Factoring Polynomials

🧩 Factoring Polynomials

Part 1 of 5 — The Greatest Common Factor

Topics in This Part

🔑 Key Concept: Factoring is un-multiplying — rewriting a sum of terms as a product. The first move in every factoring problem is to pull out the greatest common factor (GCF).

What Does "Factoring" Mean?

When you multiply, you turn a product into a sum:

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

Factoring runs that backward — it turns a sum into a product:

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

The factored form $3x(x+2)$ and the expanded form $3x^2+6x$ are equal for every value of $x$ — they're just two ways of writing the same expression.

💡 Why bother? A product equals zero only when one of its factors is zero. That's why factoring is the gateway to solving equations (Part 5) — and to simplifying, graphing, and calculus later on.

Finding the GCF of a Polynomial

The greatest common factor of a polynomial has two pieces:

1. The largest integer that divides every coefficient.
2. The lowest power of each variable shared by every term.

Example: $12x^3 + 18x^2$

• GCF of $12$ and $18$ is $6$.
• Lowest power of $x$ present in both is $x^2$.

So the GCF is $\,6x^2$, and $12x^3 + 18x^2 = 6x^2(2x + 3)$.

⚠️ Take the lowest power, not the highest. Both terms must contain the factor you pull out, so $x^2$ (not $x^3$) is the most $x$ every term shares.

Concept Check 🎯

What's Left Inside?

Once you pull out the GCF, the expression in parentheses is what remains after dividing each term by the GCF.

$\underbrace{10x^3 - 15x^2 + 5x}_{\text{GCF} = 5x} = 5x\big(\underbrace{2x^2 - 3x + 1}_{\text{each term} \,\div\, 5x}\big)$

Term by term: $\dfrac{10x^3}{5x} = 2x^2$, $\;\dfrac{-15x^2}{5x} = -3x$, $\;\dfrac{5x}{5x} = 1$.

⚠️ Don't drop the last $1$. When a term equals the GCF, dividing leaves $1$, not $0$ — that $1$ must appear inside the parentheses.

What's Inside the Parentheses? 🔽

After pulling out the GCF, choose the correct leftover factor.

Always Check by Re-Multiplying

After factoring, distribute back to confirm you recover the original:

$7x(2x^2 + x + 3) = 14x^3 + 7x^2 + 21x \;✓$

💡 This re-multiply check costs ten seconds and catches almost every GCF slip — a wrong coefficient or a dropped $1$ shows up immediately. Try the drill below, then check each by distributing.

Factor Out the GCF 🧮

Each expression equals $\text{GCF} \cdot (\ldots)$. Enter the GCF only.

1) $\;6x^2 + 9x \;\Rightarrow\;$ GCF $= \,?$ 2) $\;15y^4 - 25y^3 \;\Rightarrow\;$ GCF $= \,?$ 3) $\;14x^3 + 7x^2 + 21x \;\Rightarrow\;$ GCF $= \,?$

On to Trinomials

🔑 Habit for life: Pull out the GCF first, every single time. A leftover common factor hides inside everything that follows — and it makes the trinomials in Parts 2–3 much friendlier.

In Part 2 we tackle the heart of factoring: turning $x^2 + bx + c$ into a product of two binomials.

🧩 Factoring Polynomials

Part 2 of 5 — Trinomials with Leading Coefficient 1

🔑 The Idea: To factor $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. Those two numbers become the constants in $(x + \_)(x + \_)$.

The Sum-and-Product Method

$x^2 + bx + c = (x + p)(x + q) \quad\text{where}\quad p \cdot q = c \;\text{ and }\; p + q = b$

This works because multiplying out gives:

$(x + p)(x + q) = x^2 + (p + q)x + pq$

Worked Example: $x^2 + 7x + 12$

Find two numbers that multiply to $12$ and add to $7$.

The pair is $3$ and $4$, so $x^2 + 7x + 12 = (x + 3)(x + 4)$.

✅ Check: $(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ ✓

Reading the Signs

The signs of $b$ and $c$ tell you the signs of $p$ and $q$ before you start:

Example: $x^2 - 5x + 6$

$c = +6$ (factors same sign) and $b = -5$ (negative), so both factors are negative: $-2$ and $-3$.

$x^2 - 5x + 6 = (x - 2)(x - 3)$

Example: $x^2 + 2x - 15$

$c = -15$ (opposite signs). Need a pair differing to give $+2$: that's $+5$ and $-3$.

$x^2 + 2x - 15 = (x + 5)(x - 3)$

Predict the Signs 🔽

Before factoring, decide the signs of the two constants in $(x \,\square\, p)(x \,\square\, q)$.

Verify Every Factorization

The middle term is the quickest place a sign error shows up, so always FOIL back and watch it:

$(x + 5)(x - 3) = x^2 \underbrace{- 3x + 5x}_{+2x} - 15 = x^2 + 2x - 15 \;✓$

💡 If your inner-plus-outer terms don't recombine to the original middle term, swap a sign or reach for a different factor pair — then re-check.

Factor the Trinomial 🧮

Each factors as $(x + p)(x + q)$ with $p \le q$. Enter $p$ then $q$ (include the sign, e.g. -3).

1) $x^2 + 8x + 15 = (x + p)(x + q)$ → $p = \,?$, $\;q = \,?$ 2) $x^2 - 9x + 14 = (x + p)(x + q)$ → $p = \,?$, $\;q = \,?$

Concept Check 🎯

🧩 Factoring Polynomials

Part 3 of 5 — Trinomials when $a \ne 1$ (the AC Method)

🔑 The Challenge: When the leading coefficient $a$ isn't $1$, you can't just split the constant. The AC method (also called factoring by grouping) handles it every time.

The AC Method

To factor $ax^2 + bx + c$:

1. Multiply $a \cdot c$.
2. Find two numbers that multiply to $ac$ and add to $b$.
3. Split the middle term $bx$ using those two numbers.
4. Group in pairs and factor each pair.
5. Pull out the common binomial.

Worked Example: $6x^2 + 11x + 3$

1. $a \cdot c = 6 \cdot 3 = 18$.
2. Two numbers multiplying to $18$, adding to $11$: $\;9$ and $2$.
3. Split: $6x^2 + 9x + 2x + 3$.
4. Group: $(6x^2 + 9x) + (2x + 3) = 3x(2x + 3) + 1(2x + 3)$.
5. Common binomial $(2x+3)$: $\;(2x + 3)(3x + 1)$.

✅ Check: $(2x+3)(3x+1) = 6x^2 + 2x + 9x + 3 = 6x^2 + 11x + 3$ ✓

Find the AC Split 🧮

For each trinomial, compute $ac$, then give the two numbers (smaller first) that multiply to $ac$ and add to $b$.

1) $3x^2 + 10x + 8$: $\;ac = \,?$, then the two numbers are $\,?$ and $\,?$

A Worked Example with a Negative

Factor $4x^2 - 4x - 3$

1. $a \cdot c = 4 \cdot (-3) = -12$.
2. Two numbers multiplying to $-12$, adding to $-4$: $\;-6$ and $+2$.
3. Split: $4x^2 - 6x + 2x - 3$.
4. Group: $(4x^2 - 6x) + (2x - 3) = 2x(2x - 3) + 1(2x - 3)$.
5. Common binomial: $(2x - 3)(2x + 1)$.

⚠️ Watch the grouping signs. When the third term is negative, factor so the leftover binomials match. Here $+(2x - 3)$ comes from factoring $+1$ out of $(2x - 3)$ — not $-1$.

✅ Check: $(2x-3)(2x+1) = 4x^2 + 2x - 6x - 3 = 4x^2 - 4x - 3$ ✓

Walk the AC Method 🔽

You're factoring $2x^2 + 7x + 6$. Choose what happens at each stage.

Concept Check 🎯

🧩 Factoring Polynomials

Part 4 of 5 — Special Patterns

🔑 The Shortcut: A handful of forms factor instantly once you recognize them — no trial and error. Learn to spot a difference of squares, a perfect square trinomial, and a sum/difference of cubes.

Difference of Squares

$a^2 - b^2 = (a + b)(a - b)$

Two perfect squares separated by a minus sign. (A sum of squares, $a^2 + b^2$, does not factor over the real numbers.)

Examples

⚠️ $x^2 + 9$ is prime over the reals — there is no middle term to "lose," and a sum of squares has no real factorization.

Perfect Square Trinomials

$a^2 + 2ab + b^2 = (a + b)^2 \qquad a^2 - 2ab + b^2 = (a - b)^2$

A perfect square trinomial has a squared first term, a squared last term, and a middle term equal to twice the product of their square roots.

Examples

💡 If the middle term doesn't match $2ab$, it isn't a perfect square — fall back to the AC method.

Concept Check 🎯

Sum & Difference of Cubes

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

A handy memory device is SOAP: the binomial sign is the Same, the linear sign is Opposite, and the last sign is Always Positive.

Worked Example: $x^3 + 8$

Here $a = x$ and $b = 2$ (since $2^3 = 8$):

$x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

Worked Example: $8x^3 - 27$

Here $a = 2x$ (since $(2x)^3 = 8x^3$) and $b = 3$:

$8x^3 - 27 = (2x - 3)\big((2x)^2 + (2x)(3) + 3^2\big) = (2x - 3)(4x^2 + 6x + 9)$

💡 The quadratic factor $a^2 \mp ab + b^2$ almost never factors further — leave it as is.

Match the Pattern 🔽

Identify which special pattern fits each expression.

Apply the Cube Formula 🧮

For $x^3 - 64 = (x - b)(x^2 + bx + b^2)$, identify $b$ and fill in the quadratic factor.

1) $b = \,?$ (since $b^3 = 64$) 2) The constant term of the quadratic factor, $b^2 = \,?$

🧩 Factoring Polynomials

Part 5 of 5 — A Strategy, Solving, & Mastery Check

You can now (1) pull out a GCF, (2) factor simple trinomials, (3) use the AC method, and (4) spot special patterns. The last skill is choosing which tool to reach for — and using factoring to solve.

The Factoring Game Plan

Work through these in order, every time:

⚠️ Factor completely. After pulling a GCF, the inside often still factors. Example: $3x^2 - 27 = 3(x^2 - 9) = 3(x+3)(x-3)$ — three layers, not one.

Factoring Four Terms by Grouping

When a polynomial has four terms, group them in pairs and factor each pair, then pull out the shared binomial.

Worked Example: $x^3 + 3x^2 + 2x + 6$

$\underbrace{(x^3 + 3x^2)}_{x^2(x+3)} + \underbrace{(2x + 6)}_{2(x+3)} = x^2(x + 3) + 2(x + 3) = (x + 3)(x^2 + 2)$

✅ Check: $(x+3)(x^2+2) = x^3 + 2x + 3x^2 + 6 = x^3 + 3x^2 + 2x + 6$ ✓

Solving Equations by Factoring

The Zero-Product Property says: if $A \cdot B = 0$, then $A = 0$ or $B = 0$.

Worked Example: solve $x^2 + 3x - 10 = 0$

1. Factor: $(x + 5)(x - 2) = 0$.
2. Set each factor to zero: $x + 5 = 0$ or $x - 2 = 0$.
3. Solve: $x = -5$ or $x = 2$.

⚠️ One side must be zero first. To solve $x^2 + 3x = 10$, move everything over → $x^2 + 3x - 10 = 0$ before factoring. The Zero-Product Property only works against $0$.

Solve by Factoring 🧮

Factor, then use the Zero-Product Property. Enter the larger solution.

1) $x^2 - x - 12 = 0 \;\Rightarrow\; x = \,?$ (larger root) 2) $x^2 - 16 = 0 \;\Rightarrow\; x = \,?$ (larger root)

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.