Solving Quadratic Equations by Factoring
Solving quadratic equations by factoring is often the simplest and quickest method, as long as you know how to factor. If factoring is new to you or a little rusty, do not worry. This interactive lesson shows you how to factor each equation one step at a time, starting with simple examples and building up to harder ones.
To solve a quadratic equation by factoring, write it as a product of two factors equal to 0, like (x + )(x + ) = 0. Find the numbers that go in the boxes, then set each factor equal to 0. This works because of the zero product property: if a product equals zero, at least one of its factors must be zero.
Examples of solving quadratic equations by factoring
Reveal each solution one step at a time, or show all the steps at once.
Example 1
x² + 3x + 2 = 0
Set up the boxes.x² + 3x + 2 = (x + )(x + )The first terms must multiply to x², so both are x. The two numbers in the boxes must multiply to 2.
Which numbers can go in the boxes?1 and 2, −1 and −2These are the pairs that multiply to 2. Only one of them gives the middle term, 3x. To find it, put each pair into a small grid: the first terms, x and x, go on top, and the pair goes underneath. Then cross multiply, as shown next.
Cross multiply and add.
Multiply along the two diagonals and add. You want 3x, the middle term.
| Try | Cross multiply and add |
|---|---|
| xx12 | 2x + x = 3x ✓ |
| xx−1−2 | −2x − x = −3x |
Why is the grid arranged like that?
(x + 1)(x + 2) xx 12 This is the grid that gave 3x, shown with its two factors, (x + 1) and (x + 2). Press Next to see how they match.Fill in the boxes.(x + 1)(x + 2) = 0
See it in the multiplication.(x + 1)(x + 2) = x · x + x · 2 + 1 · x + 1 · 2
= x² + 2x + x + 2 = x² + 3x + 2 ✓The two middle products, x · 2 and 1 · x, are the diagonals of the grid.
= x² + 2x + x + 2 = x² + 3x + 2 ✓The two middle products, x · 2 and 1 · x, are the diagonals of the grid.
You don't really need the table here.The table shows why the method works, but when both first terms are x, cross multiplying just adds the two numbers: x · 2 + 1 · x = (1 + 2)x. So a quick look at the pairs is enough: which pair adds up to 3, the middle number? 1 + 2 = 3.
Set each factor equal to 0 and solve.x + 1 = 0 → x = −1
x + 2 = 0 → x = −2
x + 2 = 0 → x = −2
Check.(−1)² + 3(−1) + 2 = 1 − 3 + 2 = 0 ✓
(−2)² + 3(−2) + 2 = 4 − 6 + 2 = 0 ✓
(−2)² + 3(−2) + 2 = 4 − 6 + 2 = 0 ✓
Example 1, with a negative middle term
x² − 3x + 2 = 0
Set up the boxes.x² − 3x + 2 = (x + )(x + )Same boxes as Example 1. The numbers must still multiply to 2.
Use the same table.
This time you want −3x, so the other row works. Again, a quick look at the pairs is enough: −1 + (−2) = −3.
| Try | Cross multiply and add |
|---|---|
| xx12 | 2x + x = 3x |
| xx−1−2 | −2x − x = −3x ✓ |
Fill in the boxes.(x − 1)(x − 2) = 0
Set each factor equal to 0 and solve.x − 1 = 0 → x = 1
x − 2 = 0 → x = 2Compare with Example 1: flipping the sign of the middle term flips the signs of the solutions.
x − 2 = 0 → x = 2Compare with Example 1: flipping the sign of the middle term flips the signs of the solutions.
Check.1² − 3(1) + 2 = 1 − 3 + 2 = 0 ✓
2² − 3(2) + 2 = 4 − 6 + 2 = 0 ✓
2² − 3(2) + 2 = 4 − 6 + 2 = 0 ✓
Example 2
x² + x − 30 = 0
Set up the boxes.x² + x − 30 = (x + )(x + )Both first terms are x again. The numbers in the boxes must multiply to −30, so one is positive and one is negative.
Which numbers can go in the boxes?1 and −30, −1 and 30, 2 and −15, −2 and 15, 3 and −10, −3 and 10, 5 and −6, −5 and 6These are the pairs that multiply to −30. Test each pair in a grid, just as in Example 1, to find the one that gives the middle term, x.
Cross multiply and add.
You want 1x, which is just x. As in Example 1, each result is simply the two numbers added.
| Try | Cross multiply and add |
|---|---|
| xx1−30 | −30x + x = −29x |
| xx−130 | 30x − x = 29x |
| xx2−15 | −15x + 2x = −13x |
| xx−215 | 15x − 2x = 13x |
| xx3−10 | −10x + 3x = −7x |
| xx−310 | 10x − 3x = 7x |
| xx5−6 | −6x + 5x = −x |
| xx−56 | 6x − 5x = x ✓ |
A quick look is enough.The table is useful for seeing every pair, but you don't have to write it out. Look over the pairs and ask which one adds up to 1, the middle number: −5 + 6 = 1.
Fill in the boxes.(x − 5)(x + 6) = 0
Set each factor equal to 0 and solve.x − 5 = 0 → x = 5
x + 6 = 0 → x = −6
x + 6 = 0 → x = −6
Check.5² + 5 − 30 = 25 + 5 − 30 = 0 ✓
(−6)² + (−6) − 30 = 36 − 6 − 30 = 0 ✓
(−6)² + (−6) − 30 = 36 − 6 − 30 = 0 ✓
Example 2, with a negative middle term
x² − x − 30 = 0
Set up the boxes.x² − x − 30 = (x + )(x + )
Use the same table.
This time you want −1x, or −x. A quick look at the pairs gives it right away: 5 + (−6) = −1.
| Try | Cross multiply and add |
|---|---|
| xx1−30 | −30x + x = −29x |
| xx−130 | 30x − x = 29x |
| xx2−15 | −15x + 2x = −13x |
| xx−215 | 15x − 2x = 13x |
| xx3−10 | −10x + 3x = −7x |
| xx−310 | 10x − 3x = 7x |
| xx5−6 | −6x + 5x = −x ✓ |
| xx−56 | 6x − 5x = x |
Fill in the boxes.(x + 5)(x − 6) = 0
Set each factor equal to 0 and solve.x + 5 = 0 → x = −5
x − 6 = 0 → x = 6
x − 6 = 0 → x = 6
Check.(−5)² − (−5) − 30 = 25 + 5 − 30 = 0 ✓
6² − 6 − 30 = 36 − 6 − 30 = 0 ✓
6² − 6 − 30 = 36 − 6 − 30 = 0 ✓
Example 3, when the first term is not x²
6x² + 27x + 30 = 0
Set up the boxes.6x² + 27x + 30 =
(3x + )(2x + )
or (6x + )(x + )The first terms must multiply to 6x². With whole numbers, that means 3x and 2x, or 6x and x. The two numbers in the boxes must multiply to 30.
(3x + )(2x + )
or (6x + )(x + )The first terms must multiply to 6x². With whole numbers, that means 3x and 2x, or 6x and x. The two numbers in the boxes must multiply to 30.
What's different from Examples 1 and 2?The first terms are no longer both x. In (6x + ☐)(x + ☐), the number in the second box gets multiplied by 6x, not by x. So the middle term is no longer just the two numbers added, and the order of the numbers matters. A quick look at the pairs no longer works, so this is where the table really pays off. You cross multiply every combination, exactly as in Examples 1 and 2.
Which numbers can go in the boxes?1 and 30, 2 and 15, 3 and 10, 5 and 6These are the pairs that multiply to 30. Each pair can go in either order. Test the combinations in a grid, just as in Examples 1 and 2, with the first terms on top.
Cross multiply and add.
You want 27x. The other combinations, such as using 3 and 10 or swapping the order of a pair, give different results, none of which is 27x.
| Try | Cross multiply and add |
|---|---|
| 6xx301 | 6x + 30x = 36x |
| 6xx152 | 12x + 15x = 27x ✓ |
| 6xx65 | 30x + 6x = 36x |
| 3x2x301 | 3x + 60x = 63x |
| 3x2x152 | 6x + 30x = 36x |
| 3x2x65 | 15x + 12x = 27x ✓ |
Same grid, same multiplication.(6x + 15)(x + 2) = 6x · x + 6x · 2 + 15 · x + 15 · 2
= 6x² + 12x + 15x + 30 = 6x² + 27x + 30 ✓The columns are the factors (6x + 15) and (x + 2). The top row gives 6x², the bottom row gives 30, and the diagonals, 6x · 2 and 15 · x, give 12x + 15x = 27x.
= 6x² + 12x + 15x + 30 = 6x² + 27x + 30 ✓The columns are the factors (6x + 15) and (x + 2). The top row gives 6x², the bottom row gives 30, and the diagonals, 6x · 2 and 15 · x, give 12x + 15x = 27x.
You get two choices that work.(6x + 15)(x + 2) = 0
(3x + 6)(2x + 5) = 0Both multiply back out to 6x² + 27x + 30. Either one gives the same solutions.
(3x + 6)(2x + 5) = 0Both multiply back out to 6x² + 27x + 30. Either one gives the same solutions.
Using the first choice.6x + 15 = 0 → x = −156 = −52
x + 2 = 0 → x = −2
x + 2 = 0 → x = −2
Using the second choice.3x + 6 = 0 → x = −2
2x + 5 = 0 → x = −52
2x + 5 = 0 → x = −52
Solutions.x = −52 or x = −2
A shortcut: factor out the GCF first.3(2x² + 9x + 10) = 0 → 3(2x + 5)(x + 2) = 0All three coefficients are divisible by 3. Taking out 3 first leaves smaller numbers to try. The factor 3 can never be 0, so the solutions are the same.
Check.6(−2)² + 27(−2) + 30 = 24 − 54 + 30 = 0 ✓
6(−52)² + 27(−52) + 30 = 37.5 − 67.5 + 30 = 0 ✓
6(−52)² + 27(−52) + 30 = 37.5 − 67.5 + 30 = 0 ✓
Steps for solving a quadratic equation by factoring
Every example above follows the same steps.
Five steps
- Write the equation in standard form, ax² + bx + c = 0. Move every term to one side so the other side is 0.
- Factor out the greatest common factor (GCF), if there is one.
- Factor into two sets of boxes, like (x + )(x + ). Cross multiply the possible numbers until the middle term matches.
- Set each factor equal to 0 and solve.
- Check each solution in the original equation.
For x² + bx + c, cross multiplying just adds the two numbers, so you don't need the table. A quick look is enough: find two numbers that multiply to c and add to b. The signs follow this pattern:
| If c is… | and b is… | then the two numbers are… | Example |
|---|---|---|---|
| positive | positive | both positive | x² + 7x + 12: 3 and 4 |
| positive | negative | both negative | x² − 7x + 12: −3 and −4 |
| negative | positive or negative | one positive, one negative; the larger one (ignoring signs) has the same sign as b | x² + x − 12: 4 and −3 x² − x − 12: −4 and 3 |
The solutions have the opposite signs of the two numbers. If the numbers are 4 and −3, the factors are (x + 4)(x − 3), and the solutions are x = −4 and x = 3.
Special cases, step by step
Each tab shows a situation that often trips students up. Pick one to see it solved step by step.
Not in standard form yet
x² = 5x + 14
Is it in standard form?No. Neither side is 0. Subtract 5x and 14 from both sides.
Standard form.x² − 5x − 14 = 0
Set up the boxes.x² − 5x − 14 = (x + )(x + )The numbers in the boxes must multiply to −14.
Cross multiply and add.
You want −5x.
| Try | Cross multiply and add |
|---|---|
| xx1−14 | −14x + x = −13x |
| xx−114 | 14x − x = 13x |
| xx2−7 | −7x + 2x = −5x ✓ |
| xx−27 | 7x − 2x = 5x |
Fill in the boxes and solve.(x + 2)(x − 7) = 0 → x = −2 or x = 7
Check in the original equation.(−2)² = 4 and 5(−2) + 14 = 4 ✓
7² = 49 and 5(7) + 14 = 49 ✓
7² = 49 and 5(7) + 14 = 49 ✓
No constant term
3x² − 12x = 0
Don't divide by x.3x = 12 → x = 4Dividing both sides by x loses the solution x = 0.
No boxes needed here.With no constant term, every term contains x, so you can factor out the greatest common factor directly. Both terms are divisible by 3x.
Factor out 3x.3x(x − 4) = 0
Set each factor equal to 0 and solve.3x = 0 → x = 0
x − 4 = 0 → x = 4
x − 4 = 0 → x = 4
Check.3(0)² − 12(0) = 0 ✓
3(4)² − 12(4) = 48 − 48 = 0 ✓
3(4)² − 12(4) = 48 − 48 = 0 ✓
No middle term (difference of squares)
x² − 49 = 0
Set up the boxes.x² − 49 = (x + )(x + )There is no x term, so the middle term you want is 0x. The numbers must multiply to −49.
Cross multiply and add.
You want 0x. The two numbers are opposites.
| Try | Cross multiply and add |
|---|---|
| xx1−49 | −49x + x = −48x |
| xx−149 | 49x − x = 48x |
| xx7−7 | −7x + 7x = 0 ✓ |
| xx−77 | 7x − 7x = 0 ✓ |
Fill in the boxes and solve.(x + 7)(x − 7) = 0 → x = −7 or x = 7This pattern is called a difference of squares. It always gives two solutions, not just x = 7.
Check.(−7)² − 49 = 0 ✓ 7² − 49 = 0 ✓
Only one solution (perfect square)
x² − 10x + 25 = 0
Set up the boxes.x² − 10x + 25 = (x + )(x + )The numbers must multiply to 25.
Cross multiply and add.
You want −10x. The two numbers in the boxes are the same.
| Try | Cross multiply and add |
|---|---|
| xx125 | 25x + x = 26x |
| xx−1−25 | −25x − x = −26x |
| xx55 | 5x + 5x = 10x |
| xx−5−5 | −5x − 5x = −10x ✓ |
Fill in the boxes.(x − 5)(x − 5) = 0, or (x − 5)² = 0
Solve.x = 5Both factors give the same value, so there is only one solution. It is sometimes called a double root.
Check.5² − 10(5) + 25 = 25 − 50 + 25 = 0 ✓
First term 2x², like Example 3
2x² + 5x − 3 = 0
Set up the boxes.2x² + 5x − 3 = (2x + )(x + )The first terms must be 2x and x. The numbers in the boxes must multiply to −3.
Order matters here.As in Example 3, the first terms are not both x, so try each pair in both orders: 1 and −3, −3 and 1, −1 and 3, 3 and −1.
Cross multiply and add.
You want 5x.
| Try | Cross multiply and add |
|---|---|
| 2xx1−3 | −6x + x = −5x |
| 2xx−31 | 2x − 3x = −x |
| 2xx−13 | 6x − x = 5x ✓ |
| 2xx3−1 | −2x + 3x = x |
Fill in the boxes.(2x − 1)(x + 3) = 0
Set each factor equal to 0 and solve.2x − 1 = 0 → x = 12
x + 3 = 0 → x = −3
x + 3 = 0 → x = −3
Check.2(12)² + 5(12) − 3 = 0.5 + 2.5 − 3 = 0 ✓
2(−3)² + 5(−3) − 3 = 18 − 15 − 3 = 0 ✓
2(−3)² + 5(−3) − 3 = 18 − 15 − 3 = 0 ✓
Negative first term
−x² + 2x + 15 = 0
Multiply both sides by −1.x² − 2x − 15 = 0Factoring is easier when the x² term is positive. Multiplying by −1 changes every sign, and 0 stays 0.
Set up the boxes.x² − 2x − 15 = (x + )(x + )The numbers must multiply to −15.
Cross multiply and add.
You want −2x.
| Try | Cross multiply and add |
|---|---|
| xx1−15 | −15x + x = −14x |
| xx−115 | 15x − x = 14x |
| xx3−5 | −5x + 3x = −2x ✓ |
| xx−35 | 5x − 3x = 2x |
Fill in the boxes and solve.(x + 3)(x − 5) = 0 → x = −3 or x = 5
Check in the original equation.−(−3)² + 2(−3) + 15 = −9 − 6 + 15 = 0 ✓
−(5)² + 2(5) + 15 = −25 + 10 + 15 = 0 ✓
−(5)² + 2(5) + 15 = −25 + 10 + 15 = 0 ✓
A word problem
A rectangle is 3 meters longer than it is wide. Its area is 40 square meters. Find its width and length.
Write the equation.w(w + 3) = 40Let w be the width. The length is w + 3, and area = width × length.
Standard form.w² + 3w − 40 = 0Multiply out the left side, then subtract 40.
Set up the boxes.w² + 3w − 40 = (w + )(w + )The numbers must multiply to −40.
Cross multiply and add.
You want 3w.
| Try | Cross multiply and add |
|---|---|
| ww1−40 | −40w + w = −39w |
| ww−140 | 40w − w = 39w |
| ww2−20 | −20w + 2w = −18w |
| ww−220 | 20w − 2w = 18w |
| ww4−10 | −10w + 4w = −6w |
| ww−410 | 10w − 4w = 6w |
| ww5−8 | −8w + 5w = −3w |
| ww−58 | 8w − 5w = 3w ✓ |
Fill in the boxes and solve.(w − 5)(w + 8) = 0 → w = 5 or w = −8
Keep the answer that makes sense.A width can't be negative, so w = −8 is not an answer here.
Answer.Width 5 m, length 8 mCheck: 5 · 8 = 40 square meters ✓
Your turn
Solve this equationSolved: 0
Step
Which is the equation in standard form, with 0 on one side?
Step
What is the greatest common factor of all the terms in ?
Step
Step
Set each factor equal to 0 and solve for x. If a factor can never be 0, press Never 0. You can type a fraction like 3/2 or a decimal like 1.5.
Question 1 of 5
Score: 0
Solve by factoring. Choose the complete set of solutions.
When factoring doesn't work
Not every quadratic equation can be factored with whole numbers. Take x² + x − 1 = 0. The two numbers would have to multiply to −1 and add to 1, and the only whole-number pairs that multiply to −1 are 1 and −1, which add to 0. When no pair works, solve the equation with the quadratic formula instead. Factoring is fastest when it works, so it's worth trying first.
Common mistakes
- Factoring before the equation equals 0. In x² + 3x = 10, writing x(x + 3) = 10 doesn't help. Subtract 10 first: x² + 3x − 10 = 0, so (x + 5)(x − 2) = 0.
- Giving the numbers instead of the solutions. (x + 6)(x − 5) = 0 has the solutions x = −6 and x = 5. The signs flip when you solve each factor.
- A pair that multiplies right but adds wrong. For x² + x − 30, the pair 5 and −6 multiplies to −30 but adds to −1, not 1. Check both conditions.
- Dividing both sides by x. In 3x² = 12x, dividing by x loses the solution x = 0. Move everything to one side and factor out 3x instead.
- Only one answer for x² = 49. x² − 49 = (x + 7)(x − 7), so x = 7 and x = −7.
- Two answers for a perfect square. (x − 5)² = 0 has only one solution, x = 5.
- Keeping answers that don't make sense. In a word problem, a negative length or width is not a real answer. Check each solution against the situation.