Question

How do you find the zeroes of `f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25`?

Answer 1

Factor by grouping twice, then use the difference of squares identity twice and hence find zeros:

`+-sqrt(10)/2`, `+-sqrt(5)`, `-1/5`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

Factor by grouping:

`f(x) = 10x^5+2x^4-75x^3-15x^2+125x+25`

`= (10x^5+2x^4)-(75x^3+15x^2)+(125x+25)`

`= 2x^4(5x+1)-15x^2(5x+1)+25(5x+1)`

`= (2x^4-15x^2+25)(5x+1)`

`= (2x^4-10x^2-5x^2+25)(5x+1)`

`= ((2x^4-10x^2)-(5x^2-25))(5x+1)`

`= (2x^2(x^2-5)-5(x^2-5))(5x+1)`

`= (2x^2-5)(x^2-5)(5x+1)`

`= ((sqrt(2)x)^2-(sqrt(5))^2)(x^2-(sqrt(5))^2)(5x+1)`

`= (sqrt(2)x-sqrt(5))(sqrt(2)x+sqrt(5))(x-sqrt(5))(x+sqrt(5))(5x+1)`

Hence `f(x)` has zeros for the following values of `x`

`+-sqrt(5)/sqrt(2) = +-sqrt(10)/2`

`+-sqrt(5)`

`-1/5`