Chapter 4.5, Problem 13E

To determine

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

Answer to Problem 13E

The zeros of the given function is $x=0,x=-3$ and $x=2$ .

Explanation of Solution

Given information:

A function is given as

  $h\left(x\right)=x^4+x^3-6x^2$

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.

  $h\left(x\right)=x^4+x^3-6x^2$

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

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

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

  $x^2\left(x^2+x-6\right)=0$

Now, break the coefficient of middle term $1$ 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 $-6$ .

Such numbers will be $-2$ and $3$ .

Now, factorize left side of $x^2\left(x^2+x-6\right)=0$ as shown:

  $\begin{array}{c}{x}^2\left(x^2+x-6\right)=0\\ x^2\left(x^2+3x-2x-6\right)=0\\ x^2\left[x\left(x+3\right)-2\left(x+3\right)\right]=0\\ x^2\left(x+3\right)\left(x-2\right)=0\end{array}$

Solution can be found by equating each factor with 0.

  $x=0$

And

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

Further,

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

Therefore, the zeros of the given function is $x=0,x=-3$ and $x=2$ .

So, the graph crosses $x$ axis at $x=2,x=-3,x=0$ .

Degree is even and leading coefficient is positive.

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

Graph:

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