Interception with $x$-axis - not so trivial?

Question

I want to find the interception with the x-axis of the following function: $f(x) = \frac{1}{4}x^4-x^3+2x$.

So putting $0 = \frac{1}{4}x^4-x^3+2x$ I would get $0 = x(\frac{1}{4}x^3-x^2+2)$ but how to check for the other solutions properly?

Answer 1

$f(x)=\frac14 x^4-x^3+2x=\frac14 x(x^3-4x^2+8)$.

Let $g(x)=x^3-4x^2+8$. Since $g(2)=0$, we know that $x-2$ divides $g(x)$, so $g(x)=(x-2)(x^2+kx-4)$ for some constant $k$. Equating coefficients of $x$ gives $-2k-4=0$ so $k=-2$. Now $g(x)=(x-2)(x^2-2x-4)$.

Let $h(x)=x^2-2x-4=(x-1)^2-5$. Then the zeroes of $h(x)$ are $1+\sqrt 5$ and $1-\sqrt 5$. Hence the zeroes of $f(x)$ are $0$, $2$, $1+\sqrt 5$ and $1-\sqrt 5$.

Answer 2

Also $x=2$ is a root, so $f(x)=\frac{1}{4}(x-2)x(x^2-2x-4)$. wolframalpha is your friend.

Answer 3

One solution is $x=0$, to account for that term $x$ in the expansion. So what's left are the roots of

$$ \frac{1}{4}x^3-x^2+2 $$

This is a cubic equation of the form

$$ a x^3 + b x^2 + c x + d = 0 $$

with $a=\frac{1}{4},b=-1,c=0,d=2$.

From here, to solve this by hand, you can use the general formulas for the roots of a cubic to find the other solutions. Those formulas will give you all the roots, including the ones out in the complex plane. You can always use a tool such as WolframAlpha to check your work, as others have suggested.