Chapter 4, Problem 2CT

For the polynomial function $g\left(x\right)=2x^3+5x^2-28x-15$,

Determine the maximum number of real zeros that the function may have.

List the potential rational zeros.

Determine the real zeros of $g$. Factor $g$ over the reals.

Find the $x-$ and $y-$ intercepts of the graph of $g$.

Determine whether the graph crosses or touches the x –axis at each x –intercept.

Find the power function that the graph of $g$ resembles for large value of $\left|x\right|$.

Put all the information together to obtain the graph of $g$.

To determine

The maximum number of real zeros of the function $g\left(x\right)=2x^3+5x^2-28x-15$.

Answer to Problem 2CT

Solution:

The maximum number of real zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ is 3.

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$

Let, the function $g\left(x\right)=2x^3+5x^2-28x-15$.

The maximum number of real zeros is the degree of the polynomial.

Here, the degree of $g\left(x\right)=2x^3+5x^2-28x-15$ is 3.

Therefore, the maximum number of real zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ is 3.

To determine

The list of potential rational zeros of the function $g\left(x\right)=2x^3+5x^2-28x-15$.

Answer to Problem 2CT

Solution:

The list of potential zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ are $\pm 1,\pm \frac{1}{2},\pm 3,\pm \frac{3}{2},\pm 5,\pm \frac{5}{2},\pm 15,\pm \frac{15}{2}$.

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$

Let, the function $g\left(x\right)=2x^3+5x^2-28x-15$.

Now, to list the potential rational zeros of $g\left(x\right)$.

By rational root theorem,

The divisors of the constant term are $p=\pm 1,\pm 3,\pm 5,\pm 15$.

The divisors of the leading coefficient are $q=\pm 1,\pm 2$.

Then, possible rational zeros of the polynomial are,

$\frac{p}{q}=\pm 1,\pm \frac{1}{2},\pm 3,\pm \frac{3}{2},\pm 5,\pm \frac{5}{2},\pm 15,\pm \frac{15}{2}$.

Therefore, the list of potential zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ are $\pm 1,\pm \frac{1}{2},\pm 3,\pm \frac{3}{2},\pm 5,\pm \frac{5}{2},\pm 15,\pm \frac{15}{2}$.

To determine

The real zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ and factor $g\left(x\right)$ over the reals.

Answer to Problem 2CT

Solution:

The real zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ are $-5,-\frac{1}{2},3$ and the factor form is

$\left(2x+1\right)\left(x-3\right)\left(x+5\right)$.

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$

Let, the function $g\left(x\right)=2x^3+5x^2-28x-15$.

From part (b),

The potential rational zeros of $g\left(x\right)$ are $\pm 1,\pm \frac{1}{2},\pm 3,\pm \frac{3}{2},\pm 5,\pm \frac{5}{2},\pm 15,\pm \frac{15}{2}$.

Now, test $x=-\frac{1}{2}$ using synthetic division.

$\begin{array}{l}-\frac{1}{2}\overline{)25-28-15}\\ \underset{\_}{-1-215}\\ 24-300\end{array}$

Here, since the remainder is 0, $x=-\frac{1}{2}$ is a zero of $g$.

After taking $x+\frac{1}{2}$ as a factor, $2x^3+5x^2-28x-15=\left(x+\frac{1}{2}\right)\left(2x^2+4x-30\right)=2\left(x+\frac{1}{2}\right)\left(x^2+2x-15\right)$

Then, the depressed equation is, $x^2+2x-15$.

By quadratic formula; $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$x=\frac{-2\pm \sqrt{4+60}}{2}=\frac{-2\pm 8}{2}$

$\Rightarrow x=3\text{or}x=-5$

The factors of $x^2+2x-15$ are $\left(x-3\right)\text{and}\left(x+5\right)$.

The factor form of $g\left(x\right)$ is $2x^3+5x^2-28x-15=2\left(x+\frac{1}{2}\right)\left(x-3\right)\left(x+5\right)=\left(2x+1\right)\left(x-3\right)\left(x+5\right)$.

Hence, the real zeros of $g\left(x\right)=2x^3+5x^2-28x-15$ are $-5,-\frac{1}{2},3$ and the factor form is $\left(2x+1\right)\left(x-3\right)\left(x+5\right)$.

To determine

The $x-$ intercept and $y$-intercepts of the function, $g\left(x\right)=2x^3+5x^2-28x-15$.

Answer to Problem 2CT

Solution:

The $x-$ intercepts of the function are $-5,-\frac{1}{2}$ and 3 and the $y-$ intercept of the polynomial function is $-15$.

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$.

To find $y-$ intercepts substitute $x=0$ in the function $g\left(x\right)=2x^3+5x^2-28x-15$,

$g\left(0\right)=2{\left(0\right)}^3+5{\left(0\right)}^2-28\left(0\right)-15=-15$

Thus the $y-$ intercept of the polynomial function is $-15$

Now to find $x-$ intercept of the function substitute $g\left(x\right)=0$ in the function.

$g\left(x\right)=2x^3+5x^2-28x-15$, it gives

$0=2x^3+5x^2-28x-15$

$\Rightarrow \left(x-3\right)\left(2x+1\right)\left(x+5\right)=0$

$\Rightarrow x-3=0$ or $2x+1=0$ or $x+5=0$

$\Rightarrow x=3$ or $2x=-1$ or $x=5$

$\Rightarrow x=3$ or $x=-\frac{1}{2}$ or $x=5$

Hence the $x-$ intercepts of the function are $-5,-\frac{1}{2}$ and 3.

To determine

Whether the graph crosses or touches the $x-$ axis at each $x-$ intercepts of the $g\left(x\right)=2x^3+5x^2-28x-15$.

Answer to Problem 2CT

Solution:

Thegraph crosses the $x-$ axis at each $x-$ intercepts of the function $g\left(x\right)=2x^3+5x^2-28x-15$.

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$.

From part (c) the factors of the function $g\left(x\right)=2x^3+5x^2-28x-15$ are $x-3$, $x+\frac{1}{2}$ and $x+5$ so the zeros of the function are $-5,-\frac{1}{2}$ and 3.

$-5$ is the zero of the function with multiplicity 1 since the exponent of the factor $x+5$ is 1 .

$-\frac{1}{2}$ is the zero of the function with multiplicity 1 since the exponent of the factor $x+\frac{1}{2}$ is 1.

3 is the zero of the function with multiplicity 1 since the exponent of the factor $x-3$ is 1.

The graph of the function $g\left(x\right)$ crosses the $x-$ axis at $x=-5$ since the multiplicity of the $-5$ is 1 that is odd multiplicity and also the graph of function $g\left(x\right)$ crosses the $x-$ axis at $x=3$ and

$x=-\frac{1}{2}$ since the multiplicity of the 3 and $-\frac{1}{2}$ are 1 which is odd.

To determine

The power function that the graph of $g\left(x\right)=2x^3+5x^2-28x-15$ resembles for large values of $\left|x\right|$.

Answer to Problem 2CT

Solution:

Thegraph of the function $g\left(x\right)=2x^3+5x^2-28x-15$ resembles like $y=2x^3$ for large values of $\left|x\right|$.

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$

The polynomial function is $g\left(x\right)=2x^3+5x^2-28x-15$ .

Here the degree of the polynomial function $f\left(x\right)$ is 3.

The graph of the function $g\left(x\right)=2x^3+5x^2-28x-15$ behaves like $y=2x^3$ for large values of $\left|x\right|$.

To determine

To graph: The function, $g\left(x\right)=2x^3+5x^2-28x-15$ .

Explanation of Solution

Given Information:

The function, $g\left(x\right)=2x^3+5x^2-28x-15$

Graph:

The polynomial function is $g\left(x\right)=2x^3+5x^2-28x-15$ .

From all the above parts, the analysis of the function $g\left(x\right)=2x^3+5x^2-28x-15$ are state below:

The graph of the function $g\left(x\right)=2x^3+5x^2-28x-15$ behaves like $y=2x^3$ for large values of $\left|x\right|$.

Thezeros of the function are $-5,-\frac{1}{2}$ and 3

The $x-$ intercepts of the function are $-5,-\frac{1}{2}$ and 3 and the $y-$ intercept of the polynomial function is $-15$.

Thegraph crosses the $x-$ axis at each $x-$ intercepts of the function.

Here the degree of the polynomial function $f\left(x\right)$ is 3, the maximum number of tuning points are $3-1=2$.

Using all these information, the graph will look alike:

Now find additional points on the graph on each side of $x-$ intercept as follows

For $x=-5.5$ the value of $g\left(x\right)$ at $x=-5.5$ is $g\left(-5.5\right)=2{\left(-5.5\right)}^3+5{\left(-5.5\right)}^2-28\left(-5.5\right)-15=-42.5$

For $x=-2$ the value of $g\left(x\right)$ at $x=-2$ is $g\left(-2\right)=2{\left(-2\right)}^3+5{\left(-2\right)}^2-28\left(-2\right)-15=-45$

For $x=-0.25$ the value of $g\left(x\right)$ at $x=-0.25$ is $g\left(-0.25\right)=2{\left(-0.25\right)}^3+5{\left(-0.25\right)}^2-28\left(-0.25\right)-15=-7.71875$

For $x=2$ the value of $g\left(x\right)$ at $x=2$ is $g\left(2\right)=2{\left(2\right)}^3+5{\left(2\right)}^2-28\left(2\right)-15=-35$

For $x=3.5$ the value of $g\left(x\right)$ at $x=3.5$ is $g\left(3.5\right)=2{\left(3.5\right)}^3+5{\left(3.5\right)}^2-28\left(3.5\right)-15=34$

Now plot all these coordinates $\left(-4.1,-6.63255\right),\left(-3.5,22.78125\right),\left(-1.2,14.9072\right),\left(0.3,0.73745\right)$ and $\left(2.2,-5.3568\right)$ on the graph and join them

So the graph of the function is as follows,

Interpretation:

The graph of the function $g\left(x\right)=2x^3+5x^2-28x-15$ behaves like $y=2x^3$ for large values of $\left|x\right|$.

The $x-$ intercepts of the function are $-5,-\frac{1}{2}$ and 3 and the $y-$ intercept of the polynomial function is $-15$.

Thegraph crosses the $x-$ axis at each $x-$ intercepts of the function.

Here the degree of the polynomial function $f\left(x\right)$ is 3, the maximum number of tuning points are $3-1=2$.