7.2 Factoring Polynomials & Solving Quadratic Equations by Factoring
Key Takeaways
- Factoring decomposes a polynomial into a product of simpler irreducible factors, functioning as the exact algebraic inverse of polynomial multiplication.
- The Master Factoring Strategy begins unconditionally by factoring out the Greatest Common Factor (GCF), followed by term-count classification (Difference of Squares, product-sum trinomials, ac-grouping, or 4-term pair grouping).
- Factoring monic quadratic trinomials ($x^2 + bx + c$) requires identifying two integers $p$ and $q$ that multiply to $c$ and add to $b$, producing $(x + p)(x + q)$.
- The Zero Product Property states that if $A \cdot B = 0$, then $A = 0$ or $B = 0$; this allows quadratic equations in standard form ($ax^2 + bx + c = 0$) to be solved by factoring into linear equations and solving each independently.
Last updated: August 2026
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.
1. Factoring as the Inverse of Multiplication & The Master Strategy
Think of factoring and multiplication as inverse mathematical operations, analogous to division and multiplication:
\text{Expanded Form} & \underset{\text{Factor}}{\overset{\text{Multiply}}{\rightleftarrows}} & \text{Factored Form} \\ 2x^2 + 7x - 15 & & (2x - 3)(x + 5) \end{array}$$ ### The Master Factoring Hierarchy When presented with any polynomial factoring problem, follow this systematic flowchart: ```text Step 1: FACTOR OUT THE GREATEST COMMON FACTOR (GCF) FIRST! │ Step 2: COUNT THE REMAINING TERMS: ├── 2 Terms ──> Difference of Two Squares: a^2 - b^2 = (a + b)(a - b) │ ├── 3 Terms ──> Monic (a = 1): x^2 + bx + c ==> (x + p)(x + q) [p*q=c, p+q=b] │ ──> Non-Monic (a ≠ 1): ax^2 + bx + c ==> ac Product & Grouping │ └── 4 Terms ──> Factoring by Grouping: (ax + ay) + (bx + by) │ Step 3: CHECK IF ANY FACTORS CAN BE FACTORED FURTHER (Complete Factoring) ``` --- ## 2. Step 1: Factoring Out the Greatest Common Factor (GCF) The **Greatest Common Factor (GCF)** of a polynomial is the largest monomial that divides evenly into every term. It consists of: 1. The greatest common numerical divisor (GCD) of all coefficients. 2. The lowest exponential power of each shared variable base. **Worked Example 1:** Factor $12x^4 - 18x^3 + 6x^2$. - **Coefficients:** The greatest common divisor of $12, 18,$ and $6$ is **$6$**. - **Variables:** The variable $x$ appears in all terms; the lowest exponent is $x^2$. - **Polynomial GCF:** **$6x^2$**. - **Divide each term by the GCF:** $$\frac{12x^4}{6x^2} = 2x^2, \quad \frac{-18x^3}{6x^2} = -3x, \quad \frac{6x^2}{6x^2} = 1$$ - **Factored Expression:** **$6x^2(2x^2 - 3x + 1)$** > [!TIP] > **Negative Leading Coefficient Rule:** > If the leading term of a polynomial is negative, always factor out a negative GCF to ensure the leading term inside the parentheses is positive (e.g., $-4x^2 - 12x = -4x(x + 3)$). --- ## 3. Step 2: Factoring Binomials (Difference of Two Squares) & 4-Term Grouping ### Difference of Two Squares: $a^2 - b^2 = (a + b)(a - b)$ A two-term polynomial can be factored if and only if both terms are perfect squares separated by a **minus sign**: $$a^2 - b^2 = (a + b)(a - b)$$ > [!CAUTION] > **Sum of Squares is Prime:** > The *sum* of two squares ($a^2 + b^2$) **cannot** be factored over the set of real numbers. For example, $x^2 + 16$ is prime. **Worked Example 2 (Direct):** Factor $25x^2 - 64$. - Identify square roots: $\sqrt{25x^2} = 5x$ and $\sqrt{64} = 8$. - Factored Form: **$(5x + 8)(5x - 8)$** **Worked Example 3 (Multi-Step with GCF):** Factor $3x^3 - 48x$ completely. 1. Extract GCF first: $3x(x^2 - 16)$ 2. Factor the remaining difference of squares: $x^2 - 16 = (x + 4)(x - 4)$ 3. Complete Factorization: **$3x(x + 4)(x - 4)$** ### Factoring 4-Term Polynomials by Grouping When a polynomial has four terms, group them into two pairs, extract the GCF from each pair, and then factor out the common binomial factor. **Worked Example 4:** Factor $2x^3 + 6x^2 + 5x + 15$. 1. **Group into pairs:** $(2x^3 + 6x^2) + (5x + 15)$ 2. **Extract GCF from each pair:** $2x^2(x + 3) + 5(x + 3)$ 3. **Extract common binomial factor $(x + 3)$:** **$(2x^2 + 5)(x + 3)$** **Worked Example 5 (Watch the Negative Sign):** Factor $3x^3 - 6x^2 - 4x + 8$. 1. **Group pairs:** $(3x^3 - 6x^2) - (4x - 8)$ *(Note: factoring out $-1$ changes $+8$ to $-8$)* 2. **Extract GCFs:** $3x^2(x - 2) - 4(x - 2)$ 3. **Extract common binomial:** **$(3x^2 - 4)(x - 2)$** --- ## 4. Step 3: Factoring Monic Quadratic Trinomials ($x^2 + bx + c$) A monic quadratic trinomial has a leading coefficient of $a = 1$. It factors into the product of two linear binomials: $$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$ | Signs of $b$ and $c$ | Factor Signs | Strategy | | :--- | :---: | :--- | | $c > 0$ and $b > 0$ | Both $(+)$ | Find two positive factors of $c$ that sum to $b$. | | $c > 0$ and $b < 0$ | Both $(-)$ | Find two negative factors of $c$ that sum to $b$. | | $c < 0$ (any $b$) | One $(+)$, One $(-)$ | Factors have opposite signs; the factor with the larger absolute value shares the sign of $b$. | **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)$** --- ## 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)}$$ - **Check via FOIL:** $(3x)(2x) + (3x)(5) - 2(2x) - 2(5) = 6x^2 + 15x - 4x - 10 = 6x^2 + 11x - 10$ $\checkmark$ --- ## 6. Solving Quadratic Equations by Factoring A **quadratic equation** is a second-degree polynomial equation. Its standard form is: $$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$$ If the product of two algebraic factors is zero, at least one of the individual factors must equal zero. ```text Solving Quadratic Equations by Factoring: [ax^2 + bx + c = 0] ──> Factor ──> [(px + q)(rx + s) = 0] ──> Set Each Factor to Zero ──> Solve Linear Equations ``` ### Step-by-Step Solving Protocol 1. **Standard Form:** Move all terms to one side so the equation equals zero ($ax^2 + bx + c = 0$). 2. **Factor Completely:** Factor the polynomial side into linear binomial factors. 3. **Apply Zero Product Property:** Set each factor equal to zero. 4. **Solve Linear Equations:** Solve each resulting first-degree equation for $x$. 5. **Verify:** Check both solutions in the original equation. **Worked Example 9:** Solve $2x^2 - 5x = 12$. $$\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 $6 + 5 = \mathbf{11\text{ feet}}$.Loading diagram...
Test Your Knowledge
Factor the quadratic trinomial completely: 3x^2 + 14x - 5.
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Factor the polynomial completely: 2x^3 - 50x.
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Solve the quadratic equation for all real solutions: 2x^2 - 7x = 15.
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