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$(x-a)$ is a factor of the polynomial $P(x)$, where $a$ is an integer. If $P(x) = x^3-kx^2+2kx-8$, prove that the values of k for which P(x) has real roots are $K\le-2$ or $K\ge6$.

I tried differentiating it and using the discriminant but I couldn't that prove $K\le-2$

Thanks in advance

kjhg
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    Since x=2 is a root, you can factor out $(x-2)$ – Cornman Sep 18 '17 at 08:11
  • I suppose the question should be "for which $P(x)$ has three real roots" or "for which $P(x)$ has only real roots", because $P(x)$ has at least one real root $(x=2)$ whatever the value of $k$... – Evargalo Sep 18 '17 at 08:16

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You have $$P(x) = (x-2)(x^2+(2-k)x+4).$$

So, $P(x)$ always has a root $x=2$.

If you want $P(x)$ has three real roots, then $(2-k)^2-16 \geq 0$, ( $k\geq 6$ or $k \leq -2$).

GAVD
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