I know they aren’t the same function but what’s the issue here? What am I not taking into account? Sorry for such a basic question, it just confuses me.
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$-x^2+1=-(x^2-1)$, so you are done. They just differ by a sign. – Dietrich Burde May 30 '21 at 19:28
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1And what about the roots and the factorized form? – GZanotto May 30 '21 at 19:30
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Well, you can factorize $-x^2+1=(1-x)(x+1)$ which still gives you insight into the roots of the polynomial but takes care of the negative sign. – Lukas May 30 '21 at 19:33
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Roots are independant of the way you write the polynomial. Same for factorized form, always written as $(x-r_1)(x-r_2)(x-r_3)...$ – Ripi2 May 30 '21 at 19:34
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In factorised form it is, as I said, $-x^2+1=-(x^2-1)=-(x+1)(x-1)$. – Dietrich Burde May 30 '21 at 19:45
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$x^2-1$ and $7x^2-7$ both have the same roots.
One of them is $(x-1)(x+1)$ and the other is $7(x-1)(x+1).$
Two polynomials with the same roots, when factored, are constant multiples of each other.
In your example, the constant is $-1.$
(If you're talking about polynomials with real coefficients, then "the same roots" must be taken to mean the same complex roots, including, but not limited to, real roots. That is because $\mathbb C$ is the algebraic closure of $\mathbb R.$)
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That product does not give the same function (try expanding it out, it gives one but not the other). You are trying to use the fact that "a function is a product of its roots".
But this is slightly inaccurate, rather the function may be written as $k(x+1)(x-1)$, where $k$ is any real number (in this case). You can see that all functions of this form have the same roots, and yet multiplying the roots does not necessarily give you back the same function.
In other words, if you have a polynomial with coefficients in the real numbers, then the product of roots is a factor of your function, but not necessarily the function itself.
masiewpao
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1This is slightly incorrect. You are calling something like $x+1$ a root, but it is not. – Randall May 30 '21 at 20:11
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@Randall Yes I agree, I should be more precise and say something like the polynomial expression of $x-c$ where $c$ is the root. – masiewpao May 30 '21 at 20:48
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Ok so let’s say that I have a second degree polynomial function that it’s factorized form it’s not a k(x-1)(x+1) but other form that I don’t know but I have that function on the $ax^2 + bx + c$ form. How do I write get to the factorized form with the correct k? I don’t know if I am explaining myself right. – GZanotto May 30 '21 at 23:44