Question

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)

Answer 1

B

2[(x² + 5)(x + 3)]

Answer 2

B. 2[(x² + 5)(x + 3)]

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)