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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.${x}^{4}-2{x}^{3}+10{x}^{2}-18x+9=0$

The real solution of the equation is $1.$

See the step by step solution

## Step 1. Finding the possible number of zeros

The given equation is ${x}^{4}-2{x}^{3}+10{x}^{2}-18x+9=0$

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 $9$ are

$p:±1,±3,±9$

The factors of the leading coefficient $1$ are

$q:±1$

So, the possible rational zeros are

$\frac{p}{q}:±1,±3,±9$

## Step 3. Finding the zero

Let's test the potential zero $1$ by using substitution

Thus,

$f\left(1\right)={\left(1\right)}^{4}-2{\left(1\right)}^{3}+10{\left(1\right)}^{2}-18\left(1\right)+9\phantom{\rule{0ex}{0ex}}f\left(1\right)=0$

Since the remainder is zero, $\left(x-1\right)$ is a factor.

## Step 4. Use synthetic division

Use synthetic division to factor equation, ## Step 5. Use the zero product property

As the remaining zeros satisfy the depressed equation

So,

${x}^{3}-{x}^{2}+9x-9=0\phantom{\rule{0ex}{0ex}}{x}^{2}\left(x-1\right)+9\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\left({x}^{2}+9\right)\left(x-1\right)=0$

Use the property

${x}^{2}+9=0\phantom{\rule{0ex}{0ex}}{x}^{2}=-9$ and $x-1=0\phantom{\rule{0ex}{0ex}}x=1$

The square of a negative number is imaginary thus. it is not real. ### Want to see more solutions like these? 