Solve each equation in Exercises 1 - 14 by factoring. 10x - 1 = (2x + 1)²

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Rewrite the equation so that all terms are on one side, setting the equation equal to 0: \( 10x - 1 - (2x + 1)^2 = 0 \).

Expand the squared term \( (2x + 1)^2 \) using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \). This gives \( (2x + 1)^2 = 4x^2 + 4x + 1 \). Substitute this back into the equation.

Simplify the equation by combining like terms: \( 10x - 1 - 4x^2 - 4x - 1 = 0 \). Rearrange the terms to form a standard quadratic equation \( -4x^2 + 6x - 2 = 0 \).

Factor out \(-2\) from the entire equation to simplify: \( -2(2x^2 - 3x + 1) = 0 \).

Factor the quadratic expression \( 2x^2 - 3x + 1 \) into two binomials. Look for two numbers that multiply to \( 2 \times 1 = 2 \) and add to \(-3\). These numbers are \(-2\) and \(-1\). The factored form is \( (2x - 1)(x - 1) = 0 \). Solve each factor for \( x \) to find the solutions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Factoring

Factoring is the process of breaking down an expression into a product of simpler expressions, or factors, that when multiplied together yield the original expression. This technique is essential in solving equations, as it allows us to set each factor equal to zero to find the solutions. Understanding how to factor polynomials, especially quadratics, is crucial for solving equations like the one presented.

Quadratic Equations

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants, and a ≠ 0. In the context of the given equation, recognizing that the right side is a perfect square trinomial helps in simplifying the equation. Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula.

Zero Product Property

The Zero Product Property states that if the product of two or more factors equals zero, at least one of the factors must be zero. This principle is fundamental when solving factored equations, as it allows us to find the values of the variable that satisfy the equation. Applying this property to the factors obtained from the original equation is key to finding the solutions.

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