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Expert-verified Found in: Page 210 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # In Problems 63–72, find the real solutions of each equation.${x}^{4}+4{x}^{3}+2{x}^{2}-x+6=0$

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

See the step by step solution

## Step 1. Finding the possible number of zeros

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

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

$p:±1,±2,±3,±6$

The factors of the leading coefficient $1$ are

$q:±1$

So, the possible rational zeros are

$\frac{p}{q}:±1,±2,±3,±6$

## 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, $\left(x+2\right)$ is a factor.

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

## 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, $\left(x-3\right)$ 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 $-2and-3.$ ### Want to see more solutions like these? 