Question

Factor completely 16a3b7 + 2a6b4 – 22a4b5

Answer 1

\[Factor completely 16a ^{3}b ^{7} + 2a ^{6} b ^{4} – 22a ^{4} b ^{5}\]

Answer 2

\[16 a^3b^7 = 2 \times2\times2\times2 \times a \times a \times a \times b \times b \times b \times b \times b \times b \times b\]\[2a^6b^4 = 2 \times a \times a \times a \times a \times a \times a \times b \times b \times b \times b\]\[22a^4b^5 = 11 \times 2 \times a \times a \times a \times a \times b \times b \times b \times b \times b\]
which one are common? O.o

Answer 3

how many 2, how many a, how many b?

Answer 4

the answer is
8b^3+b-11ab^2

Answer 5

16a^3b^7+2a^6b^4–22a^4b^5 not equal 8b^3+b-11ab^2 lol

and it's wrong too :c

\[2a^3b^4(8b^7 + a^3b-11ab\]common factor

Answer 6

\[2a^3b^4(8b^3+a^3b−11ab)\]^^^^^^^^^^^^^

Answer 7

Oh that does make more sense thanks @nphuongsun93