Question

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

Answer 1

`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)`