You write the division problem in the same format you would use for numbers. If a term is missing in standard form …fill it in with a 0 coefficient.

Example: 2x4 + 3x3 + 5x – 1 = x2 – 2x + 2

1503222 2342 xxxxxx

2x4 = 2x2

x2

2x2

1503222 2342 xxxxxx

2x2

+4x2-4x32x4-( )

- 4x27x3 +5x

7x3 = 7x x2

+7x

7x3 - 14x2 +14x-( )

10x2 - 9x -1

+10

10x2 - 20x +20-( )

11x - 21

remainder

2x2 + 7x + 10 + 11x – 21 x2 – 2x + 2

Quotient + Remainder over divisor

y4 + 2y2 – y + 5 =y2 – y + 1

Answer: y2 + y + 2 + 3 y2 – y + 1

If a polynomial f(x) is divisible by (x – k), then the remainder is r = f(k).

Now you will use synthetic division (like synthetic substitution)

f(x)= 3x3 – 2x2 + 2x – 5 Divide by x - 2

Long division results in ?...... 3x2 + 4x + 10 + 15

x – 2 Synthetic Division: f(2) = 3 -2 2 -5

2

3

6

4

8

10

20

15

Which gives you: 3x2 + 4x + 10 + 15 x-2

Divide x3 + 2x2 – 6x -9 by (a) x-2 (b) x+3 (a) x-2 1 2 -6 -9

2

1

2

4

8

2

4

-5

Which is x2 + 4x + 2 + -5

x-2

(b) x+3 12 -6 -9 -3

1

-3

-1

3

-3

9

0

x2 – x - 3

A polynomial f(x) has factor x-k iff f(k)=0

note that k is a ZERO of the function because f(k)=0

Factor f(x) = 2x3 + 11x2 + 18x + 9 Given f(-3)=0

Since f(-3)=0 x-(-3) or x+3 is a factor So use synthetic division to find the others!!

211 18 9

-3

2

-6

5

-15

3

-9

0

(x + 3)(2x2 + 5x + 3)

So…. 2x3 + 11x2 + 18x + 9 factors to:

Now keep factoring (bustin ‘da ‘b’) gives you:

(x+3)(2x+3)(x+1)

Factor f(x)= 3x3 + 13x2 + 2x -8 given f(-4)=0

(x + 1)(3x – 2)(x + 4)

f(x) = x3 – 2x2 – 9x +18. One zero of f(x) is x=2 Find the others! Use synthetic div. to reduce the degree of

the polynomial function and factor completely.

(x-2)(x2-9) = (x-2)(x+3)(x-3) Therefore, the zeros are x=2,3,-3!!!

f(x) = x3 + 6x2 + 3x -10 X=-5 is one zero, find the others!

The zeros are x=2,-1,-5 Because the factors are (x-2)(x+1)(x+5)