1. 7.1 Synthetic Division & the Remainder & Factor Theorems

2. Nested Form: There is a way to express a polynomial in a different way by factoring out what we can at different places in an equation
Ex: We can write
Now, let’s try this one:
Ex 1) Write polynomial in nested form:
(Hint, start from right side
& work your way in)

3. Synthetic Division/Substitution: A process that takes up less space than long division & uses only the coefficients of each polynomial
**Must write in descending order & fill in any missing terms with 0 **
This process can tell us: (1) if a polynomial is a factor
(2) if a number is a root
(3) what the value of an expression is
Ex 2) Evaluate
(answer is the number in the last spot)
 Remainder Theorem –3 – ↓
–12 27
–81
240
–699
–3 x 4 = –12 4 –9 27
–80
233
–690
f (–3) = –690

4. Ex 3) Divide using synthetic division & then express the polynomial as a product of a divisor & a quotient with a remainder.
* Note: dividing by x – 1 makes x – 1 = 0
 x = 1
and 1 goes in the box 1 – ↓ 6 4 9 6 4 9 10
Coeff of Quotient
Remainder
Powers of x have decreased by 1
(6x2 + 4x + 9)(x – 1) + 10

5. Stemming from this example … if the result was a divisor & quotient being multiplied and there was nothing added or subtracted, the last number would be a zero!
What does this mean?!
It means if we do synthetic division / substitution and the end
result is 0, we have a factor / root
 Factor Theorem
Ex 4) Determine whether x – 4 is a factor of 4
– – –36 ↓ 4 12 36 1 3 9
 YAY!! 
YES

6. 1 1 3 –1 –3 ↓ 1 4 3 1 4 3 Yes, so (x – 1) is one factor
We can factor a polynomial completely by first using synthetic to find 1 factor & then deal with the depressed equation.
Ex 5) Express as a product of linear factors
Try 1 1
– –3 ↓ 1 4 3 1 4 3
Yes, so (x – 1) is
one factor
Now use x2 + 4x + 3 and factor it!
(x + 3)(x + 1)
g(x) = (x – 1)(x + 3)(x + 1)

7. 8 –16 16 –27 18 ↓ 12 –6 15 –18 8 –4 10 –12 Dealing with fractions
Ex 6) Divide
Factor is 2x – 3
Root is 2x – 3 = 0
8 – – ↓ 12 –6 15
–18 8 –4 10
–12
Factor out 2
So, answer is:

8. 2 2 5 k –16 ↓ 4 18 36+2k 2 9 18+k 20+2k Want this to be 0, so
Solving for an unknown inside an equation
Ex 7) Determine k so that has x – 2 as a factor
x – 2 = 0
x = 2 2
k –16 ↓ 4 18
36+2k 2 9
18+k
20+2k
Want this to be 0, so
20 + 2k = 0
20 = –2k
–10 = k

9. Summary (with proper vocab):
If in polynomial P(x), P(a) = 0, then
x = a is a zero of P(x)
x = a is a solution of the equation P(x) = 0
x – a is a factor of P(x)

10. Homework
#701 Pg #3–15 odd, 16, 23, 27, 31–37 odd, 47, 48, 51, 55, 57, 59, 61