A question about polynomials of degree 2

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SUMMARY

A second degree polynomial can be rewritten in the forms of x² - a² or x² + a² through an appropriate change of variable. The general form of a second degree polynomial is ax² + bx + c = 0, which can be transformed into (x + (b/2a))² - (b/2a)² + c/a = 0. By letting u = x + (b/2a) and k² = (b/2a)² + c/a, the polynomial can be expressed as either u² - k² or u² + k² based on the sign of k². This transformation highlights the relationship between completing the square and the factorization of polynomials.

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eric_999
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Hey!

In my calculus book they claim that a second degree polynomial always can be rewritten as x^2 - a^2 or as x^2 + a^2, if you use an appropriate change of variable. I was thinking about how this works.

Let's say we have a second degree polynomial (on the general form?) ax^2 +bx + c = 0, then I can of course rewrite it as (x + (b/2a))^2 - (b/2a)^2 + c/a = 0. My question is if they mean that (x + (b/2a)) = u, and (b/2a)^2 + c/a = k, so we always can write it like either u^2 - k^2 or u^2 + k^2 depending on if k correpsonds to a positive or negative number?

Sorry if my explanation sucks but hope you understand what I mean! Thanks!
 
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That's correct. ##u = x + \frac{b}{2a}## makes the coefficient of ##u## zero. You can then write the constant term as ##+k^2## or ##-k^2## depending on whether it is positive or negative.

You might like to think about how factorizing ##u^2 - k^2## is similar to solving the quadratic equation ##ax^2 + bx + c = 0## by completing the square, and the fact that ##u^2 + k^2## has no real roots.
 

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