Factor completely 2x3 + 6x2 + 10x + 30.

a. 2(x3 + 3x2 + 5x + 15)
b. 2[(x2 + 5)(x + 3)]
c. (2x2 + 10)(x + 3)
d. (x2 + 5)(2x + 6)

  B.  2[(x² + 5)(x + 3)]

Step-by-step explanation:

All coefficients are even, so a factor of 2 can be removed. Then we notice that pairs of terms have the same ratio of coefficients, so we can factor by grouping.

  2(x³ +3x² +5x +15) = 2(x²(x +3) +5(x +3)) = 2(x² +5)(x +3)

55

step-by-step explanation:

yes

step-by-step explanation: