1. Part 2: Synthetic Division & The Remainder Theorem

2. Synthetic Division
Synthetic Division is a process that simplifies long
division, but it can only be used when dividing a
polynomial by a linear factor of the form x – a.

3. Synthetic Division
1. Write the polynomial in standard form, including zero
coefficients where appropriate
2. Set up: use the opposite sign of a (this allows us to add
throughout the process) and write the coefficients of
the polynomial.
3. Bring down the first coefficient
4. Multiply the coefficient by the divisor. Add to the next
coefficient.
5. Continue multiplying and adding through the last
coefficient.
6. Write the quotient and remainder. The remainder will
be the last sum.

4. Example: Divide using synthetic division
(x 3
− 57 x + 56 ) ÷ ( x − 7 )

5. Example: Divide using synthetic division
(x 3
− 14 x + 51x − 54 ) ÷ ( x + 2 )
2

6. Example: Using synthetic division to solve a
problem
The polynomial x 3 + 7 x 2 − 38 x − 240
expresses the volume, in cubic inches, of the shadow
box shown.
1. What are the dimensions of
the box?
Hint: the length is greater than
the height (or depth)
2. If the width of the box is 15 in,
what are the other dimension?

7. The Remainder Theorem
The Remainder Theorem provides a quick way to
find the remainder of a polynomial long-division
problem.
If you divide a polynomial P(x) of degreen ≥ 1
by x − a then the remainder is P(a)

8. Example: Evaluating a Polynomial
Given that P ( x ) = x − 2x − x + 2
5 3 2
what is P(3)?
By the remainder theorem, P(3) is the remainder
when you divide P(x) by x – 3.

9. Example: Evaluating a Polynomial
Given that P ( x ) = x − 3 x − 28 x + 5 x + 20
5 4 3
what is P(─ 4)?
By the remainder theorem, P(─ 4) is the remainder
when you divide P(x) by x + 4.

10. Homework
P308 #21 – 39 odd, 40 – 43 all, 53 – 56 all, 57 – 61 odd