Chapter 4.5, Problem 17E

To determine

To find: the zeros of the given function. Also sketch the graph of the function.

Answer to Problem 17E

The zeros of the given function is $x=0,x=5$ and $x=-3$ .

Explanation of Solution

Given information:

A function is given as

  $g\left(x\right)=-2x^5+2x^4+40x^3$

Concept used:

Leading coefficient of a polynomial function is the coefficient of the leading term.

Degree of the polynomial is the degree of leading term or the height degree in the polynomial function.

For the polynomial $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0$ the leading coefficient is $a_n$ and the degree is $n$ .

The end behavior can describe the graph of a polynomial function as  $x$  approaches $+\infty$ or as  $x$  approaches $-\infty$ .

The end behavior of a polynomial function can be determined by the leading coefficient and the degree of the polynomial.

If degree is even and leading coefficient is negative.

  $f\left(x\right)\to -\infty$ as $x\to -\infty$ and $f\left(x\right)\to -\infty$ as $x\to +\infty$ .

If degree is odd and leading coefficient is negative.

  $f\left(x\right)\to \infty$ as $x\to -\infty$ and $f\left(x\right)\to -\infty$ as $x\to +\infty$ .

If degree is odd and leading coefficient is positive.

  $f\left(x\right)\to -\infty$ as $x\to -\infty$ and $f\left(x\right)\to \infty$ as $x\to +\infty$ .

If degree is even and leading coefficient is positive.

  $f\left(x\right)\to \infty$ as $x\to -\infty$ and $f\left(x\right)\to \infty$ as $x\to +\infty$ .

Zeros of a polynomial function can be found by equating the function with 0.

Calculation:

Consider the given function.

  $g\left(x\right)=-4x^4+8x^3+60x^2$

Now, for zeros of the function equate the function with 0.

  $\begin{array}{l}g\left(x\right)=0\\ -4x^4+8x^3+60x^2=0\end{array}$

Take common $-4x^2$ from each term of left side of the equation.

  $-4x^2\left(x^2-2x-15\right)=0$

Now, break the coefficient of middle term $-2$ as sum or difference of two numbers such that product of the numbers is equal to the product of first term and constant term that is product of $1$ and $-15$ .

Such numbers will be $-5$ and $3$ .

Now, factorize left side of $-4x^2\left(x^2-2x-15\right)=0$ as shown:

  $\begin{array}{c}-4x^2\left(x^2-2x-15\right)=0\\ -4x^2\left(x^2-5x+3x-15\right)=0\\ -4x^2\left[x\left(x-5\right)+3\left(x-5\right)\right]=0\\ -4x^2\left(x-5\right)\left(x+3\right)=0\end{array}$

Solution can be found by equating each factor with 0.

  $x=0$

And

  $\begin{array}{c}x-5=0\\ x=5\end{array}$

Further,

  $\begin{array}{c}x+3=0\\ x=-3\end{array}$

Therefore, the zeros of the given function is $x=0,x=5$ and $x=-3$ .

So, the graph crosses $x$ axis at $x=-3,x=5,x=0$ .

Degree is even and leading coefficient is negative.

  $g\left(x\right)\to -\infty$ as $x\to -\infty$ and $g\left(x\right)\to \infty$ as $x\to +\infty$ .

Graph:

The graph of the function can be sketched by the characteristics.