Q. Solve for all values of \(x\) by factoring. \[ x^{2} + 5x + 1 = 5x + 2. \]

Problem

Solve for all values of \(x\) by factoring:

\[x^{2} + 5x + 1 = 5x + 2\]

1. Gather all terms on one side to form a quadratic equal to zero.We subtract the right-hand side from both sides to isolate zero on one side. Subtracting \(5x + 2\) from both sides gives:

   \[x^{2} + 5x + 1 – (5x + 2) = 0\]

   Explain what we did: subtracting the same expression from both sides preserves equality; this puts the equation in standard quadratic form.

2. Combine like terms.Distribute the subtraction and combine the \(5x\) terms and the constant terms:

   \[x^{2} + 5x + 1 – 5x – 2 = 0\]

   The \(+5x\) and \(-5x\) cancel, and \(+1 – 2 = -1, so we obtain:

   \[x^{2} – 1 = 0\]

   Now the equation is simplified to a difference of squares.

3. Factor the quadratic.Recognize that \(x^{2} – 1\) is a difference of squares because it has the form \(a^{2} – b^{2}\) with \(a = x\) and \(b = 1\). The factorization formula is \(a^{2} – b^{2} = (a – b)(a + b)\). Applying it gives:

   \[x^{2} – 1 = (x – 1)(x + 1)\]

4. Apply the zero-product property to find solutions.The zero-product property states that if a product of factors equals zero, then at least one factor must be zero. Set each factor equal to zero:

   \[x – 1 = 0 \quad\text{or}\quad x + 1 = 0\]

   Solving each equation yields:

   \[x = 1 \quad\text{or}\quad x = -1\]

5. Check the solutions in the original equation (optional but recommended).Original equation: \(x^{2} + 5x + 1 = 5x + 2\).

   For \(x = 1\): left side is \(1^{2} + 5(1) + 1 = 1 + 5 + 1 = 7\), right side is \(5(1) + 2 = 5 + 2 = 7\). Equality holds.

   For \(x = -1\): left side is \((-1)^{2} + 5(-1) + 1 = 1 – 5 + 1 = -3\), right side is \(5(-1) + 2 = -5 + 2 = -3\). Equality holds.

Final answer

The solutions are \(x = 1\) and \(x = -1\).