Question

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

$x^4+4x^3+2x^2-x+6=0$

Short answer

The real solutions of the equation are$-2and-3.$

Step-by-step solution

Step 1. Finding the possible number of zeros 

The given equation is$x^4+4x^3+2x^2-x+6=0$

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 $6$are

$p:\pm 1,\pm 2,\pm 3,\pm 6$

The factors of the leading coefficient $1$are

$q:\pm 1$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 6$

Step 3. Finding the zero

Let's test the potential zero $1$, by using synthetic division

Since the remainder is not zero, it is not a factor.

Step 4. Finding the zero 

Now, let's test the potential zero $-2$ by using the synthetic division test

Since the remainder is zero, $(x+2)$is a factor.

The depressed equation is$x^3+2x^2-2x+3.$

Step 5. Use the synthetic division test

As the remaining zeros satisfy the depressed equation

So,

Since the remainder is$7$,$-2$is not a repeated zero of the equation.

Step 6. Finding the zero

Take a test of potential zero $-3$by using the synthetic division test

Since the remainder is zero, $(x-3)$is a factor.

The factor form can be written as$\left(x+2\right)\left(x+3\right)\left(x^2-x+1\right).$

Step 7. Use zero product property

As the remaining zeros satisfy the depressed equation

So, $x^2-x+1=0$

We cannot factor the polynomial.

Thus, the real solutions are$-2and-3.$