Factor completely 16a^3b^7 + 2a^6b^4 - 22a^4b^5
Question 2 options:
1) 2(8a^3b^7 + a^6b^4 - 11a^4b^5)
2) 2a^3b^4(8b^3 + a^3 - 11ab)
3) a^3b^4(16b^3 + 2a^3 - 22ab)
4) 8b^3 + a^3 - 11ab

Question

Factor completely 16a^3b^7 + 2a^6b^4 - 22a^4b^5
Question 2 options:
	1)	2(8a^3b^7 + a^6b^4 - 11a^4b^5)
	2)	2a^3b^4(8b^3 + a^3 - 11ab)
	3)	a^3b^4(16b^3 + 2a^3 - 22ab)
	4)	8b^3 + a^3 - 11ab

anonymous · Answer

Solution 2

anonymous · Answer

2 is the largest number to go into all 3 so it is choice 1 or 2
they all have a's and b's so 2 is the likely answer

anonymous · Answer

The problem statement is factor completely:$$2 a^6 b^4-22 a^4 b^5+16 a^3 b^7=2 a^3 b^4 \left(a^3-11 a b+8 b^3\right)  $$