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Q. 68

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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.$2{x}^{3}-11{x}^{2}+10x+8=0$

The real solutions of the equation are $-\frac{1}{2},2,and4.$

See the step by step solution

## Step 1. Finding the possible number of zeros

The given equation is $2{x}^{3}-11{x}^{2}+10x+8=0$

The given function is of degree three, so it has at most three 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

$p:±1,±2,±4,±8$

The factors of the leading coefficient $2$ are

$q:±1,±2$

So, the possible rational zeros are

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

## 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 $2{x}^{2}-7x-4.$

## Step 5. Use the zero product property

As the remaining zeros satisfy the depressed equation

So,

$2{x}^{2}-7x-4=0\phantom{\rule{0ex}{0ex}}2{x}^{2}-8x+x-4=0\phantom{\rule{0ex}{0ex}}2x\left(x-4\right)+1\left(x-4\right)=0\phantom{\rule{0ex}{0ex}}\left(2x+1\right)\left(x-4\right)=0$

Use the property

$2x+1=0\phantom{\rule{0ex}{0ex}}x=-\frac{1}{2}$ and $x-4=0\phantom{\rule{0ex}{0ex}}x=4$