Question

find the product.$(3a^{4} + 4)^{2}LXB0\bigcirc$ $9a^{8} + 16LXB1\bigcirc$ $9a^{16} + 24a^{4} + 16$done

Explanation:

Step1: Apply square of sum formula

Recall the formula $(x+y)^2 = x^2 + 2xy + y^2$, let $x=3a^4$, $y=4$.

Step2: Calculate $x^2$

$x^2 = (3a^4)^2 = 3^2 \cdot (a^4)^2 = 9a^8$

Step3: Calculate $2xy$

$2xy = 2 \cdot 3a^4 \cdot 4 = 24a^4$

Step4: Calculate $y^2$

$y^2 = 4^2 = 16$

Step5: Sum the three terms

Add the results from Steps 2-4: $9a^8 + 24a^4 + 16$

Answer:

$\boldsymbol{9a^8 + 24a^4 + 16}$ (the third option)