Evaluate Polynomials
Evaluating a Polynomial Using the Remainder Theorem
Previously, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by [latex]x - k[/latex], the remainder may be found quickly by evaluating the polynomial function at [latex]k[/latex], that is, [latex]f(k)[/latex].
the remainder theorem
If a polynomial [latex]f(x)[/latex] is divided by [latex]x−k[/latex], then the remainder is the value [latex]f(k)[/latex].
Let’s walk through the proof of the theorem.
Recall that the Division Algorithm states that, given a polynomial dividend [latex]f(x)[/latex] and a non-zero polynomial divisor [latex]d(x)[/latex], there exist unique polynomials [latex]g(x)[/latex] and [latex]r(x)[/latex] such that
[latex]f(x) = d(x)g(x) + r(x)[/latex]
and either [latex]r(x) = 0[/latex] or the degree of [latex]r(x)[/latex] is less than the degree of [latex]d(x)[/latex]. In practice divisors, [latex]d(x)[/latex] will have degrees less than or equal to the degree of [latex]f(x)[/latex]. If the divisor, [latex]d(x)[/latex], is [latex]x - k[/latex], this takes the form
[latex]f(x) = (x - k)g(x) + r[/latex]
Since the divisor [latex]x - k[/latex] is linear, the remainder will be a constant, [latex]r[/latex]. And, if we evaluate this for [latex]x = k[/latex], we have
[latex]\begin{array}{rl} f(k) & = (k - k)g(k) + r \\ & = 0 \cdot g(k) + r \\ & = r \end{array}[/latex]
In other words, [latex]f(k)[/latex] is the remainder obtained by dividing [latex]f(x)[/latex] by [latex]x - k[/latex].
- Use synthetic division to divide the polynomial by [latex]x−k[/latex].
- The remainder is the value [latex]f(k)[/latex].
Use the remainder theorem to evaluate [latex]f(x) = 6x^4 - x^3 - 15x^2 + 2x - 7[/latex] at [latex]x = 2[/latex].
Using the Factor Theorem to Solve a Polynomial Equation
The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors.
the factor theorem
According to the factor theorem, [latex]k[/latex] is a zero of [latex]f(x)[/latex] if and only if [latex](x−k)[/latex] is a factor of [latex]f(x)[/latex].
Let’s walk through the proof of the theorem.
Recall that the Division Algorithm.
[latex]f(x) = (x - k)q(x) + r[/latex]
If [latex]k[/latex] is a zero, then the remainder [latex]r[/latex] is [latex]f(k) = 0[/latex] and [latex]f(x) = (x - k)q(x) + 0[/latex] or [latex]f(x) = (x - k)q(x)[/latex].
Notice, written in this form, [latex]x - k[/latex] is a factor of [latex]f(x)[/latex]. We can conclude if [latex]k[/latex] is a zero of [latex]f(x)[/latex], then [latex]x - k[/latex] is a factor of [latex]f(x)[/latex].
Similarly, if [latex]x - k[/latex] is a factor of [latex]f(x)[/latex], then the remainder of the Division Algorithm [latex]f(x) = (x - k)q(x) + r[/latex] is [latex]0[/latex]. This tells us that [latex]k[/latex] is a zero.
This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree [latex]n[/latex] in the complex number system will have [latex]n[/latex] zeros. We can use the Factor Theorem to completely factor a polynomial into the product of [latex]n[/latex] factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.
- Use synthetic division to divide the polynomial by [latex](x-k)[/latex]
- Confirm that the remainder is [latex]0[/latex].
- Write the polynomial as the product of [latex](x-k)[/latex] and the quadratic quotient.
- If possible, factor the quadratic.
- Write the polynomial as the product of factors.
Show that [latex](x + 2)[/latex] is a factor of [latex]x^3 - 6x^2 - x + 30[/latex]. Find the remaining factors. Use the factors to determine the zeros of the polynomial.
