Long Division
Quotient
Divisor
Dividend
Remainder
66
3
9
7
7
P( x)
R
 Q( x) 
xa
xa
Long Division - A Review
Quotient
578
84 6 3 1
40
63
Divisor
56
71
64
Remainder
7
Dividend
4631
7
 578 
8
8
96
43 8 4
Division by a Binomial
P( x)
R
 Q( x) 
xa
xa
1. The polynomial
must be in descending
multiply
x +5
order of powers. Any
missing terms are to be
x + 2 x2 + 7x + 2
filled with a zero
-(x2 + 2x)
placeholder.
5x + 2
2. Only the first term is
x(x + 2) used when doing the
-(
)
5x
+
10
Divide
2 + 2x
=
x
division.
x2
-8
=x
3. Multiply your answer
x
with the entire divisor.
4. Subtract, bring down
x2  7 x  2
8
 x 5
, x  -2 the next term and repeat
x2
x2
the process.
Divide: (x2 + 7x + 2) ÷ (x + 2)

Division by a Binomial
4x2 - 3x + 2
x-2
4x - 11x + 8x + 10
3
2
NPV’s
x2
4x3 - 8x2
- 3x2 + 8x
- 3x2 + 6x
2x + 10
2x - 4
14
4 x3  11x 2  8 x  10
14
2
 4 x  3x  2 
,x  2
x2
x2
3
x
  20x  8   x  4

x2 - 4x - 4
x + 4 x3 + 0x2 - 20x + 8
x3 + 4x2
- 4x2 - 20x
- 4x2 - 16x
-4x + 8
-4x - 16
24
NPV’s
x  4
x3  20 x  8
24
2
 x  4x - 2 
,x  4
x4
x4
Synthetic Division
Divide x3 - 2x2 - 33x + 90 by (x - 5) using synthetic division.
-5
1
-2
subtract
-5
Multiply
1
3
-33 90
-15 90
-18
0
Quotient
Rem
1. Write only the constant
term of the divisor, and
the coefficients of the
dividend.
2. Bring down the first term
of the dividend.
3. Multiply 1 by -5, record
the product and subtract.
Written as x2 + 3x - 18
Using the division statement:
P(x) = (x - 5)(x2 + 3x - 18)
4. Multiply 3 by -5, record
the product and subtract.
5. Multiply -18 by -5, record
the product and subtract.
Using Synthetic Division
Divide: (x4 - 2x3 + x2 + 12x - 6) ÷ (x - 2)
-2
1
1
-2
1
12
-6
-2
0
-2
-28
0
1
14
22
x4 - 2x3 + x2 + 12x – 6 = (x - 2)(x3 + x + 14) + 22
The Remainder Theorem
Given P(x) = x3 - 4x2 + 5x + 1, determine the remainder
when P(x) is divided by x - 1.
-1
1
-4
-1
5
3
1
-2
1
-3
2
3
The remainder is 3.
NOTE: P(1) gives
Using f(x) = x3 - 4x2 + 5x + 1, determine P(1): the same answer
as the remainder
P(1) = (1)3 - 4(1)2 + 5(1) + 1
using synthetic
=1-4+5+1
division.
=3
Therefore P(1) is equal to the remainder. In other words,
when the polynomial x3 - 4x2 + 5x + 1 is divided by x - 1,
the remainder is P (1).
Remainder Theorem:
When a polynomial P(x) is divided by x - a, the
remainder is P(a). [think x - a, then x = a]
Determine the remainder when x3 - 4x2 + 5x - 1 is divided by:
a) x - 2
b) x + 1
Calculate P(-1)
Calculate P(2)
3 - 4(-1)2 + 5(-1) - 1
P(-1)
=
(-1)
3
2
P(2) = (2) - 4(2) + 5(2) - 1
= -1 - 4 - 5 - 1
= 8 - 16 + 10 - 1
= -11
=1
The remainder is 1.
Point (2, 1) is on the graph of
of f(x) = x3 - 4x2 + 5x - 1
The remainder is -11.
Point (-1, -11) is on the graph of
of f(x) = x3 - 4x2 + 5x - 1
3
x
When   kx  6  is divided by  x  4  the remainder is 30.
Determine the value of k.
P( x )  R
P(4)  30
3
 4   k (4)  6  30
64  4k  6  30
4k  28
k 7
When the polynomial 3x3 + ax2 + bx -9 is divided by
x - 2 , the remainder is -5. When the polynomial is
divided by x + 1, the remainder is -16. What are the
values of a and b?
Page 124 – 125
Low: 1 – 7
Medium: 8, 9, 10, 11, 12, 13
High:14 – 17
Using Synthetic Division
1. (4x3 - 11x2 + 8x + 6) ÷ (x - 2)
-2
4
4
-11
-8
8 6
6 -4
-3
2 10
P(x) = (x - 2)(4x2 - 3x + 2) + 10
2. (2x3 - 2x2 + 3x + 3) ÷ (x - 1)
-1
2
2
-2
3
3
-2
0
-3
0
3
6
P(x) = (x - 1)(2x2 + 3) + 6