Question

what is the completely factored form of the expression $16x^2 + 8x + 32$?
$\bigcirc\\ 4(4x^2 + 2x + 8)$
$\bigcirc\\ 4(12x^2 + 4x + 28)$
$\bigcirc\\ 8(2x^2 + x + 4)$
$\bigcirc\\ 8x(8x^2 + x + 24)$

Explanation:

Step1: Find GCD of coefficients

Find GCD of 16, 8, 32. Factors of 16: 1,2,4,8,16; 8:1,2,4,8; 32:1,2,4,8,16,32. GCD is 8.

Step2: Divide each term by GCD

Divide \(16x^2\) by 8: \( \frac{16x^2}{8} = 2x^2 \); \(8x\) by 8: \( \frac{8x}{8} = x \); 32 by 8: \( \frac{32}{8} = 4 \).

Step3: Factor out GCD

Factor out 8: \(8(2x^2 + x + 4)\). Check other options: 4(4x²+2x+8) has GCD 4 (not max), 4(12x²+4x+28) is wrong (16/4=4≠12), 8x(...) introduces extra x.

Answer:

8(2x² + x + 4) (the option with this expression, which is the third option in the list: 8(2x² + x + 4))