Comprehensive Study Guide: Algebraic Factoring and Solving Techniques

Fundamental Algebraic Principles and Factoring Methods

• Greatest Common Factor (GCF):

  • The GCF of a polynomial is the largest factor (combining the highest common numerical factor and the highest common variable power) shared by all terms in the expression.

  • Factoring out the GCF uses the distributive property in reverse:         $a \cdot b + a \cdot c = a(b + c)$

• Difference of Squares:

  • An expression of the form $a^2 - b^2$ can always be factored into:         $a^2 - b^2 = (a - b)(a + b)$

  • A sum of squares, $a^2 + b^2$, is prime over the real numbers, but factors over complex numbers as:         $a^2 + b^2 = (a - bi)(a + bi)$

• Factoring Trinomials of the Form $x^2 + bx + c$:

  • Find two numbers, $p$ and $q$, such that:         $p \times q = c$         $p + q = b$

  • The factored form is:         $(x + p)(x + q)$

• Factoring Trinomials Using the AC-Method ($ax^2 + bx + c$):

  • Multiply the coefficient of the quadratic term $a$ by the constant term $c$ to get the product $a \cdot c$.

  • Find two integers, $p$ and $q$, whose product is $a \cdot c$ and whose sum is $b$:         $p \times q = a \cdot c$         $p + q = b$

  • Rewrite the middle term $bx$ as px + qx$.\n * Factor the resulting four-term polynomial by grouping.\n\n* **Factoring by Grouping:**\n * Group terms in pairs (usually the first two and last two) that share common factors.\n * Factor out the GCF from each pair.\n * Factor out the resulting binomial GCF common to both grouped terms.\n\n# Section A.1 Factoring Review - Turn-In Homework Solutions\n\n* **Problem 1:** Factor 6x^5 - 15x^3\n * Identify the greatest common factor of numerical coefficients 6$and$15$, which is$3\n * Identify the highest power of x$common to both terms, which is$x^3\n * Factor out 3x^3:\n        6x^5 - 15x^3 = 3x^3(2x^2 - 5)\n\n* **Problem 2:** Factor 49c^2 - 9\n * Recognize this expression as a difference of two perfect squares, where (7c)^2 = 49c^2$and$3^2 = 9\n * Apply the difference of squares pattern a^2 - b^2 = (a - b)(a + b):\n        49c^2 - 9 = (7c - 3)(7c + 3)\n\n* **Problem 3:** Factor a^2 + 15a + 54\n * Identify two factors of 54$that add up to$15\n * The factor pair is 9$and$6$, since$9 \times 6 = 54$and$9 + 6 = 15\n * Factor the expression directly into two binomials:\n        a^2 + 15a + 54 = (a + 9)(a + 6)\n\n* **Problem 4:** Factor a^2 - 15a - 54\n * Identify two factors of -54$that add up to$-15\n * The factor pair is -18$and$3$, since$-18 \times 3 = -54$and$-18 + 3 = -15\n * Factor the quadratic expression:\n        a^2 - 15a - 54 = (a - 18)(a + 3)\n\n* **Problem 5:** Factor 4x^2 - 6x + 6x - 9\n * Observe that the middle terms -6x$and$6x$cancel out, leaving$4x^2 - 9$, which is a difference of squares$(2x)^2 - 3^2\n * Alternatively, use grouping directly on the four terms:\n        4x^2 - 6x + 6x - 9 = 2x(2x - 3) + 3(2x - 3)\n * Factor out the common binomial (2x - 3):\n        4x^2 - 6x + 6x - 9 = (2x + 3)(2x - 3)\n\n* **Problem 6:** Factor 9y^2 - 3y - 3y + 1\n * Group the first two terms and the last two terms:\n        9y^2 - 3y - 3y + 1 = 3y(3y - 1) - 1(3y - 1)\n * Factor out the common binomial (3y - 1):\n        9y^2 - 3y - 3y + 1 = (3y - 1)(3y - 1) = (3y - 1)^2\n\n* **Problem 7:** Factor 3a^2 + 17a + 24\n * Use the AC-method: compute a \cdot c = 3 \times 24 = 72\n * Find two numbers that multiply to 72$and sum to$17$: the numbers are$9$and$8\n * Rewrite the middle term 17a$as$9a + 8a:\n        3a^2 + 17a + 24 = 3a^2 + 9a + 8a + 24\n * Factor by grouping:\n        3a^2 + 9a + 8a + 24 = 3a(a + 3) + 8(a + 3)\n * Factor out the binomial (a + 3):\n        3a^2 + 17a + 24 = (3a + 8)(a + 3)\n\n* **Problem 8:** Factor 2x^2 + x - 10\n * Use the AC-method: compute a \cdot c = 2 \times (-10) = -20\n * Find two numbers that multiply to -20$and sum to$1$: the numbers are$5$and$-4\n * Rewrite the middle term x$as$-4x + 5x:\n        2x^2 + x - 10 = 2x^2 - 4x + 5x - 10\n * Factor by grouping:\n        2x^2 - 4x + 5x - 10 = 2x(x - 2) + 5(x - 2)\n * Factor out the binomial (x - 2):\n        2x^2 + x - 10 = (2x + 5)(x - 2)\n\n# Section A.1 Factoring Review - Practice Homework Solutions\n\n* **Problem 1: Factor the GCF**\n * **(a)** 4x^8 - 12x^4 + 2x^3\n * Find the GCF of coefficients 4$,$-12$, and$2$, which is$2\n * Find the lowest exponent of x$present in all terms, which is$x^3\n * Factor out 2x^3 from each term:\n            4x^8 - 12x^4 + 2x^3 = 2x^3(2x^5 - 6x + 1)\n * **(b)** -9a^5c^3 - 21a^4c + 12a^2c^2\n * Find the GCF of coefficients -9$,$-21$, and$12$, which is$-3\n * Find common variable powers: a^2$and$c\n * Factor out -3a^2c:\n            -9a^5c^3 - 21a^4c + 12a^2c^2 = -3a^2c(3a^3c^2 + 7a^2 - 4c)\n\n* **Problem 2: Using differences of squares to fully factor**\n * **(a)** 4x^2 - 25\n * Recognize as (2x)^2 - 5^2\n * Apply difference of squares formula:\n            4x^2 - 25 = (2x - 5)(2x + 5)\n * **(b)** 2z^5 - 98z^3\n * First factor out the GCF 2z^3:\n            2z^5 - 98z^3 = 2z^3(z^2 - 49)\n * Factor the remaining term z^2 - 49 as a difference of squares:\n            2z^5 - 98z^3 = 2z^3(z - 7)(z + 7)\n * **(c)** y^4 - 81\n * Factor as a difference of squares:\n            y^4 - 81 = (y^2 - 9)(y^2 + 9)\n * Fully factor the remaining real difference of squares y^2 - 9:\n            y^4 - 81 = (y - 3)(y + 3)(y^2 + 9)\n\n* **Problem 3: Fully factor the quadratics below**\n * **(a)** x^2 + 4x - 21\n * Find factors of -21$that sum to$4$:$7$and$-3\n * Factored result:\n            x^2 + 4x - 21 = (x + 7)(x - 3)\n * **(b)** b^2 - b - 30\n * Find factors of -30$that sum to$-1$:$-6$and$5\n * Factored result:\n            b^2 - b - 30 = (b - 6)(b + 5)\n * **(c)** n^2 + 36n + 99\n * Find factors of 99$that sum to$36$:$33$and$3\n * Factored result:\n            n^2 + 36n + 99 = (n + 33)(n + 3)\n * **(d)** z^2 - 16z + 48\n * Find factors of 48$that sum to$-16$:$-12$and$-4\n * Factored result:\n            z^2 - 16z + 48 = (z - 12)(z - 4)\n * **(e)** y^2 + 5y - 6\n * Find factors of -6$that sum to$5$:$6$and$-1\n * Factored result:\n            y^2 + 5y - 6 = (y + 6)(y - 1)\n * **(f)** y^2 - 5y + 6\n * Find factors of 6$that sum to$-5$:$-3$and$-2\n * Factored result:\n            y^2 - 5y + 6 = (y - 3)(y - 2)\n * **(g)** 3x^2 - 18x + 27\n * Factor out the GCF 3 first:\n            3x^2 - 18x + 27 = 3(x^2 - 6x + 9)\n * Recognize x^2 - 6x + 9 as a perfect square trinomial:\n            3x^2 - 18x + 27 = 3(x - 3)^2\n\n* **Problem 4: Use grouping to fully factor**\n * **(a)** 4x^3 + 10x^2 - 6x - 15\n * Group terms into pairs:\n            4x^3 + 10x^2 - 6x - 15 = 2x^2(2x + 5) - 3(2x + 5)\n * Factor out common binomial (2x + 5):\n            4x^3 + 10x^2 - 6x - 15 = (2x^2 - 3)(2x + 5)\n * **(b)** 12n^5 - 30n^4 + 6n^3 - 15n^2\n * Factor out the GCF 3n^2 first:\n            12n^5 - 30n^4 + 6n^3 - 15n^2 = 3n^2(4n^3 - 10n^2 + 2n - 5)\n * Group terms inside the brackets:\n            4n^3 - 10n^2 + 2n - 5 = 2n^2(2n - 5) + 1(2n - 5)\n * Factor out common binomial (2n - 5):\n            12n^5 - 30n^4 + 6n^3 - 15n^2 = 3n^2(2n^2 + 1)(2n - 5)\n * **(c)** y^3 - 3y^2 - 9y + 27\n * Group terms into pairs:\n            y^3 - 3y^2 - 9y + 27 = y^2(y - 3) - 9(y - 3)\n * Factor out (y - 3):\n            y^3 - 3y^2 - 9y + 27 = (y^2 - 9)(y - 3)\n * Factor the difference of squares y^2 - 9:\n            y^3 - 3y^2 - 9y + 27 = (y - 3)(y + 3)(y - 3) = (y - 3)^2(y + 3)\n\n* **Problem 5: Use the ac-method to fully factor**\n * **(a)** 3x^2 - 13x - 30\n * Compute a \cdot c = 3 \times (-30) = -90\n * Find factors of -90$that sum to$-13$:$-18$and$5\n * Rewrite middle term and group:\n            3x^2 - 18x + 5x - 30 = 3x(x - 6) + 5(x - 6)\n * Factored result:\n            3x^2 - 13x - 30 = (3x + 5)(x - 6)\n * **(b)** 4y^2 + 16y + 15\n * Compute a \cdot c = 4 \times 15 = 60\n * Find factors of 60$that sum to$16$:$10$and$6\n * Rewrite middle term and group:\n            4y^2 + 10y + 6y + 15 = 2y(2y + 5) + 3(2y + 5)\n * Factored result:\n            4y^2 + 16y + 15 = (2y + 3)(2y + 5)\n * **(c)** 6z^2 - 13z + 6\n * Compute a \cdot c = 6 \times 6 = 36\n * Find factors of 36$that sum to$-13$:$-9$and$-4\n * Rewrite middle term and group:\n            6z^2 - 9z - 4z + 6 = 3z(2z - 3) - 2(2z - 3)\n * Factored result:\n            6z^2 - 13z + 6 = (3z - 2)(2z - 3)\n * **(d)** -10x^2 - 11x + 6\n * Factor out -1 to make the leading coefficient positive:\n            -10x^2 - 11x + 6 = -(10x^2 + 11x - 6)\n * Compute a \cdot c = 10 \times (-6) = -60\n * Find factors of -60$that sum to$11$:$15$and$-4\n * Rewrite middle term and group:\n            10x^2 + 15x - 4x - 6 = 5x(2x + 3) - 2(2x + 3) = (5x - 2)(2x + 3)\n * Fully factored result:\n            -10x^2 - 11x + 6 = -(5x - 2)(2x + 3)\n\n# Fundamental Principles of Solving Algebraic Equations\n\n* **Square Root Property:**\n * For any algebraic expression u$and real number$d:\n        u^2 = d \implies u = \pm \sqrt{d}\n * If d > 0, there are two real solutions:\n        u = \sqrt{d} \quad \text{or} \quad u = -\sqrt{d}\n * If d = 0, there is one repeated real solution:\n        u = 0\n * If d < 0$, there are two complex conjugate solutions involving the imaginary unit$i = \sqrt{-1}:\n        u = \pm i\sqrt{|d|}\n\n* **Zero Product Property:**\n * If the product of two or more factors is zero, then at least one of the individual factors must equal zero:\n        A \cdot B = 0 \implies A = 0 \quad \text{or} \quad B = 0\n * This property extends to any number of linear or non-linear factors:\n        A_1 \cdot A_2 \cdot \dots \cdot A_n = 0 \implies A_i = 0 \quad \text{for some } i \in {1, 2, \dots, n}\n\n# Section A.2 Solving Review - Turn-In Homework Solutions\n\n* **Problem 1:** Solve (x - 2)^2 = -45 using the square root property.\n * Apply the square root property to isolate x - 2:\n        x - 2 = \pm \sqrt{-45}\n * Simplify the imaginary radical:\n        \sqrt{-45} = \sqrt{-1 \times 9 \times 5} = 3i\sqrt{5}\n * Add 2 to both sides:\n        x = 2 \pm 3i\sqrt{5}\n\n* **Problem 2:** Solve 5(x - 2)^2 - 45 = 0 using the square root property.\n * Isolate the squared term:\n        5(x - 2)^2 = 45\n        (x - 2)^2 = 9\n * Apply the square root property:\n        x - 2 = \pm \sqrt{9} = \pm 3\n * Solve for x in both cases:\n        x = 2 + 3 = 5 \quad \text{or} \quad x = 2 - 3 = -1\n * Solution set:\n        x \in {-1, 5}\n\n* **Problem 3:** Solve x(x + 3)(7x + 9) = 0 using the zero product property.\n * Set each factor equal to zero:\n        x = 0\n        x + 3 = 0 \implies x = -3\n        7x + 9 = 0 \implies 7x = -9 \implies x = -\frac{9}{7}\n * Solution set:\n        x \in \left{-3, -\frac{9}{7}, 0\right}\n\n* **Problem 4:** Solve y^2 - 14y + 40 = 0 using factoring and the zero product property.\n * Factor the quadratic trinomial by finding two numbers multiplying to 40$and summing to$-14$($-10$and$-4):\n        (y - 10)(y - 4) = 0\n * Set each factor to zero:\n        y - 10 = 0 \implies y = 10\n        y - 4 = 0 \implies y = 4\n * Solution set:\n        y \in {4, 10}\n\n* **Problem 5:** Solve 6z^2 + 7z - 5 = 0 using factoring and the zero product property.\n * Use the AC-method to factor 6z^2 + 7z - 5:\n        a \cdot c = 6 \times (-5) = -30\n * Factors of -30$summing to$7$are$10$and$-3:\n        6z^2 + 10z - 3z - 5 = 0\n        2z(3z + 5) - 1(3z + 5) = 0\n        (2z - 1)(3z + 5) = 0\n * Apply the zero product property:\n        2z - 1 = 0 \implies 2z = 1 \implies z = \frac{1}{2}\n        3z + 5 = 0 \implies 3z = -5 \implies z = -\frac{5}{3}\n * Solution set:\n        z \in \left{-\frac{5}{3}, \frac{1}{2}\right}\n\n# Section A.2 Solving Review - Practice Homework Solutions\n\n* **Problem 1:** Solve a^2 = -49 using the square root property.\n * Apply the square root property:\n        a = \pm \sqrt{-49}\n * Simplify using the imaginary unit i:\n        a = \pm 7i\n\n* **Problem 2:** Solve (x + 7)^2 = 25 using the square root property.\n * Apply the square root property:\n        x + 7 = \pm \sqrt{25} = \pm 5\n * Solve for x:\n        x = -7 + 5 = -2 \quad \text{or} \quad x = -7 - 5 = -12\n * Solution set:\n        x \in {-12, -2}\n\n* **Problem 3:** Solve 3(x + 4)^2 + 36 = 0 using the square root property.\n * Isolate the squared binomial term:\n        3(x + 4)^2 = -36\n        (x + 4)^2 = -12\n * Apply the square root property:\n        x + 4 = \pm \sqrt{-12}\n * Simplify the complex radical:\n        \sqrt{-12} = \sqrt{-1 \times 4 \times 3} = 2i\sqrt{3}\n * Subtract 4 from both sides:\n        x = -4 \pm 2i\sqrt{3}\n\n* **Problem 4:** Solve (a + 4)(5a + 3)(a - 4)(a - 9) = 0 using the zero product property.\n * Set each linear factor equal to zero:\n        a + 4 = 0 \implies a = -4\n        5a + 3 = 0 \implies 5a = -3 \implies a = -\frac{3}{5}\n        a - 4 = 0 \implies a = 4\n        a - 9 = 0 \implies a = 9\n * Solution set:\n        a \in \left{-4, -\frac{3}{5}, 4, 9\right}\n\n* **Problem 5:** Solve b^3 + 8b^2 - 20b = 0 using factoring and the zero product property.\n * Factor out the GCF b:\n        b(b^2 + 8b - 20) = 0\n * Factor the quadratic expression b^2 + 8b - 20$into$(b + 10)(b - 2):\n        b(b + 10)(b - 2) = 0\n * Set each factor equal to zero:\n        b = 0\n        b + 10 = 0 \implies b = -10\n        b - 2 = 0 \implies b = 2\n * Solution set:\n        b \in {-10, 0, 2}\n\n* **Problem 6:** Solve 3x^2 + 28x + 9 = 0 using factoring and the zero product property.\n * Factor using the AC-method where a \cdot c = 3 \times 9 = 27$and the sum is$28$($27$and$1):\n        3x^2 + 27x + x + 9 = 0\n        3x(x + 9) + 1(x + 9) = 0\n        (3x + 1)(x + 9) = 0\n * Set each factor to zero:\n        3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}\n        x + 9 = 0 \implies x = -9\n * Solution set:\n        x \in \left{-9, -\frac{1}{3}\right}\n\n* **Problem 7:** Solve c^4 - 81 = 0 using factoring and the zero product property.\n * Factor as a difference of squares:\n        (c^2 - 9)(c^2 + 9) = 0\n * Factor c^2 - 9 further as a difference of squares:\n        (c - 3)(c + 3)(c^2 + 9) = 0\n * Set each factor to zero to solve:\n        c - 3 = 0 \implies c = 3\n        c + 3 = 0 \implies c = -3\n        c^2 + 9 = 0 \implies c^2 = -9 \implies c = \pm 3i\n * Solution set:\n        c \in {-3, 3, -3i, 3i}$$