Question

what is a difference of squares that has a factor of x+8?
○ x²-4
○ x²-16
○ x²-64
○ x²-256

Explanation:

Step1: Recall difference of squares formula

The difference of squares formula is \(a^2 - b^2=(a + b)(a - b)\). We need a polynomial that factors into \((x + 8)(x - 8)\) (since one factor is \(x + 8\)), so the polynomial should be \(x^2-8^2=x^2 - 64\).

Step2: Check each option

• For \(x^2-4\), factor it: \(x^2 - 2^2=(x + 2)(x - 2)\), no \(x + 8\) factor.
• For \(x^2-16\), factor it: \(x^2 - 4^2=(x + 4)(x - 4)\), no \(x + 8\) factor.
• For \(x^2-64\), factor it: \(x^2 - 8^2=(x + 8)(x - 8)\), has \(x + 8\) factor.
• For \(x^2-256\), factor it: \(x^2 - 16^2=(x + 16)(x - 16)\), no \(x + 8\) factor.

Answer:

\(x^2 - 64\) (the option with \(x^2-64\))