Question

match the given trinomial to its equivalent form: $x^2 + 6x + 5$ $x^2 - 8x + 7$ $x^2 - 4x - 5$ $x^2 + 6x - 7$ $(x - 5)(x - 1)$ $(x - 5)(x + 1)$ $(x - 7)(x - 1)$ $(x + 5)(x - 1)$ $(x + 7)(x - 1)$ $(x + 7)(x + 1)$ $(x + 5)(x + 1)$ $(x - 7)(x + 1)$

Explanation:

Step1: Factor \(x^2 + 6x + 5\)

We need two numbers that multiply to \(5\) and add to \(6\). The numbers are \(5\) and \(1\). So, \(x^2 + 6x + 5=(x + 5)(x + 1)\).

Step2: Factor \(x^2 - 8x + 7\)

We need two numbers that multiply to \(7\) and add to \(-8\). The numbers are \(-7\) and \(-1\). So, \(x^2 - 8x + 7=(x - 7)(x - 1)\).

Step3: Factor \(x^2 - 4x - 5\)

We need two numbers that multiply to \(-5\) and add to \(-4\). The numbers are \(-5\) and \(1\). So, \(x^2 - 4x - 5=(x - 5)(x + 1)\).

Step4: Factor \(x^2 + 6x - 7\)

We need two numbers that multiply to \(-7\) and add to \(6\). The numbers are \(7\) and \(-1\). So, \(x^2 + 6x - 7=(x + 7)(x - 1)\).

Answer:

• \(x^2 + 6x + 5\) matches \((x + 5)(x + 1)\)
• \(x^2 - 8x + 7\) matches \((x - 7)(x - 1)\)
• \(x^2 - 4x - 5\) matches \((x - 5)(x + 1)\)
• \(x^2 + 6x - 7\) matches \((x + 7)(x - 1)\)