How do you factor completely #50a^2b^5 – 35a^4b^3 + 5a^3b^4#?
1 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
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)#
