The Remainder and Factor Theorems
and Factoring
Lesson 2.1
Division of Polynomials
1. Long Division
2. Synthetic Division
Reminders:
1. Arrange the exponents of the variables of both
dividend and the divisor in descending order.
2. If there is a missing term, insert a placeholder (0).
3. If there is a remainder, write it over the divisor
and add the fraction to the part of the quotient
previously obtained.
Division of Polynomials
Polynomial
remainder
quotient
divisor
divisor
Lesson 2.2
1. The Remainder Theorem
2. The Factor Theorem
3. Factoring
Lesson 2.2
When a polynomial is divided by a binomial,
two possible results are expected:
a quotient with or without a remainder.
If we are only after the remainder , then we
need not do long division. The Remainder
Theorem and the Factor Theorem can be
used.
Lesson 2.2
The Remainder Theorem
If a polynomial P(x) is divided by (x – c),
the remainder is the constant R(c), and
P(x) = q(x) ∙ (x – c) + R(x)
where q(x) is a polynomial with degree
one less than the degree of P(x).
Dividend equals quotient times divisor plus remainder.
Division of Polynomials
Polynomial
remainder
quotient
divisor
divisor
The Remainder Theorem:
If a polynomial P(x) is divided by x – c, then the remainder
is R = P(x).
The Remainder Theorem
Find f(3) for the following polynomial function.
f(x) = 5x2 – 4x + 3
f(3) = 5(3)2 – 4(3) + 3
f(3) = 5 ∙ 9 – 12 + 3
f(3) = 45 – 12 + 3
f(3) = 36
The Remainder Theorem
Now divide the same polynomial by (x – 3).
5x2 – 4x + 3
3
5 –4 3
15 33
5 11 36
The Remainder Theorem
f(x) = 5x2 – 4x + 3
5x2 – 4x + 3
f(3) = 5(3)2 – 4(3) + 3
f(3) = 5 ∙ 9 – 12 + 3
f(3) = 45 – 12 + 3
3
5 –4
15
5 11
3
33
36
f(3) = 36
Notice that the value obtained when evaluating the function at f(3) and the value
of the remainder when dividing the polynomial by x – 3 are the same.
Dividend equals quotient times divisor plus remainder.
5x2 – 4x + 3 = (5x + 11) ∙ (x – 3) + 36
The Remainder Theorem
Use synthetic substitution to find f(4) for the
following function.
f(x) = 5x4 – 13x3 – 14x2 – 47x + 1
4
5
5
–13 –14
20 28
7
14
–47
56
9
1
36
37
The Remainder Theorem
Use synthetic substitution to find g(–2) for the
following function.
f(x) = 5x4 – 13x3 – 14x2 – 47x + 1
–2
5
5
–13 –14 –47
1
–10 46 –64 222
–23 32 –111 223
The Remainder Theorem
Use synthetic substitution to find c(4) for the
following function.
c(x) = 2x4 – 4x3 – 7x2 – 13x – 10
4
2 –4
8
2 4
–7 –13 –10
16 36 92
9
23 82
The Factor Theorem
The binomial (x – c) is a factor
of the
polynomial f(x) if and only if
f(c) = 0.
The Factor Theorem
When a polynomial is divided by one of its
binomial factors, the quotient is called a
depressed polynomial.
If the remainder (last number in a depressed
polynomial) is zero, that means f(#) = 0. This
also means that the divisor resulting in a
remainder of zero is a factor of the polynomial.
The Factor Theorem
x3 + 4x2 – 15x – 18
x–3
3
1
1
4
3
7
–15 –18
21 18
6
0
Since the remainder is zero,
(x – 3) is a factor of
x3 + 4x2 – 15x – 18.
This also allows us to find the remaining factors of
the polynomial by factoring the depressed polynomial.
The Factor Theorem
x3 + 4x2 – 15x – 18
x–3
The factors of
3
1
1
4
3
7
–15 –18
21 18
6
0
x2 + 7x + 6
(x + 6)(x + 1)
x3 + 4x2 – 15x – 18
are
(x – 3)(x + 6)(x + 1).
The Factor Theorem
Given a polynomial and one of its factors, find the remaining
factors of the polynomial. Some factors may not be binomials.
1. x3 – 11x2 + 14x + 80
x–8
(x – 8)(x – 5)(x + 2)
2. 2x3 + 7x2 – 33x – 18
x+6
(x + 6)(2x + 1)(x – 3)