Synthetic Division Notes

Synthetic Division
Algebra 2 CP
Synthetic Division
 A trick for dividing polynomials
 Helps us solve for the roots of polynomials
 Only works when we divide by 1st
degree
(linear) polynomials
)2()11183( 24
−÷+−− xxxx
Exponent can’t be larger than 1!
Synthetic Division
)24()1152( 24
−−÷−+ xxxx
)2()11183( 24
−÷+−− xxxx
Your Turn
 On your Synthetic Division Notes packet,
complete problems 1 – 5.
 Decide if it’s possible to use synthetic division to
divide the two polynomials
1. Yes 2. Yes 3. No 4. No 5. No
Division Vocab Review
Dividend Divisor
2)3()65( 2
+=+÷++ xxxx
Quotient
Preparing for Synthetic
Division
 Can only be used when the divisor is in the
form:
 To divide, you will need the constant term of
your divisor: set your divisor equal to 0 and
solve!
x – c
Preparing for Synthetic
Division, cont.
 Polynomials need to be written in expanded,
standard polynomial form.
 For the polynomial to be expanded sequentially,
ALL terms must be listed
 AKA: If you’re missing terms, they must be filled in
with “place holders.”
Preparing for Synthetic
Division, cont.
xxx 273 35
−+
+−+++ xxx 273 35
020703 2345
+−+++ xxxxx
Your Turn
 On your Synthetic Division Notes packet, write
the dividend in expanded standard polynomial
form for problems 6 – 10.
6.
7.
8.
9.
10.
*Synthetic Division Steps
 Example Problem:
)2()11183( 24
−÷+−− xxxx
Prep Step
 Linear Divisor?
 x – 2
 Dividend in Expanded Standard Polynomial
Form?
 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
Write all the coefficients of the expanded
dividend here.
3 0 -8 -11 1
Step 3
2
“Drop” the 1st
coefficient underneath the line.
3 0 -8 -11 1
3
Step 4
2
Multiply “c” by the last value underneath the line.
Write their product just underneath the next
coefficient.
3 0 -8 -11 1
6
3
Step 5
2
Add together the numbers in that column and
write their sum underneath the line.
3 0 -8 -11 1
6
3 6
Step 6
2
Multiply “c” by the last value underneath the line.
Write their product just underneath the next
coefficient.
3 0 -8 -11 1
6 12
3 6
Step 7
2
Repeat steps 5 and 6 until a number appears in
the box underneath the last column.
3 0 -8 -11 1
6 12 8 -6
3 6 4 -3 -5
Step 8 – Naming the Quotient
2
In the last row are the coefficients of the quotient
in decreasing order. The quotient is one degree
less than the dividend.
3 0 -8 -11 1
6 12 8 -6
3 6 4 -3 -5
Step 8 – Naming the Quotient
3 6 4 -3 -5
The number in the box is the remainder.
=−÷+−− )2()11183( 24
xxxx
Your Turn
 On your Synthetic Division Notes packet,
solve for the quotient of problems 11 – 14
using synthetic division
11.
12.
13.
14.
1 von 21

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Synthetic Division Notes

  • 2. Synthetic Division  A trick for dividing polynomials  Helps us solve for the roots of polynomials  Only works when we divide by 1st degree (linear) polynomials )2()11183( 24 −÷+−− xxxx Exponent can’t be larger than 1!
  • 3. Synthetic Division )24()1152( 24 −−÷−+ xxxx )2()11183( 24 −÷+−− xxxx
  • 4. Your Turn  On your Synthetic Division Notes packet, complete problems 1 – 5.  Decide if it’s possible to use synthetic division to divide the two polynomials 1. Yes 2. Yes 3. No 4. No 5. No
  • 5. Division Vocab Review Dividend Divisor 2)3()65( 2 +=+÷++ xxxx Quotient
  • 6. Preparing for Synthetic Division  Can only be used when the divisor is in the form:  To divide, you will need the constant term of your divisor: set your divisor equal to 0 and solve! x – c
  • 7. Preparing for Synthetic Division, cont.  Polynomials need to be written in expanded, standard polynomial form.  For the polynomial to be expanded sequentially, ALL terms must be listed  AKA: If you’re missing terms, they must be filled in with “place holders.”
  • 8. Preparing for Synthetic Division, cont. xxx 273 35 −+ +−+++ xxx 273 35 020703 2345 +−+++ xxxxx
  • 9. Your Turn  On your Synthetic Division Notes packet, write the dividend in expanded standard polynomial form for problems 6 – 10. 6. 7. 8. 9. 10.
  • 10. *Synthetic Division Steps  Example Problem: )2()11183( 24 −÷+−− xxxx
  • 11. Prep Step  Linear Divisor?  x – 2  Dividend in Expanded Standard Polynomial Form?  3x4 – 8x2 – 11x + 1  3x4 + – 8x2 – 11x + 1  3x4 + 0x3 – 8x2 – 11x + 1
  • 12. Step 1 2 Write the constant value of the divisor (c) here.
  • 13. Step 2 2 Write all the coefficients of the expanded dividend here. 3 0 -8 -11 1
  • 14. Step 3 2 “Drop” the 1st coefficient underneath the line. 3 0 -8 -11 1 3
  • 15. Step 4 2 Multiply “c” by the last value underneath the line. Write their product just underneath the next coefficient. 3 0 -8 -11 1 6 3
  • 16. Step 5 2 Add together the numbers in that column and write their sum underneath the line. 3 0 -8 -11 1 6 3 6
  • 17. Step 6 2 Multiply “c” by the last value underneath the line. Write their product just underneath the next coefficient. 3 0 -8 -11 1 6 12 3 6
  • 18. Step 7 2 Repeat steps 5 and 6 until a number appears in the box underneath the last column. 3 0 -8 -11 1 6 12 8 -6 3 6 4 -3 -5
  • 19. Step 8 – Naming the Quotient 2 In the last row are the coefficients of the quotient in decreasing order. The quotient is one degree less than the dividend. 3 0 -8 -11 1 6 12 8 -6 3 6 4 -3 -5
  • 20. Step 8 – Naming the Quotient 3 6 4 -3 -5 The number in the box is the remainder. =−÷+−− )2()11183( 24 xxxx
  • 21. Your Turn  On your Synthetic Division Notes packet, solve for the quotient of problems 11 – 14 using synthetic division 11. 12. 13. 14.