Solving x^4-5x^2+4: Why & How

[mindauggas]

Homework Statement

(1)Why can't I solve $x^{4}-5x^{2}+4$
in the following way:
$(x^{4}-4x^{2}+4)-x^{2}$
...
$(x^{2}-2-x^{2})(x^{2}-2+x^{2})$
...
If there is any reason why..

(2)How to solve it if the answer to get is $(x-1)(x+1)(x-2)(x+2)$ ?

 

[mindauggas]

Solved ... sorry to bother

 

[Mark44]

Homework Statement

(1)Why can't I solve $x^{4}-5x^{2}+4$
in the following way:
$(x^{4}-4x^{2}+4)-x^{2}$

It looks like you're trying to set this up as a difference of squares, a² - b² = (a + b)(a - b).

That will work here, as x⁴ - 4x² + 4 is a perfect square, namely (x² -2)².

So the above would factor into ((x² -2)) -x)((x² -2)) + x)
= (x² -x - 2)(x² + x - 2)
= (x - 2)(x + 1)(x + 2)(x - 1).

As you can see, this works, but it is probably more difficult than factoring x⁴ - 5x² + 4 directly, realizing that it is quadratic in form.

x⁴ - 5x² + 4 = (x² - 4)(x² - 1). Each of these two factors can be broken into two linear factors.

...
$(x^{2}-2-x^{2})(x^{2}-2+x^{2})$
...
If there is any reason why..

(2)How to solve it if the answer to get is $(x-1)(x+1)(x-2)(x+2)$ ?

 

[Dickfore]

You need to further factor each quadratic trinomial:
$$x^2 \mp x + 2$$