Homework SolutionSolving x^4-5x^2+4: Why & How

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The discussion addresses the factorization of the polynomial x^4 - 5x^2 + 4. It explains that attempting to factor it as a difference of squares is not the most efficient method, as x^4 - 4x^2 + 4 is a perfect square. The correct approach is to recognize the quadratic form and factor it directly into (x^2 - 4)(x^2 - 1), which can then be further factored into linear terms (x - 2)(x + 2)(x - 1)(x + 1). The conversation emphasizes the importance of identifying the right factoring technique for efficiency. Understanding these methods is crucial for solving similar polynomial equations effectively.
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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) ?
 
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Solved ... sorry to bother
 
mindauggas said:

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.

mindauggas said:
...
(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) ?
 
You need to further factor each quadratic trinomial:
<br /> x^2 \mp x + 2<br />
 
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