1. Objective
Use long division and synthetic division to divide polynomials.

2. Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.

3. Example 1: Using Long Division to Divide a Polynomial
Divide using long division.
(–y2 + 2y3 + 25) ÷ (y – 3)
Step 1 Write the dividend in standard form, including
terms with a coefficient of 0.
Step 2 Write division in the same way you would when dividing numbers.

4. Example 1 Continued
Step 3 Divide.

5. Example 1 Continued
Step 4 Write the final answer.

6. Check It Out! Example 1a
Divide using long division.
(15x2 + 8x – 12) ÷ (3x + 1)

7. Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

9. Example 2B: Using Synthetic Division to Divide by a Linear Binomial
Divide using synthetic division.
(3x4 – x3 + 5x – 1) ÷ (x + 2)
Step 1 Find a.
Step 2 Write the coefficients and a in the synthetic division format.

10. Example 2B Continued
Step 3 Bring down the first coefficient. Then multiply and add for each column.
Step 4 Write the quotient.

11. Check It Out! Example 2a
Divide using synthetic division.
(6x2 – 5x – 6) ÷ (x + 3)

12. You can use synthetic division to evaluate polynomials
You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.

13. Example 3A: Using Synthetic Substitution
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = 2x3 + 5x2 – x + 7 for x = 2.
Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.
P(2) = 2(2)3 + 5(2)2 – (2) + 7
P(2) = 41 

14. Check It Out! Example 3a
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = x3 + 3x2 + 4 for x = –3.
Check Substitute –3 for x in P(x) = x3 + 3x2 + 4.
P(–3) = (–3)3 + 3(–3)2 + 4
P(–3) = 4 