Section 3.3

Factor Theorem and Rational Zero Theorem

Theorem 1 (Factor Theorem)

For any polynomial \(f(x)\), \(x-k\) is a factor of \(f(x)\) if and only if \(f(k)=0\).

Theorem 2 (Rational Zero Theorem)

Let \(f(x)\) be a polynomial function

\[f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0\]

If \(f(\frac{p}{q})=0\), then \(p\) is proportional to \(a_0\) and \(q\) is proportional to \(a_n\).

From the previous theorem, it is important to remember that it is \(\frac{p}{q}\) where \(p\) is associated with \(a_0\) (the constant term) and \(q\) is associated with \(a_n\) (the leading coefficient).

Example(s)

Example 53

Let \(f(x)=18x^6-21x^5-49x^4+21x^3+35x^2-4\).

List all possible rational zeros (or roots).

Solution:

Remember \(p\) is associated with the constant value, \(-4\). Also, \(q\) is associated with the leading coefficient, \(18\). Next, we will list all the factors of for \(p\) and \(q\).

\[p=-4:\pm 1, \pm 2, \pm 4\]

\[q=18: \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\]

Next, we will list all the possible rational zeros with redundant ratios.

\[\begin{alignat*}{3} \pm\frac{1}{1}, & \pm\frac{1}{2}, & \pm\frac{1}{3}, & \pm\frac{1}{6}, & \pm\frac{1}{9}, & \pm\frac{1}{18},\\ \pm\frac{2}{1}, & \pm\frac{2}{2}, & \pm\frac{2}{3}, & \pm\frac{2}{6}, & \pm\frac{2}{9}, & \pm\frac{2}{18},\\ \pm\frac{4}{1}, & \pm\frac{4}{2}, & \pm\frac{4}{3}, & \pm\frac{4}{6}, & \pm\frac{4}{9}, & \pm\frac{4}{18} \end{alignat*}\]

Eliminating redundancy we have the following as the possible rational roots.

\[\pm 1, \pm 2, \pm 4, \pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{6},\pm \frac{1}{9},\pm \frac{1}{18}\]

\[\pm \frac{2}{3},\pm \frac{2}{9},\pm \frac{4}{3}, \pm \frac{4}{9}\]

Factor \(f(x)\) into products of linear binomials.

Solution:

From the list of possible rational zeros, we will do synthetic division.

We will first try \(x=1\) (or \(k=1\)). After synthetic division we will discover \(f(1)=0\) and \(q_1(x)=18x^5-3x^4-52x^3-31x^2+4x+4\). That is,

\[\begin{align*} f(x) & = (x-1)q_1(x)\\ & = (x-1)(18x^5-3x^4-52x^3-31x^2+4x+4) \end{align*}\]

Next, we will do synthetic division on \(q_1(x)\) using \(k=-1\). We will get \(q_1(-1)=0\) and \(q_2(x)=18x^4-21x^3-31x^2+4\). That is,

\[\begin{align*} f(x) & = (x-1)q_1(x)\\ & = (x-1)\left((x+1)q_2(x)\right)\\ & = (x-1)(x+1)(18x^4-21x^3-31x^2+4) \end{align*}\]

Next, we will do synthetic division on \(q_2(x)\) using \(k=2\). We will get \(q_2(2)=0\) and \(q_3(x)=18x^3+15x^2-x-2\). That is,

\[\begin{align*} f(x) & = (x-1)(x+1)q_2(x)\\ & = (x-1)(x+1)\left((x-2)q_3(x)\right)\\ & = (x-1)(x+1)(x-2)(18x^3+15x^2-x-2) \end{align*}\]

Next, we will do synthetic division on \(q_3(x)\) using \(k=-2\). We will get \(q_3(-2)=-84\). This means \(x=-2\) is not a zero of the original function. We then move down the list of possible rational zeros. We will find

\[\begin{align*} q_3(4) & = 1386\\ q_3(-4) & = -910\\ q_3(\frac{1}{2})) & = \frac{7}{2}\\ q_3(-\frac{1}{2})) & = 0 \end{align*}\]

We finally discover that \(q_3(-\frac{1}{2})=0\). This means we will use synthetic division on \(q_3(x)\) with \(k=-\frac{1}{2}\). We will then get \(q_4(x)=18x^2+6x-4\). That is,

\[\begin{align*} f(x) & = (x-1)(x+1)(x-2)q_3(x)\\ & = (x-1)(x+1)(x-2)\left((x+\frac{1}{2})q_4(x)\right)\\ & = (x-1)(x+1)(x-2)(x+\frac{1}{2})(=18x^2+6x-4) \end{align*}\]

To find the last factors we then factor \(=18x^2+6x-4\). First, \(18\cdot(-4)=-72\), \(12\cdot (-6)=-72\), and \(12-6=6\). With that information, we can begin to factor.

\[\begin{align*} 18x^2+6x-4 & = 18x^2+12x-6x-4\\ & = 6x(3x+2)-2(3x+2)\\ & = (3x+2)(6x-2)\\ & = (6x-2)(3x+2)\\ & = 2(3x-1)(3x+2) \end{align*}\]

Putting everything together we have:

\[f(x)=2(x-1)(x+1)(x-2)(x+\frac{1}{2})(3x-1)(3x+2)\]

List all the zeros for the function \(f(x)\)

Solution:

Since we know

\[f(x)=2(x-1)(x+1)(x-2)(x+\frac{1}{2})(3x-1)(3x+2)\]

The zeros of the function \(f\) is: \(1\), \(-1\), \(2\), \(-\frac{1}{2}\), \(\frac{1}{3}\), and \(-\frac{2}{3}\).

Fundamental Theorem of Algebra and Number of Zeros Theorem

Theorem 3 (Fundamental Theorem of Algebra)

Every function defined by a polynomial of degree 1 or more has at least one complex root.

Theorem 4 (Number of Zeros Theorem)

A function defined by a polynomial of degree \(n\) has at most \(n\) distinct zeros.

Definition 39 (Multiplicity)

The number of times a zero occurs is called the multiplicity of the zero.

Example 54

Let \(f(x)=(x+1)^3(x+3)(x-2)^2\). We know that the function’s zeros are \(-1\), \(-3\), and \(2\). However, we can also say that \(f\) has a zero at \(x=-1\) with a multiplicity of \(3\), \(x=-3\) with a multiplicity of \(1\), and \(x=2\) with a multiplicity of \(2\). The power of the individual factors determines the multiplicity.

Example 55

Find a polynomial of degree three with real coefficient, zeros at \(x=-3\), \(x=-2\), and \(x=5\), and \(f(-1)=6\).

Solution:

We know that with the zeros

\[f(x)=a(x+3)(x+2)(x-5)\]

In order to find \(a\) we use \(f(-1)=6\). That is, solve the following equation.

\[\begin{align*} a(-1+3)(-1+2)(-1-5) & = 6\\ a(2)(1)(-6) & = 6\\ a & = -\frac{6}{-12} = -\frac{1}{2} \end{align*}\]

Therefore, \(f(x)=-\frac{1}{2}(x+3)(x+2)(x-5)\). In expanded form, we have:

\[f(x)=-\frac{1}{2} x^3 + \frac{19}{2} x + 15\]

Complex Numbers

Remember \(\sqrt{-1}=i\) and the following hold true.

\[\begin{align*} i & = \sqrt{-1}\\ i^2 & = -1\\ i^3 & = -i\\ i^4 & = 1 \end{align*}\]

Definition 40 (Complex Conjugate)

Let \(z=a+ib\) where \(a\) and \(b\) are real numbers and \(i=\sqrt{-1}\). Then we say \(\overline{z}=a-ib\) and \(\overline{z}\) is the conjugate of \(z\).

Property 4 (Conjugate Properties)

Let \(c\) and \(d\) be complex numbers. Then

• \(\overline{c+d} = \overline{c}+\overline{d}\)

• \(\overline{c\cdot d} = \overline{c}\cdot\overline{d}\)

• \(\overline{c^n} = \left(\overline{c}\right)^n\)

Theorem 5 (Conjugate Zero Theorem)

If \(f(x)\) defines a polynomial having only real coefficients and if \(z=a+ib\) is a zero of \(f(x)\), then \(\overline{z}=a-ib\) is also a zero.

Example 56

If \(f(x)=x^2-8x+25\). Find \(f(4+i3)\)

Solution:

\[\begin{align*} f(4+i3) & = (4+i3)^2 - 8(4+i3) + 25\\ & = 7 + i24 - 32 - i24 + 25\\ & = 0 \end{align*}\]

This means \(x=4+i3\) is a zero for the function.

Since \(f(4+i3)\) is ??. Find all the zeros for \(f\).

Solution:

Since \(x=4+i3\) is a zero by the Conjugate Zero Theorem we know \(x=4-i3\) is also a zero. That is,

\[\begin{align*} f(4-i3) & = (4-i3)^2 - 8(4-i3)+25\\ & = 7-i24-32+i24+25\\ & = 0 \end{align*}\]

Therefore, the zeros for \(f\) is \(x=4+i3\) and \(x=4-i3\).

Example 57

Find the polynomial with the least degree having only real coefficient and zeros \(x=-4\) and \(x=3-i\).

Solution:

Since the polynomial has all real coefficients and \(x=3-i\) is a zero by the conjugate zero theorem. Since \(x=-4\), \(x=3-i\), and \(x=3+i\) are the roots we have

\[f(x)=(x-(-4))(x-(3-i))(x-(3+i))\]

If we distribute we have:

\[f(x)= x^3 -2x^2 -14x +40\]

Example 58

Given \(f(2+i)=0\), find all the zeros for the function \(f(x)=x^4-x^3-17x^2+55x-50\).

Since we know \(f(2+i)=0\) we have the following from polynomial long division (using synthetic division).

Solution:

\[ f(x) = (x-(2+i))(x^3 + (1+i)x^2 + (-16+3i)x + (20-i10))\]

Next, using conjugate zero theorem we know \(f(2-i)=0\). Using synthetic division we know.

Solution:

\[ f(x)= (x-(2+i))(x-(2-i))(x^2+3x-10)\]

Next, using the remaining factor: \(x^2+3x-10\) we can find the remaining zeros.

Solution:

\[x^2+3x-10=(x+5)(x-2)\]

and \(x=-5\), \(x=2\) are the last zeros.

With all the previous work we know the zeros for \(f(x)=x^4-x^3-17x^2+55x-50\) is \(x=\{2,-5,2+i,2-i\}\).