Animated Solution for Mathematics - Differentiation: The number of distinct real roots of $x^4-4x^3+12x^2+x-1=0$ is .........

We are tasked with analyzing the quartic polynomial:

While quartic equations are notoriously difficult to solve directly, we can map the behavior of this function using the tools of calculus to determine the number of real roots.

To understand the "mood" of the function—whether it is rising or falling—we examine the first derivative:

This cubic function dictates the slope of our original quartic. To determine how many turning points exist, we must examine the second derivative:

We can simplify the second derivative by factoring:

By completing the square, we obtain:

Since $(x-1{)}^2\ge 0$ for all real $x$, the expression $f^{\prime \prime }(x)$ is strictly positive for all $x$. Because the second derivative is always positive, the slope $f^{\prime }(x)$ is strictly increasing.

This implies that $f^{\prime }(x)$ can cross the $x$-axis at most once. Consequently, the quartic function $f(x)$ possesses exactly one local minimum; it descends to a single valley and then rises indefinitely, never "wiggling" back to cross the $x$-axis again.

Now that we have established the function's shape, we test specific values to locate the roots: 1. $f(-1)=(-1{)}^4-4(-1{)}^3+12(-1{)}^2+(-1)-1=1+4+12-1-1=15$ 2. $f(0)=-1$ 3. $f(1)=(1{)}^4-4(1{)}^3+12(1{)}^2+(1)-1=1-4+12+1-1=9$

By the Intermediate Value Theorem, since $f(-1)>0$ and $f(0)<0$, there must be at least one root in the interval $(-1,0)$. Similarly, since $f(0)<0$ and $f(1)>0$, there must be at least one root in the interval $(0,1)$.

We have proven that the function has only one turning point and that it crosses the $x$-axis on both sides of that point. Because the function is strictly increasing after its single local minimum, it is impossible for it to cross the axis a third or fourth time.

The number of distinct real roots is exactly 2.