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Polynomial & Synthetic Division
MATH 109 - Precalculus
S. Rook
Overview
• Section 2.3 in the textbook:
– Polynomial long division
– Synthetic division
– Remainder & factor theorems
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Polynomial Long Division
Polynomial Long Division
• Recall that there are four main steps in long division
which cycle until the last digit is brought down:
– Division
– Multiplication
– Subtraction
– Bring Down
• Can write the result in the form: n = q · k + r where n is
the dividend (number inside), q is the quotient, k is the
divisor (number outside), and r is the remainder
• Applies to polynomial long division
– Need to remember to apply the Distributive Property
when necessary
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Polynomial Long Division
(Continued)
• Some tips for polynomial division:
– Write the polynomial in descending degree
– Fill any missing terms with placeholder zeros
– When performing the division step, use the highest term
of the divisor (polynomial outside)
• Polynomial division is helpful because in most cases
it is easier to examine a lower degree quotient than
the original polynomial
– i.e. Breaks the original polynomial apart
– E.g. (x3 – 1) / (x – 1) → (x3 – 1) = (x2 + x + 1) · (x – 1) + 0
• A remainder of 0 indicates that (x – 1) is a factor of
(x3 – 1) which means it divides evenly
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Polynomial Long Division (Example)
Ex 1: Use polynomial division:
a) 8  x  6 x3  10 x 2   2 x 2  1
b) 4 x3  11x  5  x  2
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Synthetic Division
Synthetic Division
• A faster variant of polynomial long division
• Can be used ONLY when the divisor is in the form
(x – k)
– i.e. When the divisor is linear
– Polynomial long division can be used for any
polynomial divisor
• Synthetic division has two main steps which cycle:
– Multiply
– Add
• Remember to put the dividend into descending
degree
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Synthetic Division (Continued)
• To perform a synthetic division:
– Consider just the coefficients of the dividend
• Synthetic division uses ONLY numbers
• Insert a 0 for any term that may be missing
– If given a divisor in the form of a factor (x – k), convert it
to a zero of the form x = k
• E.g. (x – 5) → x = 5 and (2x + 3) → x = -3⁄2
– Bring down the first digit of the dividend
• After performing the division:
– Write the quotient as a polynomial starting with one
degree less than the dividend
– The last column represents the remainder
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Setting Up a Synthetic Division
• The following represents how (x3 – 1) / (x – 1) would
be set up as a synthetic division:
– What do you notice?
• The dividend is
missing x2 and x
• The factor of (x – 1)
is converted to the
zero x = 1
– What is the quotient written as a polynomial?
x2 + x + 1
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Synthetic Division (Example)
Ex 2: Use synthetic division:
a) 2 x 2  10 x  12  x  4
4 x 3  16 x 2  23x  15
b)
1
x
2
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Remainder & Factor Theorems
Remainder Theorem
• Sometimes we just need to examine the remainder of a
division
• Remainder Theorem: given the polynomial function f(x),
the remainder of f(x) / (x – k) is f(k) when (x – k) is linear
– i.e. Convert x – k to the zero x = k and evaluate in f(x) to
obtain the remainder
– E.g. f(x) = x3 – 1 → remainder of x3 – 1 / x – 1 = f(1) = 0
– E.g. f(x) = 2x2 – 3x + 4 → remainder of 2x2 – 3x + 4 / x + 2 =
f(-2) = 18
• Can verify via polynomial division or synthetic division
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Factor Theorem
• Especially important is the case when the
remainder is 0
• Factor Theorem: (x – k) is a factor of the
polynomial function f(x) if f(k) = 0
– i.e. Recall that f(k) = 0 means the remainder is 0
– i.e. (x – k) divides evenly into f(x)
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Remainder & Factor Theorems
(Example)
Ex 3: a) Use the Factor Theorem to show that x
comprises a factor of f(x) b) Use synthetic division to
reduce f(x) c) Factor the resulting polynomial
completely, listing both the factored form of f(x) and
its real zeros
f(x) = x3 – 7x + 6; x = 2
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Summary
• After studying these slides, you should be able to:
– Perform polynomial division
– Synthetic division
– Apply the Remainder & Factor Theorems
• Additional Practice
– See the list of suggested problems for 2.3
• Next lesson
– Complex Numbers (Section 2.4)
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