In Problems 63–72, find the real solutions of each equation.

$x^4-x^3+2x^2-4x-8=0$

The given equation is $x^4-x^3+2x^2-4x-8=0$

The given function is of degree four, so it has at most four real zeros.

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $-8$are

$p:\pm 1,\pm 2,\pm 4,\pm 8$

The factors of the leading coefficient $1$are

$q:\pm 1$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 4,\pm 8$

Let's test the potential zero for$1,$

Since the remainder is $10$, $1$is not a zero.

Now, let's test the potential zero for$-1$ by using the synthetic division test

So, $-1$is a zero and $(x+1)$is a factor.

The factored form is$(x-1)(x^3-2x^2+4x-8).$

Let's take the test for $2$by using the synthetic division as the remaining zeros satisfy the new depressed equation

Thus, $2$is a zero and $x-2$is factor.

Now, the factor form is$(x+1)(x-2)(x^2+4).$

As the remaining zeros satisfy the depressed equation

So,

$$\begin{gathered} x^2+4=0 \\ x=\sqrt{-4} \end{gathered}$$

And $\sqrt{-4}$can't be real zero because it is an imaginary number.

Thus, there are no more zeros.