The best examples of factoring quadratic expressions: 3 examples every student should know

Instead of starting with formal definitions, let’s jump straight into the 3 core examples of factoring quadratic expressions that show up everywhere in algebra:

• Quadratics with leading coefficient 1 (like \(x^2 + 5x + 6\))
• Quadratics with leading coefficient not equal to 1 (like \(6x^2 + 11x + 3\))
• Special patterns (perfect square trinomials and difference of squares)

Once these feel natural, every other example of factoring quadratic expressions is just a twist on these patterns.

Example 1: Factoring a simple quadratic (leading coefficient 1)

Let’s start with a classic:
\[x^2 + 7x + 10\]

This is the most common example of factoring a quadratic expression that students see first. The leading coefficient (the number in front of \(x^2\)) is 1, which makes life easier.

Goal: Write it as a product of two binomials:
\[(x + _)(x + _)\]

Step-by-step:

1. Look at the constant term: 10.
   We want two numbers that multiply to 10 and add to 7 (the middle coefficient).

2. List factor pairs of 10:

   • 1 and 10 (sum = 11)
   • 2 and 5 (sum = 7)

3. Pick the pair that adds to 7: 2 and 5.

4. Write the factored form:
   \[x^2 + 7x + 10 = (x + 2)(x + 5)\]

That’s it. This is one of the simplest examples of factoring quadratic expressions, and it sets the pattern: multiply to the constant, add to the middle coefficient.

Let’s do a couple more quick ones of this same type.

More real examples of the same pattern

Take:
\[x^2 - 3x - 10\]

We want two numbers that multiply to -10 and add to -3.

• Factor pairs of -10:
  -1 and 10 (sum = 9)
  1 and -10 (sum = -9)

  -2 and 5 (sum = 3)
  2 and -5 (sum = -3) ← this pair works.

So:
\[x^2 - 3x - 10 = (x + 2)(x - 5)\]

Another one:
\[x^2 + x - 12\]

We want two numbers that multiply to -12 and add to 1.

• Factor pairs of -12:
  -1 and 12 (sum = 11)
  1 and -12 (sum = -11)

  -3 and 4 (sum = 1) ← this pair works.

So:
\[x^2 + x - 12 = (x - 3)(x + 4)\]

These are some of the best examples to practice early on, because they build your “number sense radar” for spotting the right pair quickly.

Example 2: Factoring when the leading coefficient isn’t 1

Now let’s turn up the difficulty slightly. Consider this quadratic:
\[6x^2 + 11x + 3\]

This is a classic example of factoring a quadratic expression where the leading coefficient is not 1. You can’t just jump straight to \((x + a)(x + b)\). Instead, think of it as:
\[(ax + b)(cx + d)\]

There are a couple of common methods. I’ll use the AC method (also called the “product-sum” or “grouping” method), which is still one of the most popular approaches in 2024 classrooms and online courses.

Step 1: Multiply the first and last coefficients

Multiply the coefficient of \(x^2\) (which is 6) by the constant term (which is 3):
\[6 \times 3 = 18\]

We now look for two numbers that multiply to 18 and add to 11 (the middle coefficient).

• Factor pairs of 18:
  1 and 18 (sum = 19)
  2 and 9 (sum = 11) ← this pair works.

So we’ll break the middle term, 11x, into 2x + 9x.

Step 2: Rewrite the middle term using those numbers

\[6x^2 + 11x + 3 = 6x^2 + 2x + 9x + 3\]

Step 3: Factor by grouping

Group the first two and the last two terms:
\[(6x^2 + 2x) + (9x + 3)\]

Factor each group:

• \(6x^2 + 2x = 2x(3x + 1)\)
• \(9x + 3 = 3(3x + 1)\)

Now we can factor out the common binomial \((3x + 1)\):
\[6x^2 + 11x + 3 = 2x(3x + 1) + 3(3x + 1) = (2x + 3)(3x + 1)\]

So this expression factors as:
\[6x^2 + 11x + 3 = (2x + 3)(3x + 1)\]

This is one of the best examples of factoring quadratic expressions with a non-1 leading coefficient because it shows the full AC method in action.

Another AC-method example: slightly messier numbers

Try this one:
\[8x^2 - 2x - 3\]

1. Multiply the first and last coefficients:
   \(8 \times -3 = -24\).

2. We want two numbers that multiply to -24 and add to -2.

   Factor pairs of -24:

   -1 and 24 (sum = 23)
   1 and -24 (sum = -23)

   -2 and 12 (sum = 10)
   2 and -12 (sum = -10)

   -3 and 8 (sum = 5)
   3 and -8 (sum = -5)

   -4 and 6 (sum = 2)
   4 and -6 (sum = -2) ← this pair works.

3. Rewrite the middle term using 4 and -6:
   \[8x^2 - 2x - 3 = 8x^2 + 4x - 6x - 3\]

4. Group and factor:
   \[(8x^2 + 4x) + (-6x - 3)\]

   • \(8x^2 + 4x = 4x(2x + 1)\)
   • \(-6x - 3 = -3(2x + 1)\)

5. Factor out the common binomial:
   \[8x^2 - 2x - 3 = 4x(2x + 1) - 3(2x + 1) = (4x - 3)(2x + 1)\]

Again, this is a real example of factoring a quadratic expression that looks nasty at first but falls apart nicely once you apply the AC method.

Example 3: Special patterns – the “shortcut” examples of factoring quadratic expressions

Not all quadratics are created equal. Some follow special patterns that let you factor them very quickly. These are some of the best examples of factoring quadratic expressions because once you recognize the pattern, you can save a lot of time on exams.

Perfect square trinomials

Look at this expression:
\[x^2 + 10x + 25\]

Notice:

• \(x^2 = x^2\) (obviously), and
• \(25 = 5^2\), and
• The middle term, 10x, is 2 × x × 5.

That’s exactly the pattern of a perfect square trinomial:
\[a^2 + 2ab + b^2 = (a + b)^2\]

Here, \(a = x\) and \(b = 5\), so:
\[x^2 + 10x + 25 = (x + 5)^2\]

Another example:
\[9x^2 - 12x + 4\]

• \(9x^2 = (3x)^2\)
• \(4 = 2^2\)
• Middle term: \(-12x = 2 \cdot (3x) \cdot (-2)\)

So this matches the pattern:
\[a^2 - 2ab + b^2 = (a - b)^2\]

Here, \(a = 3x\) and \(b = 2\), so:
\[9x^2 - 12x + 4 = (3x - 2)^2\]

These are very common examples of factoring quadratic expressions on standardized tests like the SAT and ACT, which often reward you for spotting the pattern instead of grinding through the AC method.

Difference of squares

Now consider:
\[16x^2 - 81\]

This expression has no middle term, which is your first clue. It’s a difference (minus sign) of two perfect squares:

• \(16x^2 = (4x)^2\)
• \(81 = 9^2\)

This fits the pattern:
\[a^2 - b^2 = (a - b)(a + b)\]

So:
\[16x^2 - 81 = (4x - 9)(4x + 9)\]

Another example:
\[25x^2 - 1\]

• \(25x^2 = (5x)^2\)
• \(1 = 1^2\)

So:
\[25x^2 - 1 = (5x - 1)(5x + 1)\]

These are some of the fastest examples of factoring quadratic expressions once you train your eye to see “perfect square – perfect square.”

Mixing it together: more real examples to practice

At this point, we’ve walked through the 3 main examples of factoring quadratic expressions: 3 examples that cover the core types:

• Simple trinomials with leading coefficient 1
• Trinomials with leading coefficient not equal to 1
• Special patterns (perfect squares and difference of squares)

Let’s look at a few more real examples that mix these ideas, so you can see how to decide which method to use.

Example: Factor \(5x^2 - 20x\)

This one is sneaky: it’s missing the constant term.

Step 1: Look for a greatest common factor (GCF) first.

Both terms share a factor of 5x:
\[5x^2 - 20x = 5x(x - 4)\]

And that’s already fully factored. Many students skip the GCF step, but it’s one of the most common real-world examples of factoring quadratic expressions in algebra and beyond.

Example: Factor \(3x^2 + 15x + 18\)

Again, start with the GCF.

All three terms share a factor of 3:
\[3x^2 + 15x + 18 = 3(x^2 + 5x + 6)\]

Now factor the quadratic inside the parentheses using the simple pattern from Example 1.

We want two numbers that multiply to 6 and add to 5:

• 2 and 3 (sum = 5) → works.

So:
\[x^2 + 5x + 6 = (x + 2)(x + 3)\]

Final answer:
\[3x^2 + 15x + 18 = 3(x + 2)(x + 3)\]

This is a good example of factoring a quadratic expression where you combine GCF and trinomial factoring.

Example: Factor \(4x^2 - 25\)

This one is a difference of squares:

• \(4x^2 = (2x)^2\)
• \(25 = 5^2\)

So:
\[4x^2 - 25 = (2x - 5)(2x + 5)\]

Example: Factor \(2x^2 + 7x + 3\)

Use the AC method again:

1. Multiply first and last coefficients: \(2 \times 3 = 6\).
2. We want two numbers that multiply to 6 and add to 7:
   1 and 6 → works.
3. Rewrite the middle term:
   \[2x^2 + 7x + 3 = 2x^2 + x + 6x + 3\]

4. Group and factor:
   \[(2x^2 + x) + (6x + 3)\]

   • \(2x^2 + x = x(2x + 1)\)
   • \(6x + 3 = 3(2x + 1)\)

5. Factor out the common binomial:
   \[2x^2 + 7x + 3 = (x + 3)(2x + 1)\]

This gives you another solid example of factoring a quadratic expression with a leading coefficient not equal to 1.

How factoring quadratics shows up in 2024–2025 math learning

If you’re studying in 2024–2025, you’re probably seeing these same patterns in:

• Online platforms like Khan Academy, IXL, and district learning portals
• College placement or entrance exams (community college, state universities)
• Standardized tests that still include algebra sections

Factoring quadratics isn’t just a “school exercise.” It connects directly to solving quadratic equations, graphing parabolas, and modeling real-world situations like projectile motion or profit functions. For instance, when you solve \(x^2 + 5x + 6 = 0\) by factoring, you’re really finding where the parabola crosses the x-axis.

If you want a deeper refresher on polynomials and factoring beyond these examples of factoring quadratic expressions, you can check out:

• Purplemath’s factoring lessons (clear, student-friendly explanations)
• Khan Academy’s factoring quadratics section
• OpenStax Algebra & Trigonometry textbook (free, open-source, widely used in US colleges)

These sources are widely used across US schools and colleges and stay current with modern course standards.

Quick mental checklist for any quadratic

When you see a quadratic expression and you’re not sure how to factor it, run through this checklist:

1. Is there a GCF you can factor out?
   If yes, do that first. It often makes the rest much easier.

2. Is it a difference of squares?
   Something like \(a^2 - b^2\) with no middle term.

3. Is it a perfect square trinomial?
   Does it look like \(a^2 ± 2ab + b^2\)?

4. If not special, is the leading coefficient 1?
   Use the “multiply to constant, add to middle” trick.

5. If the leading coefficient is not 1,
   use the AC method and factor by grouping.

Almost every example of factoring a quadratic expression you see in class, homework, or exams will fall into one of these buckets.

FAQ: Common questions about examples of factoring quadratic expressions

How do I know which method to use when I see a quadratic?

Start by scanning for the easy wins: a GCF, a difference of squares, or a perfect square trinomial. If none of those patterns fit, check whether the leading coefficient is 1. If it is, use the simple “multiply to constant, add to middle” method. If it isn’t, the AC method with grouping is usually the best option. With practice, you’ll recognize which examples of factoring quadratic expressions match which pattern almost instantly.

Can every quadratic expression be factored nicely with integers?

No. Some quadratics don’t factor over the integers, even though you can still solve them using the quadratic formula. For example, \(x^2 + x + 1\) does not factor into binomials with integer coefficients. In those cases, you can still write a factored form using irrational or complex numbers, but they aren’t the usual classroom “nice” examples of factoring quadratic expressions.

Why do teachers use the same kinds of examples over and over?

Because those patterns really do come up constantly. The best examples of factoring quadratic expressions are chosen to highlight the structures you’ll see in real problems: motion, area, revenue, or anything modeled by a parabola. Once you can factor something like \(x^2 + 7x + 10\) in your sleep, solving related equations and graphing functions becomes much easier.

Is there a quick way to check if my factoring is correct?

Yes: just multiply your factors back out (FOIL or distribution) and see if you get the original quadratic. For example, if you claim that \(6x^2 + 11x + 3 = (2x + 3)(3x + 1)\), multiply:
\[(2x + 3)(3x + 1) = 6x^2 + 2x + 9x + 3 = 6x^2 + 11x + 3\]
If the product matches the original expression, your factoring is correct.

By working through these examples of factoring quadratic expressions: 3 examples at the core plus several extra practice problems, you’ve seen the main patterns you’ll meet in algebra. Keep a small list of these types nearby when you study, and with a bit of repetition, factoring quadratics will start to feel far less mysterious and much more like a set of familiar moves.