How many real roots does $x^4 - 4x^3 + 4x^2 - 10$ have? 
How many real roots does this polynomial have?
  $$x^4 - 4x^3 + 4x^2 - 10$$

Because non-real roots come in pairs, it must have 4, 2 or 0 real roots.
Following Descartes' rules of signs, it either has one negative (real) number and one or three positive numbers.
How can I tell if it has 2 or 4 real roots?
Thank you very much in advance.
 A: The generic way is to use an intermediate value theorem.
Since this is a polynomial - it is a continuous function, therefore between two arbitrary points $x_1 < x_2$, s.t. $f(x_1) < 0 < f(x_2)$ (w.l.o.g.) there exists $x_1 < x_3 < x_2$ and $f(x_3) = 0$.
Moreover, between any two zeros of differentiable function there is a zero of its derivative.
We have : $f(0) = -10$ and obviously for some large and small $x$ $f(x) > 0$, i.e. $f(1000) > 0$ and $f(-1000) > 0$. Therefore there are at least two zeros.
$$f'(x) = 4x(x^2 - 3x + 2)$$
We can check that it has three zeros ($x = 0, x=1$ and $x =2$).
Since the function is positive at "infinities", but its derivative is a polynomial of third degree (i.e. negative at $-\infty$) we conclude, that $x=0$ is a local minimum of $f$. Subsequently $x=1$ is local maximum, and $x=2$ is local minimum once again.
By checking directly the value of the function at local extremes:
$$f(0) = -10 \\ f(1) = -9 \\ f(2) = -10$$
we establish no zeros are present in $[0, 2]$
Hence there are exactly two zeros of $f$ on $\mathbb{R}$. One on $(-\infty, 0)$ and one on $(2, \infty)$.
A: One possible way to analyze the roots is to try and plot the graph of the function and see where it goes from negative to positive.
For this purpose we first see the end behavior of the graph. Note that for very large negative and positive values, the only important term is the leading one which is $x^4$. Hence as we approach $-\infty$ or $+\infty$ the value of the function is positive.
Next we need to find the turning points of this graph, i.e., where the slope of the function goes to zero
\begin{align*}
f’(x)&= 4 x^3 -12x^2+8x \\
&= 4x (x^2-3x+2)\\
&= 4x(x-1)(x-2)=0
\end{align*}
The above equation has roots at x=0,1,2 which means the function has 3 turning points. These are the only places where our function can change its behaviour, i.e., from increasing to decreasing or decreasing to increasing.
Calculating the value of the function at these points we get
\begin{align*}
f(0)&=-10\\
f(1)&=-9\\
f(2)&= -10
\end{align*}
Now we know the behaviour of the function


*

*From $x=-\infty$ to $x=0$ the function decreases from positive to negative meaning we have one zero in this region

*From $x=0$ to $x=1$ the function starts increasing from $f(0)=-10$ to $f(1)=-9$

*From $x=1$ to $x=2$ the function again starts decreasing and goes from $f(1)=-9$ to $f(2)=-10$

*From $x=2$ to $x=+\infty$ the function increases from a negative value to a positive value and hence we have another zero here
Hence we have 2 real zeros.
A: 

$$f(x)=x^4 - 4x^3 + 4x^2 - 10$$
$$f'(x)=4x^3-12x^2+8x=4(x-0)(x-1)(x-2)$$
$$f'(x)=0 \iff 
   \left\{\begin{align}
      x&=0\\ x&=1\\ x&= 2
   \end{align}\right.$$
$$f''(x) = 12x^2 - 24x + 8=12(x-1)^2-4$$
$$f''(0)=8>0 \qquad f''(1)=-4<0 \qquad f''(2)=8>0$$


*

*The function approaches $\infty$ as $x \to -\infty$

*The function has a local minimum of $f(0) = -10$ at $x=0$.

*The function has a local maximum of $f(1) = -9$ at $x=1$.

*The function has a local minimum of $f(2) = -10$ at $x=2$.

*The function approaces $\infty$ as $x \to \infty$.

*$f(x)$ decreases from $\infty$ to $-10$ on $(-\infty, 0]$

*$f(x)$ increases from $-10$ to $-9$ on $[0, 1]$

*$f(x)$ decreases from $-8$ to $-10$ on $[1,2]$

*$f(x)$ increases from $-10$ to $\infty$ on $[2, \infty)$
So 


*

*The function has one zero on the interval $(-\infty, 0]$.

*The function has no zeros in the interval $[0, 2]$.

*The function has one zero on the interval $[2, \infty)$.

A: Hint:
Your polynomial can be factored into two quadratics using difference of squares:
$$x^2(x-2)^2-10=(x(x-2)+\sqrt{10})(x(x-2)-\sqrt{10}).$$
Can you take it from here?
A: Hint Substituting $x = u + 1$ gives the biquadratic polynomial $$u^4 - 2 u^2 - 9 = (u^2 - 1)^2 - 10 ,$$ giving the factorization $$[u^2 - (1 + a)][(u^2 - (1 - a))] , \qquad a := \sqrt{10} .$$

 Since $a > 1$ the first factor, $u^2 - (1 + a)$ has two real roots and the second factor has nonzero imaginary roots, so the polynomial in $u$---and hence the polynomial in $x$---has exactly two real roots.

A: As $x\to -\infty$ $f(x)\to +\infty>0$.  And $f(-1)= -1$ so $f(x)$ "crosses" the $x$-axis at some $x < -1$.  That's our one and only negative root.
Now $f(x) = x^2(x-2)^2 - 10$.  If $x > 0$ then $x^2$ is strictly increasing and positive and $(x-2)^2$ is strictly increasing and non-negative so $f(x)$ is strictly increasing.  As $x\to \infty$ we have $f(x)\to +\infty$ and $f(2)=-10 < 0$.  For some $x > 2$, $f(x)$ "crosses" the axis, but as $f(x)$ is strictly increasing at that point it will only "cross" once and that will be our very last root.
So all that remains is to figure out if $f(x)$ has any roots between $0$ and $2$.
but for $0 \le 1$ we have $0 \le x^2 \le 1$ and $-3 \le x-2 \le -1$ and so $1\le (x-2)^2 \le 9$ and so $f(x) < 9-10 =-1 < 0$.
And if $1\le 2$ we have $1\le x^2 \le 4$ and $-1\le x-2 \le 0$ so $0 \le (x-2)^2 \le 1$ and so $f(x) < 4-10 =-6$.
So for $0\le x \le 2$ we have $f(x) < 0$ so there are no roots there.
So there is one negative root where $x<-1$ and one positive root where $x> 2$.
A: Sometimes it can be helpful to investigate the "shape" of the function curve by making a "vertical shift" (translation) in order to "remove" the constant term of the polynomial.  For $ \ f(x) \ = \ x^4 - 4x^3 + 4x^2 - 10 \ \ , \ $ a "downward shift" by $ \ 10 \ $ units gives us a new function $ \ g(x) \ = \ x^4 - 4x^3 + 4x^2 \ = \ x^2·(x^2 - 4x + 4) $ $ = \ x^2·(x - 2)^2 \ \ . \ $  As a monic quartic polynomial, its curve "opens upward", that is, $ \ \lim_{x \ \rightarrow \ \pm \infty} \ g(x) \ = \ +\infty \ \ . \ $  This function has "double zeroes" at $ \ x \ = \ 0 \ $ and $ \ x \ = \ +2 \ \ , \ $ which means they are "turning points" (or "touching intercepts"); so these are in all four real zeroes (elsewhere, $ \ g(x) \ > \ 0 \ \ . \ ) \ $
We can calculate that $ \ g(1) \ = \ (1 - 2)^2 \ = \ +1 \ \ . \ $  It is reasonably straightforward to show that $$ \ h(x) \ = \ g(x + 1) \ = \ (x + 1)^2·( \ [x + 1] - 2)^2 \ = \ (x + 1)^2·(x - 1)^2 \ \ $$
has even symmetry  [Travis Willse shows an alternative version of this].  So the function curve has a $ \ w-$shaped "bottom" with symmetry about the line $ \ x \ = \ +1 \ \ , \ $   absolute minima at $ \ h(-1) \ = \ g(0) \ = \ h(+1) \ = \ g(+2) \ = \ 0 \ \ , \  $ and a "rise" at $ \ h(0) \ = \ g(+1) \ = \ 1 \ \ . \ $  (The application of calculus would show that $ \ h''(-1) \ = \ h''(+1) \ > \ 0 \ $ and $ \ h''(0) \ < \ 0 \ \ \ , \ $ confirming that this function curve has a relative maximum at $ \ x \ = \ 0 \ $ and minima at $ \ x \ = \ \pm 1 \ \ , \ $ as described.)
So translating $ \ h(x) \ $ "back to" $ \ f(x) \ $ places the absolute minima at $ \ (0 \ , \ -10) \ $ and $ \ (2 \ , \ -10) \ $ and the relative maximum at $ \ (1 \ , \ -9) \ \ , \ $ well "below" the $ \ x-$axis.  Only the "upward-reaching arms" of the function curve intersect the $ \ x-$axis, producing  two intercepts (visualized in Steven Alexis Gregory's graph).  Consequently $ \ f(x) \ $ has just two real zeroes.  [We can even estimate their values:  $ \ f(-1) \ = \ f(+3) \ = \ -1 \ \ , \ $ so the real zeroes of $ \ f(x) \ $ are close to $ \ x \ = \ -1 \ $ and $ \ x \ = \ +3 \ \ . \ $  (The approximate values are $ \ -1.0402 \ $ and $ \ +3.0402 \ \ . \ ) \ ] $
