Question
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6. Factor completely: \( x^{2}-x-6 \) ? A. \( (x-2)(x+3) \) B. \( (x-1)(x-6) \) C. \( (x+2)(x-3) \) D. \( (x+1)(x-6) \)

Ask by Christensen Marsh. in the United States
Feb 19,2025

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Answer

The answer is \( (x+2)(x-3) \), choice C.

Solution

Factor the expression \( x^{2}-x-6 \). Factor the expression by following steps: - step0: Factor: \(x^{2}-x-6\) - step1: Rewrite the expression: \(x^{2}+\left(2-3\right)x-6\) - step2: Calculate: \(x^{2}+2x-3x-6\) - step3: Rewrite the expression: \(x\times x+x\times 2-3x-3\times 2\) - step4: Factor the expression: \(x\left(x+2\right)-3\left(x+2\right)\) - step5: Factor the expression: \(\left(x-3\right)\left(x+2\right)\) The factored form of \(x^{2}-x-6\) is \((x-3)(x+2)\). Therefore, the correct answer is C. \( (x+2)(x-3) \).

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To factor the quadratic expression \( x^{2}-x-6 \), we are looking for two numbers that multiply to -6 (the constant term) and add to -1 (the coefficient of the linear term). Those numbers are -3 and 2. Thus, we can express the quadratic as \( (x-3)(x+2) \). Now, let's look at its historical context. Quadratic equations date back to ancient civilizations, including the Babylonians, who employed geometrical methods to solve them long before the algebraic approach emerged. The quadratic formula we know and love today was developed over centuries, with contributions from various cultures, culminating in the more generalized algebraic manipulations we use now! For real-world applications, consider how factoring quadratic equations can help in optimizing various scenarios. For instance, if you're an architect designing a parabolic arch, being able to factor equations helps quickly determine points of intersections and critical dimensions to ensure stability and aesthetic appeal in your designs!

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