3.6 Zeros of polynomial functions  (Page 6/14)

Begin by determining the number of sign changes.

There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine $f(-x)$ to determine the number of negative real roots.

$$\begin{array}{l}f(-x)=-{(-x)}^4-3{(-x)}^3+6{(-x)}^2-4(-x)-12\hfill \\ f(-x)=-x^4+3x^3+6x^2+4x-12\hfill \end{array}$$

Again, there are two sign changes, so there are either 2 or 0 negative real roots.

There are four possibilities, as we can see in [link] .

Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by $V=lwh.$ We were given that the length must be four inches longer than the width, so we can express the length of the cake as $l=w+4.$ We were given that the height of the cake is one-third of the width, so we can express the height of the cake as $h=\frac{1}{3}w.$ Let’s write the volume of the cake in terms of width of the cake.

Substitute the given volume into this equation.

Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are $\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351,$ and $\pm 1053.$ We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check $x=1.$

Since 1 is not a solution, we will check $x=3.$

Since 3 is not a solution either, we will test $x=9.$

Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.

The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.