91Ó°ÊÓ

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.

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

The factors of the constant term $6$are

$p:Â\pm 1,Â\pm 2,Â\pm 3,Â\pm 6$

The factors of the leading coefficient $1$are

$q:Â\pm 1$

So, the possible rational zeros are

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

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

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

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$x^3+2x^2−2x+3.$

As the remaining zeros satisfy the depressed equation

So,

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

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$\left(x+2\right)\left(x+3\right)\left(x^2-x+1\right).$

As the remaining zeros satisfy the depressed equation

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

We cannot factor the polynomial.

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