Factoring Simple Polynomials

Factoring is the reverse of multiplying. When you multiply \((x + 3)(x + 2)\), you get \(x^2 + 5x + 6\). When you factor \(x^2 + 5x + 6\), you're finding those original binomials: \((x + 3)(x + 2)\).

Factoring is useful for solving equations, simplifying expressions, and understanding the structure of polynomials. It's like finding what ingredients went into making a cake—instead of baking (multiplying), you're figuring out the recipe (factoring).

There are several factoring techniques. In this lesson, we'll focus on the most fundamental ones: factoring out the greatest common factor (GCF) and factoring by grouping.

Finding the Greatest Common Factor

The GCF is the largest expression that divides evenly into all terms of a polynomial. To factor out the GCF, you pull it out front and divide each term by it.

Example: Factor \(6x + 9\).

Both 6 and 9 are divisible by 3. So 3 is the GCF.

$$6x + 9 = 3(2x + 3)$$

You can verify this by distributing: \(3(2x + 3) = 6x + 9\) ✓

Example: Factor \(4x^2 + 8x\).

The GCF of 4 and 8 is 4. Both terms also have at least one \(x\), so \(x\) is part of the GCF too.

GCF: \(4x\)

$$4x^2 + 8x = 4x(x + 2)$$

Check: \(4x(x + 2) = 4x^2 + 8x\) ✓

Example: Factor \(15y^3 - 10y^2 + 5y\).

The coefficients 15, 10, and 5 all have GCF of 5. All three terms have at least one \(y\).

GCF: \(5y\)

$$15y^3 - 10y^2 + 5y = 5y(3y^2 - 2y + 1)$$

Check: \(5y(3y^2 - 2y + 1) = 15y^3 - 10y^2 + 5y\) ✓

When There's a Negative GCF

Sometimes it's helpful to factor out a negative number if the leading coefficient is negative.

Example: Factor \(-6x - 12\).

You could factor out 6: \(6(-x - 2)\)

But it's often cleaner to factor out -6: \(-6(x + 2)\)

Both are correct, but the second form is more standard because the leading coefficient inside the parentheses is positive.

Example: Factor \(-3a^2 + 9a\).

Factor out \(-3a\):

$$-3a^2 + 9a = -3a(a - 3)$$

Factoring by Grouping

When a polynomial has four terms, you can sometimes factor it by grouping pairs of terms.

Example: Factor \(ax + ay + bx + by\).

Group the first two terms and the last two terms:

$$(ax + ay) + (bx + by)$$

Factor out the GCF from each group:

$$= a(x + y) + b(x + y)$$

Now notice that \((x + y)\) appears in both terms. Factor it out:

$$= (x + y)(a + b)$$

Example: Factor \(6x^2 + 9x + 4x + 6\).

Group: $$(6x^2 + 9x) + (4x + 6)$$

Factor GCF from each group: $$= 3x(2x + 3) + 2(2x + 3)$$

Factor out \((2x + 3)\): $$= (2x + 3)(3x + 2)$$

Check by multiplying: \((2x + 3)(3x + 2) = 6x^2 + 4x + 9x + 6 = 6x^2 + 13x + 6\)

Wait, that's not right! Let me recalculate. The original was \(6x^2 + 9x + 4x + 6 = 6x^2 + 13x + 6\). Let me verify the factoring.

\((2x + 3)(3x + 2)\):

• First: \(6x^2\)
• Outer: \(4x\)
• Inner: \(9x\)
• Last: \(6\)

Total: \(6x^2 + 13x + 6\) ✓

Example: Factor \(2x^3 - 6x^2 + 5x - 15\).

Group: $$(2x^3 - 6x^2) + (5x - 15)$$

Factor each group: $$= 2x^2(x - 3) + 5(x - 3)$$

Factor out \((x - 3)\): $$= (x - 3)(2x^2 + 5)$$

When Grouping Doesn't Work Right Away

Sometimes you need to rearrange terms to make grouping work.

Example: Factor \(xy + 3y + 2x + 6\).

If we group the first two and last two, we get: $$(xy + 3y) + (2x + 6)$$ $$= y(x + 3) + 2(x + 3)$$ $$= (x + 3)(y + 2)$$

That worked! But we could also rearrange first: $$xy + 2x + 3y + 6$$ $$(xy + 2x) + (3y + 6)$$ $$= x(y + 2) + 3(y + 2)$$ $$= (y + 2)(x + 3)$$

Same result, just in different order.

Combining GCF with Other Factoring

Always factor out the GCF first before trying other factoring methods.

Example: Factor \(3x^2 + 12x + 12\).

GCF is 3: $$= 3(x^2 + 4x + 4)$$

Now we could factor \(x^2 + 4x + 4\) further if we knew how to factor trinomials (which you'll learn in Algebra II), but we'll stop here for now.

Checking Your Factoring

Always check by multiplying the factors back together. If you get the original polynomial, you factored correctly.

Example: Did I factor \(10a^2 - 15a\) correctly as \(5a(2a - 3)\)?

Check: \(5a(2a - 3) = 10a^2 - 15a\) ✓

Yes!

Give These a Try

Factor completely:

1. \(12x + 18\)
2. \(5y^2 - 10y\)
3. \(-4m^2 + 8m\)
4. \(ax + bx + ay + by\)
5. \(3x^2 + 6x + 2x + 4\)
6. \(8n^3 - 12n^2 + 10n - 15\)

Solutions:

1. GCF is 6: \(6(2x + 3)\)
2. GCF is \(5y\): \(5y(y - 2)\)
3. GCF is \(-4m\): \(-4m(m - 2)\)
4. Group: \(x(a + b) + y(a + b) = (a + b)(x + y)\)
5. Group: \(3x(x + 2) + 2(x + 2) = (x + 2)(3x + 2)\)
6. Group: \(4n^2(2n - 3) + 5(2n - 3) = (2n - 3)(4n^2 + 5)\)

What Trips People Up

Not factoring out the GCF completely is wasteful. If you have \(8x^2 + 12x\) and factor out \(2x\) to get \(2x(4x + 6)\), you're not finished—there's still a GCF of 2 inside those parentheses: \(2x \cdot 2(2x + 3) = 4x(2x + 3)\).

Forgetting to factor out the variable part of the GCF causes incomplete answers. In \(6x^3 + 9x^2\), the GCF isn't just 3—it's \(3x^2\). Both terms have at least \(x^2\), so factor it out.

Sign errors when factoring out a negative happen frequently. When you factor \(-1\) out of \(-x + 5\), you get \(-1(x - 5)\), not \(-1(x + 5)\). Every term's sign flips when you pull out that negative.

Not checking your work by multiplying back is asking for trouble. Multiply your factors together and make sure you get the original polynomial. If you don't, you made a mistake somewhere.

Trying to factor prime polynomials wastes time. Not every polynomial factors nicely using integers. If you've looked for a GCF and tried grouping (when applicable) and nothing works, the polynomial might just be prime. That's okay.