How do you factor completely #50a^2b^5 – 35a^4b^3 + 5a^3b^4#?

1 Answer
Jun 16, 2016

Answer:

#50a^2b^5-35a^4b^3+5a^3b^4#

#=5a^2b^3(10b^2-7a^2+ab)#

#=50a^2b^3(b+(1-sqrt(281))/20a)(b+(1+sqrt(281))/20a)#

Explanation:

Note that all three terms are divisible by #5#, #a^2# and #b^3#, so therefore by #5a^2b^3#.

Separate that out as a factor:

#50a^2b^5-35a^4b^3+5a^3b^4=5a^2b^3(10b^2-7a^2+ab)#

Note that the remaining term is an homogeneous quadratic which we can factor by completing the square:

#10b^2+ab-7a^2#

#=10(b^2+2*1/20ab+1/20^2a^2-1/20^2a^2-7/10a^2)#

#=10((b+1/20a)^2-1/400a^2-280/400a^2)#

#=10((b+1/20a)^2-281/20^2a^2)#

#=10((b+1/20a)^2-(sqrt(281)/20a)^2)#

#=10((b+1/20a)-(sqrt(281)/20a))((b+1/20a)+(sqrt(281)/20a))#

#=10(b+(1-sqrt(281))/20a)(b+(1+sqrt(281))/20a)#

Putting it all together:

#50a^2b^5-35a^4b^3+5a^3b^4#

#=50a^2b^3(b+(1-sqrt(281))/20a)(b+(1+sqrt(281))/20a)#