### 2 3 Polynomial Division

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P.159: 5-16 (Some)
P. 159: 19-36 (Some)
P. 160: 46
P. 160: 49-56 (Some)
P. 160: 57-64 (Some)
P. 161: 79
Homework Supplement
Use long division to divide 5 into 3462.
without decimals.
2. What would it
mean if the
remainder was
zero?
692
5 3462
30
46
45
12
10
2
Use long division to divide 5 into 3462.
Divisor
692
5 3462
30
46
45
12
10
2
Quotient
Dividend
Remainder
Use long division to divide 5 into 3462.
Dividend
Divisor
3462
2
 692 
5
5
Quotient
Remainder
Divisor
If you are lucky enough to get a remainder of
zero when dividing, then the divisor divides
evenly into the dividend.
This means that the divisor is a factor of the
dividend.
For example, when dividing 3 into 192, the
remainder is 0. Therefore, 3 is a factor of 192.
Objectives:
1. To divide
polynomials using
long and synthetic
division
2. To apply the Factor
and Remainder
Theorems to find
real zeros of
polynomial functions
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Assignment:
P.159: 5-16 (Some)
P. 159: 19-36 (Some)
P. 160: 46
P. 160: 49-56 (Some)
P. 160: 57-64 (Some)
P. 161: 79
Homework
Supplement
As a class, use your
vast mathematical
knowledge to
define each of
these words
without the aid of
Quotient
Remainder
Dividend
Divisor
Divides
Evenly
Factor
Dividing polynomials works just like long
division. In fact, it is called long division!
Before you start dividing:
1. Make sure the divisor and dividend are in
standard form (highest to lowest powers).
with a coefficient of 0 as a place holder.
2x  x  3
3
2x  0x  x  3
3
2
Divide x + 1 into x2 + 3x + 5
x 2
x  1 x22  3x  5
−x  − x
2x  5
−2 x  − 2
3
How many times
does x go into x2?
Multiply x by x + 1
Multiply 2 by x + 1
Line up the first term of the quotient with the
term of the dividend with the same degree.
Divide x + 1 into x2 + 3x + 5
Quotient
x 2
x  1 x22  3x  5
−x  − x
2x  5
−2 x  − 2
Divisor
3
Dividend
Remainder
Divide x + 1 into x2 + 3x + 5
Dividend
x  3x  5
3
 x2
x 1
x 1
2
Quotient
Divisor
Remainder
Divisor
Divide 6x3 – 16x2 + 17x – 6 by 3x – 2
Use long division to divide x4 – 10x2 + 2x + 3 by
x–3
When your divisor is of the form x  k, where k
is a constant, then you can perform the
division quicker and easier using just the
coefficients of the dividend.
This is called fake division. I mean, synthetic
division.
Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following
pattern.
k a b c d
ka
a
= Multiply by k
Remainder
Coefficients of Quotient (in decreasing order)
Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following
pattern.
k a b c d
ka
= Multiply by k
a
Important Note: You are always adding columns using synthetic
division, whereas you subtracted columns in long division.
Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following
pattern.
k a b c d
ka
= Multiply by k
a
Important Note: k can be positive or negative. If you divide by x
+ 2, then k = -2 because x + 2 = x – (-2).
Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following
pattern.
k a b c d
ka
= Multiply by k
a
Important Note: Add a coefficient of zero for any missing terms!
Use synthetic division to divide x4 – 10x2 + 2x + 3 by
x+3
Evaluate f (−3) for f (x) = x4 – 10x2 + 2x + 3.
If a polynomial f (x) is divided by x – k, the
remainder is r = f (k).
This means that you could use synthetic division
the remainder.
Divide 2x3 + 9x2 + 4x + 5 by x + 3 using synthetic
division.
Use synthetic division to divide
f(x) = 2x3 – 11x2 + 3x + 36 by x – 3.
Since the remainder is zero when dividing f(x) by
x – 3, we can write:
f ( x)
 2 x 2  5 x  12,
x3
so f ( x)  ( x  3)(2 x 2  5x  12)
This means that x – 3 is a factor of f(x).
A polynomial f(x) has a factor x – k if and only if
f(k) = 0.
This theorem can be used to help factor/solve a
polynomial function if you already know one
of the factors.
Factor f(x) = 2x3 – 11x2 + 3x + 36 given that x – 3
is one factor of f(x). Then find the zeros of f(x).
Given that x – 4 is a factor of x3 – 6x2 + 5x + 12, rewrite
x3 – 6x2 + 5x + 12 as a product of two polynomials.
Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42
given that f(7) = 0.
Objectives:
1. To divide
polynomials using
long and synthetic
division
2. To apply the Factor
and Remainder
Theorems to find
real zeros of
polynomial functions
•
•
•
•
•
•
•
Assignment:
P.159: 5-16 (Some)
P. 159: 19-36 (Some)
P. 160: 46
P. 160: 49-56 (Some)
P. 160: 57-64 (Some)
P. 161: 79
Homework
Supplement
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