To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[/latex].

[latex]\begin{array}{l}\\ 2\overline{)\begin{array}{lllllllll}6\hfill & -1\hfill & -15\hfill & 2\hfill & -7\hfill \\ \hfill & \text{ }12\hfill & \text{ }\text{ }\text{ }22\hfill & 14\hfill & \text{ }\text{ }32\hfill \end{array}}\\ \begin{array}{llllll}\hfill & \text{}6\hfill & 11\hfill & \text{ }\text{ }\text{ }7\hfill & \text{ }\text{ }16\hfill & \text{ }\text{ }25\hfill \end{array}\end{array}[/latex]

The remainder is [latex]25[/latex]. Therefore, [latex]f\left(2\right)=25[/latex].

Analysis of the Solution

We can check our answer by evaluating [latex]f\left(2\right)[/latex].

[latex]\begin{array}{lll}f\left(x\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\ f\left(2\right) & =6{\left(2\right)}^{4}-{\left(2\right)}^{3}-15{\left(2\right)}^{2}+2\left(2\right)-7 \\ f\left(2\right) & =25\hfill \end{array}[/latex]

The only possible rational zeros of [latex]f\left(x\right)[/latex] are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2.

The constant term is –4; the factors of –4 are [latex]p=\pm 1,\pm 2,\pm 4[/latex].

The leading coefficient is 2; the factors of 2 are [latex]q=\pm 1,\pm 2[/latex].

If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2.

[latex]\begin{array}{l}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2}\text{ }& \frac{p}{q}=\pm \frac{2}{1},\pm \frac{2}{2}\text{ }& \frac{p}{q}=\pm \frac{4}{1},\pm \frac{4}{2}\end{array}[/latex]

Note that [latex]\frac{2}{2}=1[/latex] and [latex]\frac{4}{2}=2[/latex], which have already been listed, so we can shorten our list.

[latex]\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}}=\pm 1,\pm 2,\pm 4,\pm \frac{1}{2}[/latex]

The Rational Zero Theorem tells us that if [latex]\frac{p}{q}[/latex] is a zero of [latex]f\left(x\right)[/latex], then p is a factor of 1 and q is a factor of 2.

[latex]\begin{array}{l}\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}}\hfill \\ \text{}\frac{p}{q}=\frac{\text{Factors of 1}}{\text{Factors of 2}}\hfill \end{array}[/latex]

The factors of 1 are [latex]\pm 1[/latex] and the factors of 2 are [latex]\pm 1[/latex] and [latex]\pm 2[/latex]. The possible values for [latex]\frac{p}{q}[/latex] are [latex]\pm 1[/latex] and [latex]\pm \frac{1}{2}[/latex]. These are the possible rational zeros for the function.

We can use synthetic division to show that [latex]\left(x+2\right)[/latex] is a factor of the polynomial.

The remainder is zero, so [latex]\left(x+2\right)[/latex] is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:

[latex]\left(x+2\right)\left({x}^{2}-8x+15\right)[/latex]

We can factor the quadratic factor to write the polynomial as

[latex]\left(x+2\right)\left(x - 3\right)\left(x - 5\right)[/latex]

By the Factor Theorem, the zeros of [latex]{x}^{3}-6{x}^{2}-x+30[/latex] are –2, 3, and 5.

The Rational Zero Theorem tells us that if [latex]\frac{p}{q}[/latex] is a zero of [latex]f\left(x\right)[/latex], then p is a factor of –1 and q is a factor of 4.

[latex]\begin{array}{l}\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}}\hfill \\ \text{}\frac{p}{q}=\frac{\text{Factors of -1}}{\text{Factors of 4}}\hfill \end{array}[/latex]

The factors of –1 are [latex]\pm 1[/latex] and the factors of 4 are [latex]\pm 1,\pm 2[/latex], and [latex]\pm 4[/latex]. The possible values for [latex]\frac{p}{q}[/latex] are [latex]\pm 1,\pm \frac{1}{2}[/latex], and [latex]\pm \frac{1}{4}[/latex].
These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.

Dividing by [latex]\left(x - 1\right)[/latex] gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as [latex]\left(x - 1\right)\left(4{x}^{2}+4x+1\right)[/latex].

The quadratic is a perfect square. [latex]f\left(x\right)[/latex] can be written as [latex]\left(x - 1\right){\left(2x+1\right)}^{2}[/latex].

We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.

[latex]\begin{array}{l}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{array}[/latex]

The zeros of the function are 1 and [latex]-\frac{1}{2}[/latex] with multiplicity 2.

Analysis of the Solution

Look at the graph of the function f. Notice, at [latex]x=-0.5[/latex], the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero –0.5. At [latex]x=1[/latex], the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero [latex]x=1[/latex].

The Rational Zero Theorem tells us that if [latex]\frac{p}{q}[/latex] is a zero of [latex]f\left(x\right)[/latex], then p is a factor of 3 and q is a factor of 3.

[latex]\begin{array}{l}\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factor of the leading coefficient}}\hfill \\ \text{}\frac{p}{q}=\frac{\text{Factors of 3}}{\text{Factors of 3}}\hfill \end{array}[/latex]

The factors of 3 are [latex]\pm 1[/latex] and [latex]\pm 3[/latex]. The possible values for [latex]\frac{p}{q}[/latex], and therefore the possible rational zeros for the function, are [latex]\pm 3, \pm 1, \text{and} \pm \frac{1}{3}[/latex]. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3.

Dividing by [latex]\left(x+3\right)[/latex] gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as [latex]\left(x+3\right)\left(3{x}^{2}+1\right)[/latex].

We can then set the quadratic equal to 0 and solve to find the other zeros of the function.

[latex]\begin{array}{l}3{x}^{2}+1=0\hfill \\ \text{ }{x}^{2}=-\frac{1}{3}\hfill \\ \text{ }x=\pm \sqrt{-\frac{1}{3}}=\pm \frac{i\sqrt{3}}{3}\hfill \end{array}[/latex]

The zeros of [latex]f\left(x\right)[/latex] are –3 and [latex]\pm \frac{i\sqrt{3}}{3}[/latex].

Analysis of the Solution

Look at the graph of the function f. Notice that, at [latex]x=-3[/latex], the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[/latex]. Also note the presence of the two turning points. This means that, since there is a 3^rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of [latex]x=-3[/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[/latex] is three. Either way, our result is correct.

Because [latex]x=i[/latex] is a zero, by the Complex Conjugate Theorem [latex]x=-i[/latex] is also a zero. The polynomial must have factors of [latex]\left(x+3\right),\left(x - 2\right),\left(x-i\right)[/latex], and [latex]\left(x+i\right)[/latex]. Since we are looking for a degree 4 polynomial and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.

[latex]\begin{array}{l}f\left(x\right)=a\left(x+3\right)\left(x - 2\right)\left(x-i\right)\left(x+i\right)\\ f\left(x\right)=a\left({x}^{2}+x - 6\right)\left({x}^{2}+1\right)\\ f\left(x\right)=a\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\right)\end{array}[/latex]

We need to find a to ensure [latex]f\left(-2\right)=100[/latex]. Substitute [latex]x=-2[/latex] and [latex]f\left(2\right)=100[/latex]
into [latex]f\left(x\right)[/latex].

[latex]\begin{array}{l}100=a\left({\left(-2\right)}^{4}+{\left(-2\right)}^{3}-5{\left(-2\right)}^{2}+\left(-2\right)-6\right)\hfill \\ 100=a\left(-20\right)\hfill \\ -5=a\hfill \end{array}[/latex]

So the polynomial function is:

[latex]f\left(x\right)=-5\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\right)[/latex]

or

[latex]f\left(x\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[/latex]

Analysis of the Solution

We found that both i and –i were zeros, but only one of these zeros needed to be given. If i is a zero of a polynomial with real coefficients, then –i must also be a zero of the polynomial because –i is the complex conjugate of i.

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 [latex]f\left(-x\right)[/latex] to determine the number of negative real roots.

[latex]\begin{array}{l}f\left(-x\right)=-{\left(-x\right)}^{4}-3{\left(-x\right)}^{3}+6{\left(-x\right)}^{2}-4\left(-x\right)-12\hfill \\ f\left(-x\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12\hfill \end{array}[/latex]

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

There are four possibilities, as we can see below.

Analysis of the Solution

We can confirm the numbers of positive and negative real roots by examining a graph of the function. We can see from the graph that the function has 0 positive real roots and 2 negative real roots.

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

[latex]\begin{array}{l}V=\left(w+4\right)\left(w\right)\left(\frac{1}{3}w\right)\\ V=\frac{1}{3}{w}^{3}+\frac{4}{3}{w}^{2}\end{array}[/latex]

Substitute the given volume into this equation.

[latex]\begin{array}{l}\text{ }351=\frac{1}{3}{w}^{3}+\frac{4}{3}{w}^{2}\hfill & \text{Substitute 351 for }V.\hfill \\ 1053={w}^{3}+4{w}^{2}\hfill & \text{Multiply both sides by 3}.\hfill \\ \text{ }0={w}^{3}+4{w}^{2}-1053 \hfill & \text{Subtract 1053 from both sides}.\hfill \end{array}[/latex]

Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351[/latex], and [latex]\pm 1053[/latex]. 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 [latex]x=1[/latex].

Since 1 is not a solution, we will check [latex]x=3[/latex].

Since 3 is not a solution either, we will test [latex]x=9[/latex].

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.

[latex]l=w+4=9+4=13\text{ and }h=\frac{1}{3}w=\frac{1}{3}\left(9\right)=3[/latex]

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