Polynomial and Synthetic Division Techniques

Polynomial and
Synthetic Division
 
What you should learn
 
How to use long division to divide polynomials by other
polynomials
How to use synthetic division to divide polynomials by
binomials of the form (
x
k
)
How to use the Remainder Theorem and the Factor Theorem
 
1. x goes into x
3
?
 
x
2 
times.
 
2. Multiply (x-1) by x
2
.
 
4. Bring down 4x.
 
5. x goes into 2x
2
?
 
2x
 
times.
 
6. Multiply (x-1) by 2x.
 
8. Bring down -6.
 
9. x goes into 6x?
 
3. Change sign, Add.
 
7. Change sign, Add
 
6
 
times.
 
11. Change sign, Add .
 
10. Multiply (x-1) by 6.
Long Division.
Check
Divide.
 
Long Division.
Check
Example
Check
=
Review Long Division.
Check
Synthetic Division -
 
To use synthetic division:
There must be a coefficient for every
possible power of the variable.
The divisor must have a leading coefficient
of 1.
 
divide a polynomial by a polynomial
Step #1: Write the terms of the  
  
  
polynomial so the degrees are in
  
descending order.
 
 
Step #2: Write the constant 
r
 of the 
  
  
divisor 
x-r
 to the left and write 
 
  
down the coefficients.
 
Since the divisor is 
x
-3,   
r=
3
 
Step #3: Bring down the first 
  
coefficient, 5.
 
 
 
15
 
Step #5: Repeat process multiplying 
 
 
      the sum, 15, by 
r
; 
   
 
      and place this number under the
 
      next coefficient, then add.
 
41
 
Step #5 cont.: Repeat the same procedure.
 
124
 
378
 
Where did 123 and 372 come from?
Step #6: Write the quotient.
 
The numbers along the bottom are coefficients
of the power of 
x
 in descending order, starting
with the power that is one less than that of the
dividend.
 
The quotient is:
 
Remember to place the
remainder over the divisor.
 
Synthetic Division
 
Divide x
4 
– 10x
2
 – 2x + 4 by x + 3
 
1
 
0
 
-10
 
-2
 
4
 
-3
 
1
 
-3
 
-3
 
+9
 
-1
 
3
 
1
 
-3
 
1
 
1
 
-2
 
-8
 
3
 
1
 
3
 
1
 
3
 
-5
 
The Remainder Theorem
If a polynomial 
f(x)
 is divided by 
x – k
, the remainder is 
r
 = 
f(k).
The Factor Theorem
 
2
 
7
 
-4
 
-27
 
-18
 
+2
 
2
 
4
 
11
 
22
 
18
 
36
 
9
 
18
 
0
Uses of the Remainder in Synthetic
Division
 
The remainder 
r
, obtained in synthetic
division of 
f(x)
 by (
x – k
), provides
the following information.
1.
r
 = 
f(k)
2.
If 
r
 = 0 then (
x – k
) is a factor of 
f(x).
3.
If 
r
 = 0 then (
k
, 0) is an x intercept of
the graph of 
f
.
Theorems About
Roots of Polynomial
Equations
Rational Roots
Consider the following . . .
 
x
3
 – 5x
2
 – 2x + 24 = 0
 
This equation factors to:
(x+2)(x-3)(x-4)= 0
 
The roots therefore are: -2, 3, 4
Take a closer look at the original
equation and our roots:
 
x
3
 – 5x
2
 – 2x + 24 = 0
 
The roots therefore are: -2, 3, 4
 
What do you notice?
 
-2, 3, and 4 all go into the last term, 24!
Spooky! Let’s look at another
Take a closer look at the original
equation and our roots:
This leads us to the
Rational Root Theorem
1. For polynomial 
Possible roots are ___________________________________
 
Here p = -3 and q = 1
 
Factors of -3
Factors of 1
 
 
2. For polynomial 
Possible roots are ______________________________________________
 
Here p = 12 and q = 3
 
Factors of 12
Factors of 3
 
 
 
Or  3,-3, 1, -1
Where did all of these come from?
Let’s look at our solutions
 
Note that 
+
 2 is listed
twice; we only
consider it as one
answer
Note that 
+
 1 is listed
twice; we only
consider it as one
answer
That is where our 9 possible answers come from!
Note that 
+
 4 is listed
twice; we only
consider it as one
answer
Let’s Try One
Find the POSSIBLE roots of    5x
3
-24x
2
+41x-20=0
Let’s Try One
5x
3
-24x
2
+41x-20=0
 
The possible roots are:
That’s a lot of answers!
Obviously 5x
3
-24x
2
+41x-20=0 does not have all of those roots
as answers.
Remember: these are only POSSIBLE roots. We take these
roots and figure out what answers actually WORK.
Step 1 – find p and q
 
p = -3
q = 1
Step 2 – by RRT, the
only rational root is of
the form…
 
Factors of p
 
Factors of q
 
 
 
Step 3 – factors
 
Factors of -3 = 
±3, ±1
 
Factors of 1 = 
± 1
Step 4 – possible roots
 
-3, 3, 1, and -1
 
 
Step 5 – Test each root
Step 6 – synthetic
division
 
 
 
 
X    X³ + X² – 3x – 3
 
-3
 3
 1
-1
 
(-3)³ + (-3)² – 3(-3) – 3 = -12
 
(3)³ + (3)² – 3(3) – 3 = 24
 
(1)³ + (1)² – 3(1) – 3 = -4
 
(-1)³ + (-1)² – 3(-1) – 3 = 0
THIS IS YOUR ROOT 
BECAUSE WE ARE LOOKING
FOR WHAT ROOTS WILL 
MAKE THE EQUATION =0
-1
 
1     1     -3     -3
 
0
 
1
 
-3
 
3
 
0
 
 
-
1
 
0
 
1x²  + 0x    -3
Step 7 – Rewrite
 
x³ + x² - 3x - 3
 = (x + 1)(x² – 3)
Step 1 – find p and q
 
p = -6
q = 1
Step 2 – by RRT, the
only rational root is of
the form…
 
Factors of p
 
Factors of q
 
 
 
Step 3 – factors
 
Factors of -6 = 
±1, ±2, ±3, ±6
 
Factors of 1 = 
±1
Step 4 – possible
roots
 
-6, 6, -3, 3, -2, 2, 1,
and -1
 
 
Step 5 – Test each root
Step 6 – synthetic division
 
 
 
 
X
 
 
 
 
x
³
 
 
 
5
x
²
 
+
 
8
x
 
 
6
 
  -6
   6
   3
  -3
   2
  -2
   1
  -1
 
THIS IS YOUR ROOT
3
 
1     -5     8     -6
 
-6
 
1
 
 2
 
6
 
-2
 
  3
 
0
 
1x²  + -2x  +  2
 
  -450
   78
   0
  -102
   -2
  -50
   -2
  -20
Step 7 – Rewrite
 
x³  – 5x² + 8x – 6
 = (x - 3)(x² – 2x + 2)
 
Quadratic Formula
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Learn how to perform polynomial division using long division and synthetic division methods. Understand how to divide polynomials by other polynomials or binomials, utilize the Remainder Theorem and Factor Theorem, and apply these concepts through detailed examples.


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  1. Polynomial and Synthetic Division

  2. What you should learn How to use long division to divide polynomials by other polynomials How to use synthetic division to divide polynomials by binomials of the form (x k) How to use the Remainder Theorem and the Factor Theorem

  3. 1. x goes into x3? x2 times. 2. Multiply (x-1) by x2. 3. Change sign, Add. + + 2 x 2 6 x + + + 3 2 1 4 6 x x x x x x x x 4. Bring down 4x. 3 3 2 2 5. x goes into 2x2? 2xtimes. 6. Multiply (x-1) by 2x. 7. Change sign, Add + + + 2 4 2 2 x x x 0 2 2 x 2 2 x 0 + 2 x 8. Bring down -6. 9. x goes into 6x? 10. Multiply (x-1) by 6. 6times. + 6 6 6 6 6 0 x x x 6 11. Change sign, Add .

  4. Long Division. x + 5 + + + 2 3 8 3 3 15 x x x x x x x 2+ 2 Check + + ( 3 )( 5 ) 5 3 x x x 5 + x 5 + x 5 x 15 15 0 15 = = + + + + + 2 15 x x 2 8 15 x x

  5. + + 3x 9 2x Divide. + + + 3 2 3 0 3 3 3 3 3 0 27 x x x x x x x x x x x 3 3 2 2 3 27 3 x + + 2 0 9 9 9 9 9 x x x x x x x 2 2 27 27 27 0 3 3 27 x x +

  6. Long Division. x + 2 + 2 4 2 4 4 8 x x x x x x x 2 2 Check + 2 ( 2 )( 4 ) 4 2 x x x 2 x 2 x 2 x 8 8 0 + 8 = = + 8 x x 2 2 8 x x

  7. Example 2 + p + 44 + 2 20 p = p +p 4 + 6 6 p + + + 2 6 2 6 20 p p p p p 2 2+ p 6 p Check 44 p + + + + + + 4 4 4 20 24 44 ( 6 )( ) 4 ( ) 6 p p p p p p + 6 24 = + + 2 4 6 24 44 p p p = + + 2 2 20 p p

  8. Review Long Division. x + 2 + 2 4 2 4 4 8 x x x x x x x 2 2 Check + 2 ( 2 )( 4 ) 4 2 x x x 2 x 2 x 2 x 8 8 0 + 8 = = + 8 x x 2 2 8 x x

  9. Synthetic Division - divide a polynomial by a polynomial To use synthetic division: There must be a coefficient for every possible power of the variable. The divisor must have a leading coefficient of 1. ( ( ) ) 6: + + + + 4 2 Ex x x x x 5 4 6 ( 3)

  10. ( ( ) ) 6 + + + + 4 2 x x x x 5 4 ( 3) Step #1: Write the terms of the polynomial so the degrees are in descending order. 5x4+ 0x3 4x2+x +6 Since the numerator does not contain all the powers of x, you must include a 0 for thex3.

  11. ( ( ) ) 6 + + + + 4 2 x x x x 5 4 ( 3) Step #2: Write the constant r of the divisor x-r to the left and write down the coefficients. 5x4+ 0x3 4x2+x +6 3 -4 6 1 5 0 Since the divisor is x-3, r=3

  12. ( ( ) ) + + + + 4 2 x x x x 5 4 6 ( 3) Step #3: Bring down the first coefficient, 5. 3 5 0 -4 1 6 5

  13. ( ( ) ) 6 4 2 + + + + x x x x 5 4 ( 3) Step #4: Multiply the first coefficient by r, so 3 5 and place under the second coefficient then add. = = 15 3 5 0 -4 1 6 15 15 5

  14. ( ( ) ) 6 4 2 + + + + x x x x 5 4 ( 3) Step #5: Repeat process multiplying the sum, 15, by r; and place this number under the next coefficient, then add. = = 15 3 45 3 5 0 -4 1 6 15 45 41 15 5

  15. ( ( ) ) 6 4 2 + + + + x x x x 5 4 ( 3) Step #5 cont.: Repeat the same procedure. Where did 123 and 372 come from? 3 5 0 -4 1 6 15 45 41 123 124 372 378 15 5

  16. ( ( ) ) 6 4 2 + + + + x x x x 5 4 ( 3) Step #6: Write the quotient. The numbers along the bottom are coefficients of the power of x in descending order, starting with the power that is one less than that of the dividend. 3 5 0 -4 1 6 15 45 41 123 124 372 378 15 5

  17. ( ( ) ) 6 4 2 + + + + x x x x 5 4 ( 3) The quotient is: 5x3+15x2+ 41x +124 +378 x 3 Remember to place the remainder over the divisor.

  18. Synthetic Division Divide x4 10x2 2x + 4 by x + 3 1 0 -10 -2 4 -3 -3 +9 -3 3 -1 1 1 1 -3 3 + 4 2 1 + 10 2 4 x x x + 3 2 +x = 3 1 x x x + 3 x

  19. x + 1 1 -2 -8 3 2 3 2 3 3 8 x x x x x x x 3 3 2 2+ -5 1 1 8 3 5 x x x + 3 = = = = 2 2 ( ) 2 8 f x x x ) 3 ( f ) 3 ( 9 ) 3 ( 2 8 8 5 6

  20. The Remainder Theorem If a polynomial f(x) is divided by x k, the remainder is r = f(k). x + 1 2 3 2 3 3 8 x x x x x x x = = = = 2 2 ( ) 2 8 f x x x 2 2+ ) 3 ( f ) 3 ( 9 ) 3 ( 2 8 8 5 8 3 5 x x 6 x + 3

  21. The Factor Theorem A polynomial f(x) has a factor (x-k) if and only if f(k)=0. Show that (x-2) is a factor of: ? ? = 2?4+ 7?3 4?2 27? 18 ? 2 = 2(24) + 7 23 4 22 27 2 18 ? 2 = 32 + 56 16 26 18 ? 2 = 0 2 7 -4 -27 -18 +2 4 22 18 36 9 0 2 11 18

  22. Uses of the Remainder in Synthetic Division The remainder r, obtained in synthetic division of f(x) by (x k), provides the following information. 1. r = f(k) 2. If r = 0 then (x k) is a factor of f(x). 3. If r = 0 then (k, 0) is an x intercept of the graph of f.

  23. Theorems About Roots of Polynomial Equations Rational Roots

  24. Consider the following . . . x3 5x2 2x + 24 = 0 This equation factors to: (x+2)(x-3)(x-4)= 0 The roots therefore are: -2, 3, 4

  25. Take a closer look at the original equation and our roots: x3 5x2 2x + 24 = 0 The roots therefore are: -2, 3, 4 What do you notice? -2, 3, and 4 all go into the last term, 24!

  26. Spooky! Lets look at another 24x3 22x2 5x + 6 = 0 This equation factors to: (2x+1)(3x-2)(4x-3)= 0 The roots therefore are: 1 2,2 3,3 4

  27. Take a closer look at the original equation and our roots: 24x3 22x2 5x + 6 = 0 This equation factors to: (2x+1)(3x-2)(4x-3)= 0 The roots therefore are: 1 What do you notice? 2,2 3,3 4 The numerators 1, 2, and 3 all go into the last term, 6! The denominators 2, 3, and 4 all go into the first term, 24!

  28. This leads us to the Rational Root Theorem = + + + + n n 1 Y a x a x ... a x a 0 1 n 1 n For a polynomial, If ? ? is a root of the polynomial, then p is a factor of ?? and q is a factor of ?0

  29. + = 3 2 x x 3 x 3 0 1. For polynomial Here p = -3 and q = 1 Factors of -3 Factors of 1 3, 1 1 Or 3,-3, 1, -1 Possible roots are ___________________________________ + + + = 3 2 3 x 9 x 4 x 12 0 2. For polynomial Here p = 12 and q = 3 Factors of 12 Factors of 3 12, 6 , 4 , 3 , 2 1 3, 1 Possible roots are ______________________________________________ ?? 12, 6, 4, 3, 2, 1, 4 3, 2 3, 1 3 Where did all of these come from?

  30. Lets look at our solutions 12, 6 , 4 , 3 , 2 1 3, 1 12 12 = = 4 12 3 1 Note that + 2 is listed twice; we only consider it as one answer 6 6 = 2 = 6 3 1 3 3 = 1 = 3 3 1 2 2 2 = = 2 Note that + 1 is listed twice; we only consider it as one answer 3 3 1 1 1 1 = = 1 3 3 1 4 4 4 = = 4 3 3 1 Note that + 4 is listed twice; we only consider it as one answer That is where our 9 possible answers come from!

  31. Lets Try One Find the POSSIBLE roots of 5x3-24x2+41x-20=0

  32. Lets Try One 5x3-24x2+41x-20=0 ? = 20 ??????? ?? 20 ??? 20, 10, 5, 4, 2, 1 ? = 5 ??????? ?? 5 ??? 5, 1 The possible roots are: 20, 10, 5, 4 , 2, 1, 4 5, 2 5, 1 5

  33. Thats a lot of answers! Obviously 5x3-24x2+41x-20=0 does not have all of those roots as answers. Remember: these are only POSSIBLE roots. We take these roots and figure out what answers actually WORK.

  34. Step 1 find p and q Step 2 by RRT, the only rational root is of the form p = -3 q = 1 Factors of p Factors of q

  35. Step 3 factors Step 4 possible roots Factors of -3 = 3, 1 Factors of 1 = 1 -3, 3, 1, and -1

  36. Step 5 Test each root Step 6 synthetic division X X + X 3x 3 1 1 -3 -3 -1 -3 (-3) + (-3) 3(-3) 3 = -12 (3) + (3) 3(3) 3 = 24 3 3 -1 0 (1) + (1) 3(1) 3 = -4 1 1 0 -3 0 (-1) + (-1) 3(-1) 3 = 0 -1 THIS IS YOUR ROOT BECAUSE WE ARE LOOKING FOR WHAT ROOTS WILL MAKE THE EQUATION =0 1x + 0x -3

  37. Step 7 Rewrite Step 8 factor more and solve x + x - 3x - 3 = (x + 1)(x 3) (x + 1)(x 3) (x + 1)(x 3)(x + 3) Roots are -1, 3

  38. Step 1 find p and q Step 2 by RRT, the only rational root is of the form p = -6 q = 1 Factors of p Factors of q

  39. Step 3 factors Step 4 possible roots Factors of -6 = 1, 2, 3, 6 Factors of 1 = 1 -6, 6, -3, 3, -2, 2, 1, and -1

  40. Step 6 synthetic division Step 5 Test each root X x 5x + 8x 6 -450 -6 1 -5 8 -6 3 78 6 0 THIS IS YOUR ROOT 3 6 3 -6 -102 -3 1 -2 2 0 -2 2 -50 -2 -2 1 1x + -2x + 2 -20 -1

  41. Step 7 Rewrite Step 8 factor more and solve (x - 3)(x 2x + 2) x 5x + 8x 6 = (x - 3)(x 2x + 2) Quadratic Formula ? = ? ? = ? ? Roots are 3, 1 ?

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