Factoring x^4-5x^2+4

  • Thread starter mindauggas
  • Start date
  • #1
127
0

Homework Statement


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

(2)How to solve it if the answer to get is [itex](x-1)(x+1)(x-2)(x+2)[/itex] ???
 

Answers and Replies

  • #2
127
0
Solved ... sorry to bother
 
  • #3
34,675
6,386

Homework Statement


(1)Why can't I solve [itex]x^{4}-5x^{2}+4[/itex]
in the following way:
[itex](x^{4}-4x^{2}+4)-x^{2}[/itex]
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.

...
[itex](x^{2}-2-x^{2})(x^{2}-2+x^{2})[/itex]
...
If there is any reason why..

(2)How to solve it if the answer to get is [itex](x-1)(x+1)(x-2)(x+2)[/itex] ???
 
  • #4
2,967
5
You need to further factor each quadratic trinomial:
[tex]
x^2 \mp x + 2
[/tex]
 

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