7.2 Factoring Polynomials & Solving Quadratic Equations by Factoring

7.2 Factoring Polynomials & Solving Quadratic Equations by Factoring

Factoring is one of the most powerful and frequently tested algebraic skills on the TABE 13&14 assessment. While polynomial multiplication combines simple factors into an expanded polynomial, factoring reverses this process, breaking an expanded polynomial down into a product of simpler, irreducible factors. Mastering factoring allows you to simplify rational expressions, determine parabolic roots, and solve quadratic equations with speed and confidence.

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1. Factoring as the Inverse of Multiplication & The Master Strategy

Think of factoring and multiplication as inverse mathematical operations, analogous to division and multiplication:

$$\begin{array}{ccc} \text{Expanded Form} & \underset{\text{Factor}}{\overset{\text{Multiply}}{\rightleftarrows}} & \text{Factored Form} \\ 2x^2 + 7x - 15 & & (2x - 3)(x + 5) \end{array}$$

The Master Factoring Hierarchy

When presented with any polynomial factoring problem, follow this systematic flowchart:

Step 1: FACTOR OUT THE GREATEST COMMON FACTOR (GCF) FIRST!
   │
Step 2: COUNT THE REMAINING TERMS:
   ├── 2 Terms ──> Difference of Two Squares: a^2 - b^2 = (a + b)(a - b)
   │
   ├── 3 Terms ──> Monic (a = 1): x^2 + bx + c  ==> (x + p)(x + q) [p*q=c, p+q=b]
   │           ──> Non-Monic (a ≠ 1): ax^2 + bx + c ==> ac Product & Grouping
   │
   └── 4 Terms ──> Factoring by Grouping: (ax + ay) + (bx + by)
   │
Step 3: CHECK IF ANY FACTORS CAN BE FACTORED FURTHER (Complete Factoring)

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2. Step 1: Factoring Out the Greatest Common Factor (GCF)

The Greatest Common Factor (GCF) of a polynomial is the largest monomial that divides evenly into every term. It consists of:

1. The greatest common numerical divisor (GCD) of all coefficients.
2. The lowest exponential power of each shared variable base.

Worked Example 1: Factor $12x^4 - 18x^3 + 6x^2$.

• Coefficients: The greatest common divisor of $12, 18,$ and $6$ is $6$.
• Variables: The variable $x$ appears in all terms; the lowest exponent is $x^2$.
• Polynomial GCF: $6x^2$.
• Divide each term by the GCF: $$\frac{12x^4}{6x^2} = 2x^2, \quad \frac{-18x^3}{6x^2} = -3x, \quad \frac{6x^2}{6x^2} = 1$$
• Factored Expression: $6x^2(2x^2 - 3x + 1)$

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3. Step 2: Factoring Binomials (Difference of Two Squares) & 4-Term Grouping

Difference of Two Squares: $a^2 - b^2 = (a + b)(a - b)$

A two-term polynomial can be factored if and only if both terms are perfect squares separated by a minus sign:

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

Worked Example 2 (Direct): Factor $25x^2 - 64$.

• Identify square roots: $\sqrt{25x^2} = 5x$ and $\sqrt{64} = 8$.
• Factored Form: $(5x + 8)(5x - 8)$

Worked Example 3 (Multi-Step with GCF): Factor $3x^3 - 48x$ completely.

1. Extract GCF first: $3x(x^2 - 16)$
2. Factor the remaining difference of squares: $x^2 - 16 = (x + 4)(x - 4)$
3. Complete Factorization: $3x(x + 4)(x - 4)$

Factoring 4-Term Polynomials by Grouping

When a polynomial has four terms, group them into two pairs, extract the GCF from each pair, and then factor out the common binomial factor.

Worked Example 4: Factor $2x^3 + 6x^2 + 5x + 15$.

1. Group into pairs: $(2x^3 + 6x^2) + (5x + 15)$
2. Extract GCF from each pair: $2x^2(x + 3) + 5(x + 3)$
3. Extract common binomial factor $(x + 3)$: $(2x^2 + 5)(x + 3)$

Worked Example 5 (Watch the Negative Sign): Factor $3x^3 - 6x^2 - 4x + 8$.

1. Group pairs: $(3x^3 - 6x^2) - (4x - 8)$ (Note: factoring out $-1$ changes $+8$ to $-8$)
2. Extract GCFs: $3x^2(x - 2) - 4(x - 2)$
3. Extract common binomial: $(3x^2 - 4)(x - 2)$

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4. Step 3: Factoring Monic Quadratic Trinomials ($x^2 + bx + c$)

A monic quadratic trinomial has a leading coefficient of $a = 1$. It factors into the product of two linear binomials:

$$x^2 + bx + c = (x + p)(x + q)$$

where the numbers $p$ and $q$ must simultaneously satisfy two conditions:

1. Product Condition: $p \cdot q = c$
2. Sum Condition: $p + q = b$

Sign Rules for $p$ and $q$

Worked Example 6: Factor $x^2 - 7x + 12$.

• $c = +12$ and $b = -7 \implies$ both factors must be negative.
• Factor pairs of $12$: $(-1)(-12) \to -13$; $(-2)(-6) \to -8$; $(-3)(-4) \to -7$.
• Selected values: $p = -3, q = -4$.
• Factored Form: $(x - 3)(x - 4)$

Worked Example 7: Factor $x^2 + 3x - 28$.

• $c = -28$ and $b = +3 \implies$ opposite signs, positive factor has larger magnitude.
• Factor pairs of $-28$: $(+28, -1) \to +27$; $(+14, -2) \to +12$; $(+7, -4) \to +3$.
• Selected values: $p = +7, q = -4$.
• Factored Form: $(x + 7)(x - 4)$

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5. Step 4: Factoring Non-Monic Quadratic Trinomials ($ax^2 + bx + c, a \neq 1$)

When the leading coefficient $a \neq 1$ and cannot be divided out via a common GCF, use the $ac$ Product & Grouping Method:

The 4-Step $ac$ Algorithm

1. Multiply the leading coefficient $a$ by the constant $c$ to find the target product $ac$.
2. Find two integers $p$ and $q$ that multiply to $ac$ and add to $b$ ($p \cdot q = ac$ and $p + q = b$).
3. Rewrite the middle term $bx$ as the sum $px + qx$, turning the 3-term trinomial into a 4-term polynomial.
4. Factor the 4-term polynomial by grouping.

Worked Example 8: Factor $6x^2 + 11x - 10$.

• Step 1 (Find $ac$): $a = 6, c = -10 \implies ac = 6 \cdot (-10) = -60$.
• Step 2 (Find $p, q$): Find two factors of $-60$ that add to $b = +11$.
  • Testing factor pairs: $(+15) \cdot (-4) = -60$ and $15 + (-4) = +11$. Values are $+15$ and $-4$.
• Step 3 (Split middle term): Rewrite $11x$ as $15x - 4x$: $$6x^2 + 15x - 4x - 10$$
• Step 4 (Factor by grouping): $$(6x^2 + 15x) + (-4x - 10) = 3x(2x + 5) - 2(2x + 5) = \mathbf{(3x - 2)(2x + 5)}$$
• Check via FOIL: $(3x)(2x) + (3x)(5) - 2(2x) - 2(5) = 6x^2 + 15x - 4x - 10 = 6x^2 + 11x - 10$ $\checkmark$

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6. Solving Quadratic Equations by Factoring

A quadratic equation is a second-degree polynomial equation. Its standard form is:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

The Zero Product Property

Solving quadratic equations by factoring relies on a fundamental property of real arithmetic:

$$\text{If } A \cdot B = 0, \quad \text{then } A = 0 \quad \text{or} \quad B = 0$$

If the product of two algebraic factors is zero, at least one of the individual factors must equal zero.

Solving Quadratic Equations by Factoring:
[ax^2 + bx + c = 0] ──> Factor ──> [(px + q)(rx + s) = 0] ──> Set Each Factor to Zero ──> Solve Linear Equations

Step-by-Step Solving Protocol

1. Standard Form: Move all terms to one side so the equation equals zero ($ax^2 + bx + c = 0$).
2. Factor Completely: Factor the polynomial side into linear binomial factors.
3. Apply Zero Product Property: Set each factor equal to zero.
4. Solve Linear Equations: Solve each resulting first-degree equation for $x$.
5. Verify: Check both solutions in the original equation.

Worked Example 9: Solve $2x^2 - 5x = 12$.

$$\begin{aligned} \text{Step 1 (Set to 0):} &\quad 2x^2 - 5x - 12 = 0 \\ \text{Step 2 (Factor via } ac \text{):} &\quad ac = 2(-12) = -24; \quad p = -8, q = +3 \\ &\quad 2x^2 - 8x + 3x - 12 = 0 \\ &\quad 2x(x - 4) + 3(x - 4) = 0 \\ &\quad (2x + 3)(x - 4) = 0 \\ \text{Step 3 (Set to 0):} &\quad 2x + 3 = 0 \quad \text{or} \quad x - 4 = 0 \\ \text{Step 4 (Solve):} &\quad 2x = -3 \implies x = -\frac{3}{2} \quad \text{and} \quad x = 4 \end{aligned}$$

Solution Set: $x = \left\{-\frac{3}{2}, 4\right\}$

Applied Workplace Word Problem: Dimensions of a Carpentry Workshop

A carpenter designs a rectangular workbench where the length is $5\text{ feet}$ longer than the width ($w$). If the total surface area of the benchtop is $66\text{ square feet}$, find its exact physical dimensions.

$$\begin{aligned} \text{Equation:} &\quad w(w + 5) = 66 \\ \text{Expand and Set to 0:} &\quad w^2 + 5w - 66 = 0 \\ \text{Factor:} &\quad (w + 11)(w - 6) = 0 \\ \text{Zero Product Property:} &\quad w = -11 \quad \text{or} \quad w = 6 \end{aligned}$$

Domain Constraint Analysis: A physical workbench width cannot be negative ($w > 0$). We reject the extraneous solution $w = -11$. Therefore, the workbench has a width of $6\text{ feet}$ and a length of $6 + 5 = \mathbf{11\text{ feet}}$.