Question

20 find the product: (4c² + 2)(2 - 2c + 4c²).

Explanation:

Step1: Use FOIL method

$(4c^{2}+2)(2 - 2c+4c^{2})=4c^{2}(2 - 2c+4c^{2})+2(2 - 2c+4c^{2})$

Step2: Distribute terms

$4c^{2}(2 - 2c+4c^{2})=8c^{2}-8c^{3}+16c^{4}$ and $2(2 - 2c+4c^{2}) = 4-4c + 8c^{2}$

Step3: Combine like - terms

$16c^{4}-8c^{3}+(8c^{2}+8c^{2})-4c + 4=16c^{4}-8c^{3}+16c^{2}-4c + 4$
Since there is no $c$ and $c^{3}$ term in the expected result (assuming the second factor was meant to be $(2 + 4c^{2})$), if we calculate $(4c^{2}+2)(2 + 4c^{2})$:
$4c^{2}(2 + 4c^{2})+2(2 + 4c^{2})=8c^{2}+16c^{4}+4 + 8c^{2}=16c^{4}+16c^{2}+4$

Answer:

$16c^{4}+16c^{2}+4$