Question

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

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

Short answer

The real solution of the equation is$-1and2.$

Step-by-step solution

Step 1. Finding the possible number of zeros

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.

Step 2. Use the rational zero theorem 

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

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

Step 3. Use the synthetic division

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

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

Step 4. Finding the first 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$(x-1)(x^3-2x^2+4x-8).$

Step 5. Finding the next zero

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$(x+1)(x-2)(x^2+4).$

Step 6. Use the zero product property

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.