Question

How do you factor `x^3-x^2-4x+5`?

Answer 1

Factoring by grouping does not work with this polynomial, but read on...

Explanation:

Factoring by grouping does not immediately help with this cubic polynomial. If it were `x^3-x^2-4x+4` then it would be much easier:

`x^3-x^2-4x+4`

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

`= x^2(x-1)-4(x-1)`

`= (x^2-4)(x-1)`

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

Was there a typo in the question?

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Can we factor `f(x) = x^3-x^2-4x+5` ?

The only possible rational zeros are `+-1`, `+-5`, neither of which work.

This cubic turns out to have one Real zero and two Complex zeros.

It is possible, if a little lengthy, to find them using Cardano's method, which will give you:

Real zero:

`x_1 = 1/3(1-root(3)((97-3sqrt(69))/2)-root(3)((97+3sqrt(69))/2))`

Complex zeros:

`x_2 = 1/3(1-omega root(3)((97-3sqrt(69))/2)-omega^2 root(3)((97+3sqrt(69))/2))`

`x_3 = 1/3(1-omega^2 root(3)((97-3sqrt(69))/2)-omega root(3)((97+3sqrt(69))/2))`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

Then:

`x^3-x^2-4x+5=(x-x_1)(x-x_2)(x-x_3)`