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ALG-3 Solutions: Factoring Techniques

Solutions Reference · ← Back to Lesson ALG-3

Section 5 — Guided Practice Solutions

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Problem 1 — Identify and Extract the GCF (C1)

Variant 0 — Factor \( 6x^3 + 9x^2 \)

GCF: \( \gcd(6,9) = 3 \); lowest power of \( x \) in both terms is \( x^2 \). So GCF \( = 3x^2 \).

Factored form: \( 3x^2(2x + 3) \).

Verify: \( 3x^2 \cdot 2x = 6x^3 \) ✓ and \( 3x^2 \cdot 3 = 9x^2 \) ✓

Variant 1 — Factor \( 8x^4 - 12x^2 \)

GCF: \( \gcd(8,12) = 4 \); lowest power of \( x \) is \( x^2 \). So GCF \( = 4x^2 \).

Factored form: \( 4x^2(2x^2 - 3) \).

Verify: \( 4x^2 \cdot 2x^2 = 8x^4 \) ✓ and \( 4x^2 \cdot 3 = 12x^2 \) ✓

Variant 2 — Factor \( 15x^3 - 10x \)

GCF: \( \gcd(15,10) = 5 \); lowest power of \( x \) is \( x^1 \) (from \( -10x \)). So GCF \( = 5x \).

Factored form: \( 5x(3x^2 - 2) \).

Verify: \( 5x \cdot 3x^2 = 15x^3 \) ✓ and \( 5x \cdot 2 = 10x \) ✓

Variant 3 — Factor \( 4x^3 + 6x^2 - 2x \)

GCF: \( \gcd(4,6,2) = 2 \); lowest power of \( x \) is \( x^1 \) (from \( -2x \)). So GCF \( = 2x \).

Factored form: \( 2x(2x^2 + 3x - 1) \).

Verify: \( 2x \cdot 2x^2 = 4x^3 \) ✓, \( 2x \cdot 3x = 6x^2 \) ✓, \( 2x \cdot 1 = 2x \) ✓

Note: \( 2x^2 + 3x - 1 \) does not factor over \( \mathbb{Z} \) (discriminant \( = 9 + 8 = 17 \)), so this is the final answer.

Variant 4 — Factor \( 12x^5 - 8x^3 + 4x^2 \)

GCF: \( \gcd(12,8,4) = 4 \); lowest power of \( x \) is \( x^2 \) (from \( 4x^2 \)). So GCF \( = 4x^2 \).

Factored form: \( 4x^2(3x^3 - 2x + 1) \).

Verify: \( 4x^2 \cdot 3x^3 = 12x^5 \) ✓, \( 4x^2 \cdot 2x = 8x^3 \) ✓, \( 4x^2 \cdot 1 = 4x^2 \) ✓

Problem 2 — Choose the Correct Technique (C1–C5)

Expression A: \( x^2 - 81 \) → Difference of Squares

\( x^2 - 81 = x^2 - 9^2 = (x-9)(x+9) \). Both terms are perfect squares with a minus sign between them.

Expression B: \( x^2 + 9x + 14 \) → Quadratic Trinomial (Product-Sum)

Find \( p, q \) with \( pq = 14 \) and \( p + q = 9 \). Try \( (2, 7) \): \( 2 \cdot 7 = 14 \) ✓, \( 2 + 7 = 9 \) ✓. Answer: \( (x+2)(x+7) \).

Expression C: \( x^3 - 3x^2 + 2x - 6 \) → Factoring by Grouping

Group: \( (x^3 - 3x^2) + (2x - 6) = x^2(x-3) + 2(x-3) = (x^2+2)(x-3) \).

Note: \( x^2 + 2 \) is a sum of squares — it cannot be factored further over \( \mathbb{R} \).

Expression D: \( 9x^2 + 16 \) → Cannot be factored over ℝ

This is a sum of squares \( (3x)^2 + 4^2 \). The sum of squares \( a^2 + b^2 \) has no real factorization. (Factoring would require the complex number \( i \).)

Problem 3 — Factor a Difference of Squares (C3)

Variant 0 — Factor \( x^2 - 9 \)

\( x^2 - 9 = x^2 - 3^2 = (x-3)(x+3) \)

Variant 1 — Factor \( 4x^2 - 25 \)

\( 4x^2 - 25 = (2x)^2 - 5^2 = (2x-5)(2x+5) \)

Variant 2 — Factor \( x^4 - 1 \) completely

First application: \( x^4 - 1 = (x^2)^2 - 1^2 = (x^2-1)(x^2+1) \)

Second application on \( x^2 - 1 \): \( x^2 - 1 = (x-1)(x+1) \)

\( x^2 + 1 \) is a sum of squares — does not factor over \( \mathbb{R} \).

Final answer: \( (x-1)(x+1)(x^2+1) \)

Variant 3 — Factor \( 9x^2 - 4y^2 \)

\( 9x^2 - 4y^2 = (3x)^2 - (2y)^2 = (3x-2y)(3x+2y) \)

Variant 4 — Factor \( x^2 - \tfrac14 \)

\( x^2 - \tfrac{1}{4} = x^2 - \left(\tfrac{1}{2}\right)^2 = \left(x - \tfrac{1}{2}\right)\left(x + \tfrac{1}{2}\right) \)

Problem 4 — Factor \( 6x^2 - 7x - 3 \) Step by Step (C4)

Step 1 — Compute \( ac \): \( a = 6 \), \( c = -3 \), so \( ac = 6 \cdot (-3) = -18 \).

Step 2 — Find \( p, q \) with \( pq = -18 \) and \( p + q = -7 \):

Systematically check pairs: \( (1,-18): 1+(-18)=-17 \) ✗; \( (-1,18): -1+18=17 \) ✗; \( (2,-9): 2+(-9)=-7 \) ✓. Use \( p=2 \), \( q=-9 \).

Step 3 — Split the middle term and factor by grouping:

\[ 6x^2 + 2x - 9x - 3 = 2x(3x+1) - 3(3x+1) = (2x-3)(3x+1) \]

Check: \( (2x-3)(3x+1) = 6x^2 + 2x - 9x - 3 = 6x^2 - 7x - 3 \) ✓

Section 6 — Independent Practice Solutions

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Problems 1 & 2 — Generator Problems (C1, C4)

These problems generate with random parameters. The worked examples below illustrate the method for typical instances.

Problem 1 — GCF Extraction method: To factor \( g \cdot x^{n+1} \pm b \cdot x^n \) where \( b = g \cdot k \):

  1. Identify the numeric GCF of the coefficients.
  2. Identify the lowest power of \( x \) present in all terms.
  3. Factor out their product and divide each term.

Example: \( 3x^4 + 12x^3 \). GCF of coefficients: \( \gcd(3,12) = 3 \). Lowest power of \( x \): \( x^3 \). Factor out \( 3x^3 \): \( 3x^3(x + 4) \).

Problem 2 — Monic Quadratic Trinomial method: To factor \( x^2 + bx + c \):

  1. Find integers \( p, q \) such that \( pq = c \) AND \( p + q = b \).
  2. Write \( (x + p)(x + q) \).

Example: \( x^2 - 2x - 15 \). Need \( pq = -15 \), \( p + q = -2 \). Try \( (3, -5) \): \( 3 \cdot (-5) = -15 \) ✓, \( 3 + (-5) = -2 \) ✓. Answer: \( (x+3)(x-5) \).

Problem 3 — Two-Step: GCF then Difference of Squares (C1, C3)

Variant 0 — Factor \( 3x^2 - 75 \) completely

Step 1 (GCF): \( 3x^2 - 75 = 3(x^2 - 25) \)

Step 2 (DoS): \( x^2 - 25 = (x-5)(x+5) \)

Answer: \( 3(x-5)(x+5) \)

Variant 1 — Factor \( 5x^2 - 20 \) completely

Step 1 (GCF): \( 5x^2 - 20 = 5(x^2 - 4) \)

Step 2 (DoS): \( x^2 - 4 = (x-2)(x+2) \)

Answer: \( 5(x-2)(x+2) \)

Variant 2 — Factor \( 2x^4 - 2x^2 \) completely

Step 1 (GCF): \( 2x^4 - 2x^2 = 2x^2(x^2 - 1) \)

Step 2 (DoS): \( x^2 - 1 = (x-1)(x+1) \)

Answer: \( 2x^2(x-1)(x+1) \)

Variant 3 — Factor \( 4x^3 - 4x \) completely

Step 1 (GCF): \( 4x^3 - 4x = 4x(x^2 - 1) \)

Step 2 (DoS): \( x^2 - 1 = (x-1)(x+1) \)

Answer: \( 4x(x-1)(x+1) \)

Variant 4 — Factor \( 7x^2 - 63 \) completely

Step 1 (GCF): \( 7x^2 - 63 = 7(x^2 - 9) \)

Step 2 (DoS): \( x^2 - 9 = (x-3)(x+3) \)

Answer: \( 7(x-3)(x+3) \)

Problem 4 — Factor a Cubic Using the Factor Theorem (C5)

Variant 0 — Factor \( x^3 - 6x^2 + 11x - 6 \) completely

Test \( x = 1 \): \( 1 - 6 + 11 - 6 = 0 \) ✓ → \( (x-1) \) is a factor.

Synthetic division:

 1 | 1  -6   11  -6
   |     1   -5   6
   ─────────────────
   1  -5    6   0

Quotient: \( x^2 - 5x + 6 \). Product-Sum: \( (-2)(-3) = 6 \), \( (-2)+(-3) = -5 \) → \( (x-2)(x-3) \).

Answer: \( (x-1)(x-2)(x-3) \)

Variant 1 — Factor \( x^3 + 2x^2 - x - 2 \) completely

Test \( x = 1 \): \( 1 + 2 - 1 - 2 = 0 \) ✓ → \( (x-1) \) is a factor.

Synthetic division:

 1 | 1   2  -1  -2
   |     1   3   2
   ─────────────────
   1   3   2   0

Quotient: \( x^2 + 3x + 2 \). Product-Sum: \( (1)(2) = 2 \), \( 1+2 = 3 \) → \( (x+1)(x+2) \).

Answer: \( (x-1)(x+1)(x+2) \)

Variant 2 — Factor \( x^3 - x^2 - 4x + 4 \) completely

Test \( x = 1 \): \( 1 - 1 - 4 + 4 = 0 \) ✓ → \( (x-1) \) is a factor.

Synthetic division:

 1 | 1  -1  -4   4
   |     1   0  -4
   ─────────────────
   1   0  -4   0

Quotient: \( x^2 - 4 = (x-2)(x+2) \) (difference of squares).

Answer: \( (x-1)(x-2)(x+2) \)

Variant 3 — Factor \( x^3 + x^2 - 4x - 4 \) completely

Test \( x = -1 \): \( -1 + 1 + 4 - 4 = 0 \) ✓ → \( (x+1) \) is a factor.

Synthetic division:

-1 | 1   1  -4  -4
   |    -1   0   4
   ─────────────────
   1   0  -4   0

Quotient: \( x^2 - 4 = (x-2)(x+2) \).

Answer: \( (x+1)(x-2)(x+2) \)

Variant 4 — Factor \( x^3 - 3x^2 - x + 3 \) completely

Test \( x = 1 \): \( 1 - 3 - 1 + 3 = 0 \) ✓ → \( (x-1) \) is a factor.

Synthetic division:

 1 | 1  -3  -1   3
   |     1  -2  -3
   ─────────────────
   1  -2  -3   0

Quotient: \( x^2 - 2x - 3 \). Product-Sum: \( (-3)(1) = -3 \), \( (-3)+1 = -2 \) → \( (x-3)(x+1) \).

Answer: \( (x-1)(x-3)(x+1) \)

Problem 5 — Fractional Exponent Factoring (C6)

Factor \( 5x^{3/2} + 10x^{1/2} - 5x^{-1/2} \) completely.

Step 1 — Identify the smallest exponent: The exponents are \( \tfrac{3}{2}, \tfrac{1}{2}, -\tfrac{1}{2} \). Smallest = \( -\tfrac{1}{2} \).

Step 2 — Identify the numeric GCF: \( \gcd(5, 10, 5) = 5 \).

Step 3 — Factor out \( 5x^{-1/2} \):

\[5x^{-1/2} \cdot x^{3/2 - (-1/2)} = 5x^{-1/2} \cdot x^2 \quad \Rightarrow \quad 5x^{3/2} \checkmark\]

\[5x^{-1/2} \cdot 2x^{1/2 - (-1/2)} = 5x^{-1/2} \cdot 2x^1 \quad \Rightarrow \quad 10x^{1/2} \checkmark\]

\[5x^{-1/2} \cdot (-1) \cdot x^0 \quad \Rightarrow \quad -5x^{-1/2} \checkmark\]

Result: \( 5x^{-1/2}(x^2 + 2x - 1) \)

Can \( x^2 + 2x - 1 \) be factored further? Discriminant \( = 4 + 4 = 8 \) — not a perfect square, so it doesn't factor over \( \mathbb{Z} \). Final answer: \( 5x^{-1/2}(x^2 + 2x - 1) \), or equivalently \( \dfrac{5(x^2 + 2x - 1)}{\sqrt{x}} \).

Common mistake check: A student might factor out \( 5x^{1/2} \) (positive exponent) instead of \( 5x^{-1/2} \). The result would be \( 5x^{1/2}(x + 2 - x^{-1}) \), which still contains a negative exponent inside — messier, not simpler. Always factor out the most negative (smallest) exponent.