Question

How do you find all the zeros of `f(x)=4x^3-20x^2-3x+15`?

Answer 1

Factor by grouping and by using the difference of squares identity to find:

`f(x) =(2x-sqrt(3))(2x+sqrt(3))(x-5)`

hence has zeros `x = +-sqrt(3)/2` and `x = 5`

Explanation:

Factor by grouping, then use the difference of squares identity:

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

with `a=2x` and `b=sqrt(3)`, as follows:

`f(x) = 4x^3-20x^2-3x+15`

`=(4x^3-20x^2)-(3x-15)`

`=4x^2(x-5)-3(x-5)`

`=(4x^2-3)(x-5)`

`=((2x)^2-(sqrt(3))^2)(x-5)`

`=(2x-sqrt(3))(2x+sqrt(3))(x-5)`

So the zeros of `f(x)` are `x = +-sqrt(3)/2` and `x = 5`