Example: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros

Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, i, such that [latex]f\left(-2\right)=100[/latex].

Answer: 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.

Example: Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for [latex]f\left(x\right)=-{x}^{4}-3{x}^{3}+6{x}^{2}-4x - 12[/latex].

Answer: 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.

Positive Real Zeros	Negative Real Zeros	Complex Zeros	Total Zeros
2	2	0	4
2	0	2	4
0	2	2	4
0	0	4	4

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.

Example: Solving Polynomial Equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

Answer: 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.