Question

How do you find all zeros with multiplicities of `f(x)=x^3-7x^2+x-7`?

Answer 1

The zeros of `f(x)` are `7`, `i`, `-i`, all with multiplicity `1`.

Explanation:

Given:

`f(x) = x^3-7x^2+x-7`

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

`x^3-7x^2+x-7 = (x^3-7x^2)+(x-7)`

`color(white)(x^3-7x^2+x-7) = x^2(x-7)+1(x-7)`

`color(white)(x^3-7x^2+x-7) = (x^2+1)(x-7)`

Note that `x^2+1` has no linear factors with real coefficients since `x^2+1 > 0` for all real values of `x`. We can factor it as a difference of squares using complex coefficients:

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

So the zeros of `f(x)` are `7`, `i`, `-i`, all with multiplicity `1`.