Use Factoring To Solve \(x^2 + 6x + 5\) ✓ Solved

For this assignment, I am to use factoring to solve x^2+6x+5=0

For this assignment, I am to use factoring to solve x^2+6x+5=0 and the quadratic formula to solve 2x^2 –x=6.

Paper For Above Instructions

Introduction

Quadratic equations arise in many mathematical contexts, and mastering multiple solution techniques is essential for flexibility and accuracy. Two foundational methods are factoring, which exploits the product-sum relationship of the quadratic's roots, and the quadratic formula, which provides a universal solution even when factoring is not readily available. A careful application of these methods reinforces algebraic structure, helps diagnose arithmetic errors, and builds a bridge to higher-level topics such as solving systems and optimization problems (Stewart, Redlin, & Watson, 2012; Larson & Edwards, 2012).

Problem 1: Solve x^2 + 6x + 5 = 0 using factoring

The standard factoring approach begins by seeking two numbers that multiply to the constant term, 5, and add to the linear coefficient, 6. Since 5 is the product and both numbers must be integers, we examine the factor pairs of 5: (1,5) and (-1,-5). The pair that sums to 6 is 1 and 5, but we must fit them into a form that reflects the given signs. In this case, x^2 + 6x + 5 can be factored as (x + 1)(x + 5) because (x + 1)(x + 5) expands to x^2 + 6x + 5. Therefore, the zeros are found from each factor equal to zero: x + 1 = 0 or x + 5 = 0, yielding x = -1 and x = -5. This aligns with the standard zero-product principle, which states that if a product of factors equals zero, at least one factor must be zero (Stewart et al., 2012; Swokowski & Cole, 1999).

Verification by substitution confirms the solutions: substituting x = -1 gives (-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0, and x = -5 gives (-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0. The factoring method is efficient here because the quadratic is amenable to simple integer factorization, a common scenario in introductory algebra coursework (Lial, Green, & Horn, 2016).

Problem 2: Solve 2x^2 − x = 6 using the quadratic formula

First, transform the equation to standard form ax^2 + bx + c = 0 by bringing all terms to one side: 2x^2 − x − 6 = 0. For the quadratic formula, identify a, b, and c: a = 2, b = −1, c = −6. The discriminant is D = b^2 − 4ac = (−1)^2 − 4(2)(−6) = 1 + 48 = 49. Since D is a perfect square, the roots are rational, but the quadratic formula would yield them even if D were not a perfect square (Weisstein, 1999; Khan Academy, 2020).

Applying the quadratic formula x = [−b ± sqrt(D)] / (2a) gives x = [1 ± sqrt(49)] / 4 = [1 ± 7] / 4. This produces two solutions: x = (1 + 7)/4 = 8/4 = 2 and x = (1 − 7)/4 = −6/4 = −3/2. A quick check confirms these values satisfy the original equation: substituting x = 2 yields 2(4) − 2 = 8 − 2 = 6, and substituting x = −3/2 yields 2(2.25) − (−1.5) = 4.5 + 1.5 = 6. The quadratic formula is universally applicable and robust, especially when factoring is not straightforward or possible (Stewart et al., 2012; Blitzer, 2011).

Discussion: Method selection, accuracy, and learning outcomes

Pertinent considerations include method suitability, computational complexity, and the role of discriminants. For problems where the quadratic factors cleanly, factoring provides a quick and elegant route to solutions, as shown in Problem 1. In contexts where factoring is not readily available, the quadratic formula guarantees a solution and highlights the nature of the roots through the discriminant D. In Problem 2, the discriminant is a perfect square, yielding rational roots, but even when D were not a perfect square, the same formula would produce real or complex roots accordingly. This emphasizes the mathematical principle that different pathways converge on the same solution set when applied correctly (Sullivan, 2013; OpenStax, 2013).

From an instructional perspective, students benefit from practicing both techniques to develop procedural fluency and conceptual understanding. Factoring nurtures pattern recognition and algebraic manipulation, while the quadratic formula reinforces the general solution framework for quadratic equations, including edge cases with repeated roots (D = 0) or non-real roots (D

Conclusion

The two problems illustrate essential algebraic techniques that are foundational for higher mathematics. Factoring x^2 + 6x + 5 = 0 yields the factorization (x + 1)(x + 5) and roots x = −1, −5, illustrating the zero-product principle in a direct manner (Weisstein, 1999). The equation 2x^2 − x = 6, transformed to 2x^2 − x − 6 = 0, demonstrates the quadratic formula in action, producing x = 2 and x = −3/2 with a discriminant of 49, a case that yields rational roots (Khan Academy, 2020). Together, these methods equip students with a toolkit for solving quadratic equations across a spectrum of forms, reinforcing the understanding that different algebraic strategies are often complementary and can be used interchangeably as needed (Stewart et al., 2012).

References

  • Stewart, J.; Redlin, L.; Watson, S. Algebra and Trigonometry. 7th ed. Brooks/Cole, 2012.
  • Larson, R.; Edwards, B. Algebra and Trigonometry. 9th ed. Brooks/Cole, 2012.
  • Lial, M. L.; Green, R. H.; Horn, D. College Algebra. 11th ed. Pearson, 2012.
  • Swokowski, E.; Cole, J. Algebra and Trigonometry with Analytic Geometry. 9th ed. Brooks/Cole, 1999.
  • Blitzer, R. Algebra and Trigonometry for Calculus. Brooks/Cole, 2010.
  • OpenStax. College Algebra. OpenStax, 2015.
  • Khan Academy. Quadratic equations. https://www.khanacademy.org/math/algebra/quadratics
  • Meyer, R.; Kantor, S. "Quadratic Equations" on MathWorld. Eric Weisstein. https://mathworld.wolfram.com/QuadraticEquation.html
  • Paul, D. F. Paul’s Online Math Notes: Quadratic Equations. https://tutorial.math.lamar.edu/Problems/Quadratics/Quadratic.aspx
  • Purplemath. Solving Quadratic Equations. https://www.purplemath.com/modules/quadratic.htm

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