<s>[INST] Prove $x^8-4x^6+14x^4+44x^2+25$ has three turning points Find a polynomial $f(x)$ with real coefficients that satisfies the following conditions:

*

*$f(x)$ has three turning points

*zeros are $i$ and $2 + i$

*the coefficient of the highest power of $x$ is $1$.

$f(x)$ has zeros $i,2+i$, hence $-i$ and $2-i$ are also zeros of that polynomial. But
$$\left(x-2-i\right)\left(x+2+i\right)\left(x^2+1\right)=\left(x^4-2x^2-3\right)+i\left(-4-4x^2\right)$$
contain complex coefficient. I try to remove the complex coefficient by
$$\left(\left(x^4-2x^2-3\right)+i\left(-4-4x^2\right)\right)\left(\left(x^4-2x^2-3\right)-i\left(-4-4x^2\right)\right)$$
$$
=x^8-4x^6+14x^4+44x^2+25$$
Which have roots $-i,i,2+i,2-i,-2+i,-2-i$ with the coefficient of the highest power of $x$ is $1$. But how to prove it has three turning points? I can show the existence of two turning points as $i$ and $-i$ has even multiplity $\implies$ $2\times 1$ bounce.
 [/INST] Your original approach works best:
$$p(x)=\left(x-(2-i)\right)\left(x-(2+i)\right)\left(x^2+1\right) = (x^2-4x+5)(x^2+1) =x^4-4x^3+6x^2-4x+5$$
and multiplying this by a conjugate $q(x) = x^4 - 4x^3 + 6x^2 - 4x - 5$, we get:
$$f(x) = x^8 - 8 x^7 + 28 x^6 - 56 x^5 + 68 x^4 - 48 x^3 + 16 x^2 - 25$$
Multiplying by the conjugate gives three turning points! This is true as $f'(x) = p'(x) q(x) + p(x) q'(x)$, and since $p'(x) = q'(x)$ as only the constant term differs, we have $p'(x)(p(x) + q(x))$. $p'(x)$ must have at least one root as it is a cubic ($p'(x) = 4(x-1)^3$), $x$ is a factor of $p(x) + q(x)$ as the constant term is missing, and we are left with another cubic, whose root is not $0$ or $1$, so $f'(x)$ has at least three roots.
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