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Objectives: Assignment: To divide P.159: 5-16 (Some)

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Objectives: Assignment: To divide P.159: 5-16 (Some)
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
As a class, use your
vast mathematical
knowledge to
define each of
these words
without the aid of
your textbook.
Quotient
Remainder
Dividend
Divisor
Divides
Evenly
Factor
Use long division to divide 5 into 3462.
6 92
5 3462
30
46
45
12
10
2
Use long division to divide 5 into 3462.
Divisor
6 92
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.
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).
2. If your polynomial is missing a term, add it in
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 x 2  3x  5
2
−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
x 2
x  1 x 2  3x  5
2
−x  − x
2x  5
−2 x  − 2
Divisor
3
Quotient
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.
= Add terms
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.
= Add terms
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.
= Add terms
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.
= Add terms
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
to evaluate f (5) or f (−2). Your answer will be
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 x2  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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