 6 = 0
2x² + 4x − 3x − 6 = 0
ac = −12: split x into 4x and −3x
2x(x + 2) − 3(x + 2) = 0
group in pairs
(2x − 3)(x + 2) = 0
the common bracket comes out
x = 32 or x = −2
check: 2(4) − 2 − 6 = 0 ✓
The AC method finds two numbers that multiply to a × c and add to b, then factors in pairs.
Expect a quadratic to solve that is not yet in standard form — x on both sides, or a bracket equal to a number other than zero — and a word problem on area or consecutive integers where one root must be rejected. The fastest route is zero on one side, factor, then each bracket to zero; with no x term, isolate the square and take ±. The 10-second check: substitute each root into the original equation, not the factored one.
Solve for x
(x + 1)(x − 4) = 6
The right side is not zero, so expand first: x² − 3x − 4 = 6, giving x² − 3x − 10 = 0, which factors as (x − 5)(x + 2) = 0. So x = 5 or −2. B sets each original bracket to zero, which is only valid when the product equals zero — here it equals 6, and neither −1 nor 4 works: (−1 + 1)(−1 − 4) = 0, not 6.
Hint: the zero product property needs a zero. Expand, move the 6 across, then factor the result.
