Lesson Objectives
• Demonstrate an understanding of synthetic division
• Learn how to evaluate a polynomial function using the remainder theorem

## How to Evaluate a Polynomial Function Using the Remainder Theorem

As we progress through the remainder of this section, we will encounter various techniques for working with polynomial functions. In many cases, we will need to rely on our knowledge of synthetic division.

### Review of Synthetic Division

Synthetic Division is a shortcut for dividing polynomials when the divisor is of the form x - k. Notice that x - k is a binomial where the coefficient of x is 1 and the exponent on x is also 1.
1. Write the dividend in standard form (descending powers)
• Use 0 as the coefficient for any missing terms, including a missing constant term
2. Write the divisor in the form of x - k, obtain the value for k
• Ex: x - 3, here k = 3
• Ex: x + 7, rewrite: x - (-7), here k = -7
3. Set up the division, we will refer to this as row 1
• k will go to the left of the long division symbol
• Write the coefficients of the dividend underneath the long division symbol
• Set up row 2, directly below row 1, and place a horizontal line below
4. Bring down the leading coefficient of the dividend into row 3, beneath the horizontal line
5. Multiply k by the value just written in row 3
6. Write the product in the next column (moving right) in row 2
7. Add the values in the new column and write the sum in row 3
8. Repeat steps 5 - 7 until all the columns are completed
9. We will use the numbers from row 3 to write our answer
• The degree of the first term of the quotient is one degree less than the degree of the dividend
• The final value is the remainder, which could be zero
Let's use this process with an example.
Example #1: Use synthetic division to perform the division. $$(x^3 - 13x^2 + 35x + 25) \, ÷ \, (x - 5)$$ 1) Write the dividend in standard form (descending powers), and use 0 as the coefficient for any missing terms. Here, we already have our dividend in standard form and we are not missing any terms.
2) Write the divisor in the form of x - k and obtain the value for k. Here, we already have this form, we can see that x - 5, gives us a k-value of 5.
3) Set up the division: From the image above, we can see that we placed 5, which is k, to the left of the long division symbol. Additionally, we wrote the coefficients of the dividend (1, -13, 35, and 25) underneath the long division symbol. Now, we will set up row 2 and place a horizontal line below. 4) Bring down the leading coefficient of the dividend into row 3, beneath the horizontal line. 5) Multiply k by the value just written in row 3. Here, k is 5, and the value just written into row 3 is 1. $$5 \cdot 1 = 5$$ 6) Write the product in the next column (moving right) in row 2. The product of 5 is going directly underneath the -13. 7) Add the values in the new column and write the sum in row 3. $$-13 + 5 = -8$$ 8) Repeat steps 5 - 7 until all the columns are completed. $$5 \cdot -8 = -40$$ $$35 + (-40) = -5$$ $$5 \cdot -5 = -25$$ $$25 + (-25) = 0$$ 9) We will use the numbers from row 3 to write our answer. We have the numbers 1, -8, -5, and 0 in that order. The degree of the first term of the quotient is one degree less than the degree of the dividend. Our original dividend had a degree of 3, so our quotient will have a degree of 2. This means the first number, which is 1, is the coefficient for x2, followed by -8, which is the coefficient for x, and then -5, which is the constant term. The 0 is the remainder here, which means we don't have a remainder. $$\frac{x^3 - 13x^2 + 35x + 25}{x - 5} = x^2 - 8x - 5$$ Remember, we can always check our answer with multiplication. When there is no remainder, the quotient times the divisor will give us the dividend back. $$(x^2 - 8x - 5)(x - 5) = x^3 - 13x^2 + 35x + 25 \, ✓$$ Example #2: Use synthetic division to perform the division. $$(x^3 + 9x^2 - 1) \, ÷ \,(x + 9)$$ Let's rewrite our problem as: $$(x^3 + 9x^2 + 0x - 1) \, ÷ \, (x - (-9))$$ Notice that we used 0 as the coefficient for x, since this is a missing term. Additionally, we rewrote x + 9 as x - (-9), where k will be -9. Looking at row 3, we can see that we have 1 as the coefficient for x2, 0 as the coefficient for x, 0 as our constant, and -1 as the remainder. $$\frac{x^3 + 9x^2 - 1}{x + 9} = x^2 - \frac{1}{x + 9}$$ To check, we multiply the quotient (x2) by the divisor (x + 9), and then add the remainder (-1) to the result. If our answer is correct, we will get the dividend (x3 + 9x2 - 1) as a result. $$(x^2)(x + 9) + (-1) = x^3 + 9x^2 - 1 \, ✓$$

### Remainder Theorem

The remainder theorem gives us a shortcut for evaluating a polynomial function at a given value. For any polynomial f(x) and any complex number k, there exists a unique polynomial q(x) and a number r such that: $$f(x) = (x - k)q(x) + r$$ When our polynomial function is written in this format, it is obvious that f(k) = r: $$f(x)=(x - k)q(x) + r$$ $$f(k)=(k - k)q(k) + r =0 + r=r$$ This result is known as the remainder theorem. This tells us that if the polynomial function f(x) is divided by (x - k), the remainder is the same as f(k). We can use this to quickly evaluate a polynomial function for a given value. Let's look at an example.
Example #3: Evaluate f(x) at k. $$f(x)=11x^3 + 6x^2 - 14$$ $$k=-1$$ First, let's plug in a -1 for x and see what we get: $$f(-1)=11(-1)^3 + 6(-1)^2 - 14$$ $$=-11 + 6 - 14=-19$$ Now, we will see that we get the same result as the remainder from synthetic division. Notice how the remainder is -19. This tells us that f(-1) = -19.

### Using the Remainder Theorem to Test Potential Zeros

A zero of a polynomial function is some number k such that f(k) = 0. Using the remainder theorem, we can quickly decide whether a number k is a zero of a polynomial function. A zero of f(x) is referred to as a zero, a root, or a solution to f(x) = 0. All we need to do is set up a synthetic division to find f(k), if the remainder is zero, then we can conclude that k is a zero of f(x). Let's look at a few examples.
Example #4: Determine if k is a zero of f(x). $$f(x) = x^3 + 2x^2 - 2x + 2$$ $$k = -3$$ From the synthetic division, we can see that f(-3) = -1, which is not zero. Therefore, we can conclude that our k-value of -3 is not a zero.
Example #5: Determine if k is a zero of f(x). $$f(x) = x^3 + 3x^2 - 20x - 100$$ $$k = -4 + 2i$$ From the synthetic division, we can see that f(-4 + 2i) = 0. Therefore, we can conclude that our k-value of -4 + 2i is a zero.

#### Skills Check:

Example #1

Evaluate f(x) at k. $$f(x)=13x^3 + 7x^2 - x - 6$$ $$k=1$$

A
$$f(k)=-5$$
B
$$f(k)=9$$
C
$$f(k)=10$$
D
$$f(k)=15$$
E
$$f(k)=13$$

Example #2

Evaluate f(x) at k. $$f(x)=x^3 + 19x^2 + 74x + 56$$ $$k=-2$$

A
$$f(k)=8$$
B
$$f(k)=10$$
C
$$f(k)=-11$$
D
$$f(k)=-24$$
E
$$f(k)=19$$

Example #3

Determine if k is a zero. $$f(x)=2x^4 + 5x^2 - 12$$ $$k= 2i$$