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Q. 42

Expert-verified
Found in: Page 209

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Find the real zeros of f. Use the real zeros to factor f. $f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+12$

The real zeros of the function are $-\frac{3}{2},2,2$.

And it is factored as $f\left(x\right)=\left(2x+3\right)\left(x-2\right)\left(x-2\right)$.

See the step by step solution

## Step 1. Find the possible number of zeros

The given function $f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+12$ is of degree three, so it has at most three real zeros.

## Step 2. Use the rational zero theorem

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

The factors of the constant term $12$ are

$p:±1,±2,±3,±4,±6,±12$

The factors of the leading coefficient $2$ are

$q:±1,±2$

So the possible rational zeros are

$\frac{p}{q}:±1,±2,±3,±4,±6,±12,±\frac{1}{2},±\frac{3}{2}$

## Step 3. Graph the function

The graph of the function is given as

The graph has two turning points, so it will have three real zeros.

## Step 4. Find the first factor

From the graph, it appears that $2$ is a zero of the function. On performing synthetic division we get

As the remainder is zero so $2$ is a zero of the function.

And the quotient $2{x}^{2}-x-6$ is the depressed polynomial. So the given function is factored as

$f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+12\phantom{\rule{0ex}{0ex}}f\left(x\right)=\left(x-2\right)\left(2{x}^{2}-x-6\right)$

## Step 5. Factor the depressed equation

Equating the depressing polynomial with zero and factoring by grouping we get

$\begin{array}{rcl}2{x}^{2}-x-6& =& 0\\ 2{x}^{2}-4x+3x-6& =& 0\\ 2x\left(x-2\right)+3\left(x-2\right)& =& 0\\ \left(x-2\right)\left(2x+3\right)& =& 0\end{array}$

So the other zeros are

$\begin{array}{rcl}x-2& =& 0\\ x& =& 2\end{array}$

and

$\begin{array}{rcl}2x+3& =& 0\\ 2x& =& -3\\ x& =& -\frac{3}{2}\end{array}$

## Step 6. Factor the function

All the three real zeros of the function $f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+12$ are $-\frac{3}{2},2,2$.

And it can be written in factored form as $f\left(x\right)=\left(2x+3\right)\left(x-2\right)\left(x-2\right)$.