Transcript Synthetic Division

Synthetic Division
1 March 2011
Synthetic Division
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A trick for dividing polynomials
Helps us solve for the roots of polynomials
Only works when we divide by 1st degree
(linear) polynomials
(3x  8 x  11x  1)  ( x  2)
4
2
My degree can’t be larger than 1!
Synthetic Division
(3x  8 x  11x  1)  ( x  2)
4
2
(2 x  5x  11)  ( x  4 x  2)
4
2
Your Turn
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On the Synthetic Division – Guided Notes
handout, complete problems 1 – 5. You will:
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Decide if it’s possible to use synthetic division to
divide the two polynomials
Division Vocab Review
( x  5x  6)  ( x  3)  x  2
2
Dividend
Divisor
Quotient
Preparing for Synthetic
Division
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Can only be used when the divisor is in the
form of (linear)
x–c
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If the divisor isn’t in the form x – c, then you
must convert the expression to include
subtraction.
X + 5 change to x – (-5)
Preparing for Synthetic
Division, cont.
x 5
x  11
x  11  x  (11)
Preparing for Synthetic
Division, cont.
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Polynomials need to be written in expanded,
standard polynomial form.
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Translation: If you’re missing terms, then you
need to write them out as 0 times (*) the variable.
Preparing for Synthetic
Division, cont. Missing some
terms
3x  7 x  2 x
5
3
3x   7 x   2 x 
5
3
3x  0 x  7 x  0 x  2 x  0
5
4
3
2
Your Turn
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On Synthetic Division - Guided Notes handout,
write the dividend in expanded standard
polynomial form for problems 6 – 10.
Write the divisor in the form x – c.
(8 x  2 x)  ( x  2)
3
(8 x  0 x  2 x  0)  ( x  (2))
3
2
*Synthetic Division Steps
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Example Problem:
(3x  8 x  11x  1)  ( x  2)
4
2
Prep Step
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Divisor x – c?
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x–2
Dividend in Expanded Standard Polynomial
Form?
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3x4 – 8x2 – 11x + 1
3x4 +
– 8x2 – 11x + 1
3x4 + 0x3 – 8x2 – 11x + 1
Step 1
2
Write the constant value of the divisor (c) here.
Step 2
2
3
0
-8
-11
1
Write all the coefficients of the expanded
dividend here.
Step 3
2
3
0
-8
-11
1
3
“Drop” the 1st coefficient underneath the line.
Step 4
2
3
0
-8
-11
1
6
3
Multiply “c” by the last value underneath the line.
Write their product just underneath the next
coefficient.
Step 5
2
3
0
-8
-11
1
6
3
6
Add together the numbers in that column and
write their sum underneath the line.
Step 6
2
3
3
0
-8
6
12
-11
1
6
Multiply “c” by the last value underneath the line.
Write their product just underneath the next
coefficient.
Step 7
2
3
3
0
-8
-11
1
6
12
8
-6
6
4
-3
-5
Repeat steps 5 and 6 until a number appears in
the box underneath the last column.
Step 8 – Naming the Quotient
2
3
3
0
-8
-11
1
6
12
8
-6
6
4
-3
-5
In the last row are the coefficients of the quotient
in decreasing order. The quotient is one degree
less than the dividend.
Step 8 – Naming the Quotient
3
6
4
-3
-5
The number in the box is the remainder.
5
x2
(3x  8x  11x  1)  ( x  2) 
4
2
3
3x
+
2
6x
+ 4x – 3
5
x2
Synthetic Division and the
Factor Theorem
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Conclusions:
Your Turn:
So What’s Next?
( x  7 x  6)  ( x  2)
3
* To get the
remaining roots,
set the expression
equal to 0, factor,
and solve.
x  2x  3
2
Your Turn:
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On the Synthetic Division Practice handout,
solve for the remaining roots for problems
1 – 4 and 10 – 12
Rewriting the Original
Polynomial
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We can use the roots and linear factors to
rewrite the polynomial
This form is called the product of linear
factors
If you multiplied all the linear factors together,
then you’d get the original polynomial
Reminder: Roots vs. Linear
Factors
2
f ( x )  x  2x  3
2
0  x  2x  3
Linear
Factors
0  ( x  3)(x  1)
0  x 3
3 x
Roots
0  x 1
1  x
Product of Linear Factors
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2
Product = Multiply
f ( x )  x  2x  3
Product of linear factors =
2
0  x  2x  3
Multiply all the linear
factors
0  ( x  3)(x  1)
 Translation: Rewrite all
0  x 3
0  x 1
the linear factors with
3 x
1  x
parentheses around
each factor
Product of Linear
Helpful format for
Factors
graphing polynomials