TBIL-Precal Zeros of Polynomial Functions (PR5)

Section 4.5 Zeros of Polynomial Functions (PR5)

Objectives

Determine the zeros of a polynomial function with real coefficients.
🔗

🔗

Subsection 4.5.1 Activities

Remark 4.5.1.

Recall that to find the \(x\)-intercepts of a function \(f(x)\text{,}\) we need to find the values of \(x\) that make \(f(x)=0\text{.}\) We saw in Section 1.5 that the zero product property (Definition 1.5.3) was helpful when \(f(x)\) is a polynomial that we can factor. Let’s consider how we approach polynomials which are not in factored form or easily factorable.

🔗

Activity 4.5.2.

Consider the function \(f(x)=x^3 - 7 x^2 + 7 x + 15\text{.}\)

🔗

(a)

Use polynomial division from Section 4.4 to divide \(f(x)\) by \(x-2\text{.}\) What is the remainder?

🔗

Answer.

\(9\)

🔗

(b)

Find \(f(2)\text{.}\) What do you notice?

🔗

Answer.

\(f(2)=9\) Remainder and value are the same

🔗

Theorem 4.5.3. Remainder Theorem.

If a polynomial \(p(x)\) is divided by \(x-a\) then the remainder is equal to \(p(a)\text{.}\)

🔗

Activity 4.5.4.

Again consider the function \(f(x)=x^3 - 7 x^2 + 7 x + 15\text{.}\)

🔗

(a)

Divide \(f(x)\) by \(x-3\text{.}\) What is the remainder?

🔗

Answer.

\(0\)

🔗

(b)

Find \(f(3)\text{.}\) What do you notice?

🔗

Answer.

\(f(3)=0\) When value is 0 there is no remainder

🔗

Theorem 4.5.5. Factor Theorem.

A number \(a\) is a zero of a polynomial function \(p(x)\) (that is, \(p(a)=0\)) if and only if \(x-a\) is a factor of \(p(x)\text{.}\)

🔗

Remark 4.5.6.

If we know one zero, then we can divide by \(x-a\) where \(a\) is a zero. After this, the quotient will have smaller degree and we can work on factoring the rest. We can “chip away” at the polynomial one zero at a time.

🔗

Activity 4.5.7.

One more time consider the function \(f(x)=x^3 - 7 x^2 + 7 x + 15\text{.}\)

🔗

(a)

We already know from Activity 4.5.4 that \(x-3\) is a factor of the polynomial \(f(x)\text{.}\) Use division to express \(f(x)\) as \((x-3)\cdot q(x)\text{,}\) where \(q(x)\) is a quadratic function.

🔗

\(\displaystyle q(x)=x^2-2x-3\)

🔗
\(\displaystyle q(x)=x^2-10x-37\)

🔗
\(\displaystyle q(x)=x^2-4x-5\)

🔗
\(\displaystyle q(x)=x^2+4x-5\)

🔗

Answer.

🔗

(b)

Notice that \(q(x)\) is something we can factor. Factor this quadratic and find the remaining zeros.

🔗

\(\displaystyle -5\)

🔗
\(\displaystyle 5\)

🔗
\(\displaystyle 4\)

🔗
\(\displaystyle -1\)

🔗
\(\displaystyle 3\)

🔗
\(\displaystyle 1\)

🔗

Answer.

B and F

🔗

Remark 4.5.8.

We were able to find all the zeros of the polynomial in Activity 4.5.7 because we were given one of the zeros. If we don’t have a zero to help us get started (or need more than one zero for a function of higher degree), we have a couple of options.

🔗

Activity 4.5.9.

Consider the function \(f(x) = 18 x^4 + 67 x^3 - 81 x^2 - 202 x + 168\text{.}\)

🔗

(a)

Graph the function. According to the graph, what value(s) seem to be zeros?

🔗

Answer.

\(-4\) and \(-2\)

🔗

(b)

Use the Remainder Theorem to confirm that your guesses are actually zeros.

🔗

Answer.

\(f(-4)=0\) and \(f(-2)=0\)

🔗

(c)

Now use these zeros along with polynomial division to rewrite the function as \(f(x)=(x-a)(x-b)q(x)\) where \(a\) and \(b\) are zeros and \(q(x)\) is the remaining quadratic function.

🔗

Answer.

\((x+4)(x+2)(18x^2-41x+21)\)

🔗

(d)

Solve the quadratic \(q(x)\) to find the remaining zeros.

🔗

Answer.

\(\dfrac{3}{2}\) and \(\dfrac{7}{9}\)

🔗

(e)

List all zeros of \(f(x)\text{.}\)

🔗

Answer.

\(-4, -2, \dfrac{7}{9}, \dfrac{3}{2}\)

🔗

(f)

Rewrite \(f(x)\) as a product of linear factors.

🔗

Answer.

\((x+4)(x+2)(2x-3)(9x-7)\)

🔗

Remark 4.5.10.

Using the graph to find an initial zero can be helpful, but they may not always be easy to identify.

🔗

Activity 4.5.11.

Consider the quadratic function \(f(x)=(2x-5)(3x-8)=6x^2-31x+40\text{.}\)

🔗

(a)

What are the roots of this quadratic?

🔗

Answer.

\(\dfrac{5}{2}\) and \(\dfrac{8}{3}\)

🔗

(b)

What do you notice about these roots in relation to the factors of \(a=6\) and \(c=40\) in \(f(x)=6x^2-31x+40\text{?}\)

🔗

Answer.

The numerator is a factor of \(c\) and the denominator is a factor of \(a\text{.}\)

🔗

Remark 4.5.12.

In Activity 4.5.11 we found that the roots were both factors of the constant term divided by factors of the leading coefficient. This can be extended to polynomials of larger degree.

🔗

Theorem 4.5.13. Rational Root Theorem.

If a polynomial \(p(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_1x+a_0 \) has integer coefficients, then the rational zeros have the form \(\dfrac{p}{q}\) where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\text{.}\)

🔗

Activity 4.5.14.

Consider the polynomial \(f(x)=5x^3-2x^2+20x-8\text{.}\)

🔗

(a)

List the factors of the constant term.

🔗

Answer.

\(\pm 8, \pm 4, \pm 2 \pm 1\)

🔗

(b)

List the factors of the leading coefficient.

🔗

Answer.

\(\pm 5, \pm 1\)

🔗

(c)

Use parts (a) and (b) to list all the possible rational roots.

🔗

Answer.

\(\pm 8, \pm 4, \pm 2 \pm 1, \pm \dfrac{8}{5}, \pm \dfrac{4}{5}, \pm \dfrac{2}{5}, \pm \dfrac{1}{5}\)

🔗

(d)

Use the Remainder Theorem to determine at least one root of \(f(x)\text{.}\)

🔗

Answer.

\(\dfrac{2}{5}\)

🔗

Activity 4.5.15.

Consider the polynomial \(f(x)=6x^4+5x^3-6x-5\)

🔗

(a)

Use the graph and the Rational Root Theorem (Theorem 4.5.13) to find the rational zeros of \(f(x)\text{.}\)

🔗

Answer.

\(1\) and \(-\dfrac{5}{6}\)

🔗

(b)

Use the roots, along with the Factor Theorem, to simplify the polynomial into linear and quadratic factors.

🔗

Answer.

\((6x+5)(x-1)(x^2+x+1)\)

🔗

(c)

Find the zeros of the quadratic factor.

🔗

Answer.

\(\dfrac{-1\pm \sqrt{3}}{2}\)

🔗

(d)

List the roots of the polynomial.

🔗

Answer.

\(\dfrac{-1+ \sqrt{3}}{2}, \dfrac{-1- \sqrt{3}}{2}, 1, \) and \(-\dfrac{5}{6}\)

🔗

Remark 4.5.16.

Notice that the zeros of the quadratic factor were imaginary and are related. This also occured in Activity 1.5.21.

🔗

Theorem 4.5.17. Conjugate Zeros Theorem.

Let \(p(x)\) be a polynomial function with real coefficients. If \(a+bi\) is a complex zero of the function, then the conjugate \(a-bi\) is also a zero of the function. These two zeroes are called conjugate zeros, or a conjugate pair of zeros.

🔗