Lesson 5.5 : Zeros of Polynomial Functions

New Concept

This lesson equips you to find all zeros of a polynomial. You'll use theorems to test for rational zeros ($x=\frac{p}{q}$), factor polynomials, and apply the Fundamental Theorem of Algebra to account for all real and complex roots.

What’s next

Soon, you will dive into interactive examples and practice cards to apply these powerful theorems and find all the zeros of polynomial functions.

Property

If a polynomial $f(x)$ is divided by $x - k$, then the remainder is the value $f(k)$.

To evaluate $f(x)$ at $x=k$ using the Remainder Theorem:

1. Use synthetic division to divide the polynomial by $x - k$.
2. The remainder is the value $f(k)$.

Examples

• To evaluate $f(x) = x^3 - 2x^2 + 5x - 4$ at $x=3$, we divide by $x-3$. Using synthetic division, the remainder is $14$, so $f(3) = 14$.

Property

According to the Factor Theorem, $k$ is a zero of $f(x)$ if and only if $(x - k)$ is a factor of $f(x)$.

To use the Factor Theorem:

1. Use synthetic division to divide the polynomial by $(x - k)$.
2. Confirm that the remainder is 0.
3. Write the polynomial as the product of $(x - k)$ and the quadratic quotient.
4. If possible, factor the quadratic.
5. Write the polynomial as the product of factors.

Examples

• To show that $(x - 1)$ is a factor of $f(x) = x^3 + 2x^2 - x - 2$, synthetic division with $k=1$ yields a remainder of 0. The remaining factors are $(x+1)$ and $(x+2)$.

Property

The Rational Zero Theorem states that, if the polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ has integer coefficients, then every rational zero of $f(x)$ has the form $\frac{p}{q}$ where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$.

When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.

Examples

• For $f(x) = 3x^3 + 2x^2 - 7x + 2$, the factors of the constant term 2 are $p = \pm 1, \pm 2$. The factors of the leading coefficient 3 are $q = \pm 1, \pm 3$. The possible rational zeros $\frac{p}{q}$ are $\pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}$.

Property

To find the zeros of a polynomial function $f$:

1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
4. Find the zeros of the quadratic function by factoring or using the quadratic formula.

Examples

• To find the zeros of $f(x) = x^3 - 4x^2 + x + 6$, we test possible rational zeros. We find $x = 2$ is a zero. The quotient is $x^2 - 2x - 3$, which factors to $(x-3)(x+1)$. The zeros are $2, 3, -1$.

• Find all zeros of $f(x) = 2x^3 - x^2 - 13x - 6$. Possible rational zeros include $\pm 1, \pm 2, \dots$. Testing shows $x=-2$ is a zero. The remaining quadratic is $2x^2-5x-3=0$, which gives $x=3$ and $x=-\frac{1}{2}$.

Property

The Fundamental Theorem of Algebra states that, if $f(x)$ is a polynomial of degree $n > 0$, then $f(x)$ has at least one complex zero.

We can use this theorem to argue that, if $f(x)$ is a polynomial of degree $n > 0$, and $a$ is a non-zero real number, then $f(x)$ has exactly $n$ linear factors

Property

According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form $(x - c)$, where $c$ is a complex number.

If the polynomial function $f$ has real coefficients and a complex zero in the form $a + bi$, then the complex conjugate of the zero, $a - bi$, is also a zero. This is the Complex Conjugate Theorem.

Examples

• Find a polynomial with real coefficients and zeros at 2 and $3i$. By the Complex Conjugate Theorem, $-3i$ must also be a zero. So, $f(x) = a(x-2)(x-3i)(x+3i) = a(x-2)(x^2+9)$.

Property

According to Descartes’ Rule of Signs, if we let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ be a polynomial function with real coefficients:

• The number of positive real zeros is either equal to the number of sign changes of $f(x)$ or is less than the number of sign changes by an even integer.
• The number of negative real zeros is either equal to the number of sign changes of $f(-x)$ or is less than the number of sign changes by an even integer.

Examples

• For $f(x) = x^3 - 2x^2 + 3x - 4$, the signs are $+, -, +, -$. There are 3 sign changes, so there are 3 or 1 positive real zeros.

• For $f(x) = x^4 + 2x^3 + x^2 + 5x + 1$, the signs are all positive. There are 0 sign changes, so there are no positive real zeros.