Factor completely.
$2x^2+9x+9$

• A. $\(\left\)(2x+3\(\right\))\(\left\)(x+3\(\right\))$
• B. $\(\left\)(2x+3\(\right\))\(\left\)(x-3\(\right\))$
• C. $\(\left\)(x+3\(\right\))\(\left\)(2x-3\(\right\))$
• D. $\(\left\)(x-3\(\right\))\(\left\)(2x-3\(\right\))$

Answer: (A)

Verified step by step guidance

1. Identify the quadratic expression to factor: \(2x^2 + 9x + 9\).
2. Multiply the coefficient of \(x^2\) (which is 2) by the constant term (which is 9) to get \(2 \times 9 = 18\).
3. Find two numbers that multiply to 18 and add up to the middle coefficient, 9. These numbers are 6 and 3.
4. Rewrite the middle term \$9x$ as \(6x + 3x\), so the expression becomes \(2x^2 + 6x + 3x + 9\).
5. Group the terms in pairs and factor each group: \(2x^2 + 6x\) factors to \(2x(x + 3)\), and \(3x + 9\) factors to \(3(x + 3)\). Then factor out the common binomial \((x + 3)\) to get \((2x + 3)(x + 3)\).