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I try to compute the roots of $$p_4 = x^4 - 4 x^ 3 + 8 x^2 - 8 x + 4$$

Now Wolfram Alpha tells me this expression is the same as $$p_4=((x - 2) x + 2)^2$$

How do I see this? Is there any way to obtain this?

Edit: I know it is $p_4=(x^2 - 2 x + 2)^2$, but I would never obtain this by myself. I'm interested in knowing a way to see this without "guessing".

user26857
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MarcE
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  • @dxiv how do I see that there are no real roots? – MarcE Apr 27 '17 at 22:04
  • The first derivative $p_4^{,'}(x)$ has easy-to-find roots. P.S. Took down my previous comment since it wasn't too helpful in this case. – dxiv Apr 27 '17 at 22:10
  • @dxiv Ok thank you, I don't understand. When is it the case that the roots of the derivatives are roots for the polynomial? – MarcE Apr 27 '17 at 22:13
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    When is it the case that the roots of the derivatives are roots for the polynomial? A common root of a polynomial and its derivative is a root of multiplicity at least 2 of the polynomial. But that's not what I meant. The roots of the derivative give the local extrema of the function. In this case, the only real root of $p_4^{'}$ is $x=1$ which turns out to be a minimum for $p_4,$. But $p_4(1) \gt 0,$, therefore $p_4 \gt 0$ for all real $x,$, in other words $p_4$ has no real roots. – dxiv Apr 27 '17 at 22:20
  • @dxiv I see. Thank you. – MarcE Apr 27 '17 at 22:24

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Factor out $x^2$ in the first three terms: \begin{align} x^4-4x^3+8x^2-8x+4&=x^2(x^2-4x+4)+4x^2-8x+4\\ &=\bigl(x(x-2)\bigr)^2+4x(x-2)+4\\ &=\bigl(x(x-2)+2\bigr)^2. \end{align}

Bernard
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