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𝐏𝐓𝐒 πŸ‘
Bridge to Calculus Workshop
Summer 2020
Lesson 22
Polynomial Long
Division
"Perfect numbers like perfect men
are very rare.β€œ – Descartes -
Lehman College, Department of Mathematics
Factoring Binomials (1 of 1)
Example 1. Factor the following expression completely:
Solution. Note that the greatest common integer factor
is 6 and the highest common power of π‘₯ is:
24π‘₯4
βˆ’ 54π‘₯8
π‘₯4
24π‘₯4
βˆ’ 54π‘₯8
= 6π‘₯4
4 βˆ’ 9π‘₯4
= 6π‘₯4
22
βˆ’ 3π‘₯2 2
= 6π‘₯4
2 βˆ’ 3π‘₯2
2 + 3π‘₯2
= 6π‘₯4
2 βˆ’ π‘₯ 3 2 + π‘₯ 3 2 + 3π‘₯2
Lehman College, Department of Mathematics
Factoring Binomials (1 of 1)
Example 2. Factor the following expression completely:
Solution. Note that the greatest common integer factor
is 1 and the highest common power of π‘₯ is:
Here, π‘₯2
+ 1 is an irreducible quadratic.
Is π‘₯4 + 1 irreducible?
No, but it is the product of two irreducible quadratics.
π‘₯9
βˆ’ π‘₯
π‘₯
π‘₯9
βˆ’ π‘₯ = π‘₯ π‘₯8 βˆ’ 1
= π‘₯ π‘₯4
βˆ’ 1 π‘₯4
+ 1
= π‘₯ π‘₯2
βˆ’ 1 π‘₯2
+ 1 π‘₯4
+ 1
= π‘₯ π‘₯ βˆ’ 1 π‘₯ + 1 π‘₯2 + 1 π‘₯4 + 1
Lehman College, Department of Mathematics
Rational Numbers (1 of 2)
Let us look at integer multiplication. For example, if:
then we say 2 and 3 are integer factors of 6. Now, we
introduce a new operation called integer division. If we
have the product 2 β‹… 3 = 6, then we define the quotients:
Let us extend the concept to quotients, such as:
Since there are no integers π‘š and 𝑛, such that:
Then the above quotients are not integers.
2 β‹… 3 = 6
6
3
= 2
6
2
= 3and
5
3
6
4
and
3 β‹… 𝑛 = 5 4 β‹… π‘š = 6and
Lehman College, Department of Mathematics
Rational Numbers (2 of 2)
Let 𝑝 and π‘ž be integers with π‘ž β‰  0, then the quotient:
Is called a rational number. The adjective rational
comes from the word ratio, meaning quotient.
Why is division by zero not allowed? Suppose we have:
Where π‘Ÿ is a rational number. Then π‘Ÿ β‹… 0 = 2, but the
product of any number with zero is zero, so no such
number π‘Ÿ can exist, and the quotient is thus undefined.
How about the quotient
0
0
? Suppose
0
0
= π‘Ÿ, so π‘Ÿ β‹… 0 = 0.
In this case, π‘Ÿ could be any real, and is thus undefined.
𝑝
π‘ž
2
0
= π‘Ÿ
Lehman College, Department of Mathematics
Integer Division (1 of 6)
Example 1. Perform the following operation:
Solution. Since:
Then:
The number 1 is called the quotient, and 2 is the
remainder of integer division.
5
3
3 β‹… 1 = 3 3 β‹… 2 = 6and< 5 > 5
5
3
=
3 + 2
3
=
3
3
+
2
3
= 1 +
2
3
Lehman College, Department of Mathematics
Rational Function Definition (1 of 1)
Let 𝑝(π‘₯) and π‘ž(π‘₯) be polynomials with π‘ž(π‘₯) β‰  0, then
the quotient:
is called a rational function. For example, the following
are all rational functions:
𝑝(π‘₯)
π‘ž(π‘₯)
1
π‘₯
π‘₯ + 1
π‘₯2 + 3π‘₯ βˆ’ 4
5π‘₯3
+ 2π‘₯ + 3
3π‘₯2 + 2π‘₯ βˆ’ 1
(a) (b) (c)
5(d)
π‘₯ + 1
π‘₯ βˆ’ 1
(e) 4π‘₯2 + 2π‘₯ + 5(f)
Lehman College, Department of Mathematics
Polynomial Long Division (1 of 8)
Example 2. Perform the following operation:
Solution. We distribute the product:
Similar to integer division, it follows that:
(3π‘₯ + 5)(π‘₯ + 2)
3π‘₯ + 5 π‘₯ + 2 = 3π‘₯ π‘₯ + 2 + 5 π‘₯ + 2
= 3π‘₯2 + 6π‘₯ + 5π‘₯ + 10
= 3π‘₯2
+ 11π‘₯ + 10
3π‘₯2
+ 11π‘₯ + 10
π‘₯ + 2
= 3π‘₯ + 5
3π‘₯2
+ 11π‘₯ + 10
3π‘₯ + 5
= π‘₯ + 2
Lehman College, Department of Mathematics
Polynomial Long Division (2 of 8)
Example 3: In a rational function, if the numerator
polynomial is the same or of higher degree than the
denominator, we can perform polynomial long-division:
It follows that:
3π‘₯2
+ 11π‘₯ + 10π‘₯ + 2
βˆ’(3π‘₯2 + 6π‘₯)
5π‘₯
3π‘₯
βˆ’ (5π‘₯ + 10)
0
+ 5
+ 10
3π‘₯2 + 11π‘₯ + 10
π‘₯ + 2
= 3π‘₯ + 5
Quotient
Remainder
Lehman College, Department of Mathematics
Polynomial Long Division (3 of 8)
Example 4. Perform the following operation:
Solution. We distribute the product:
Similar to integer division, it follows that:
(π‘₯ βˆ’ 1)(π‘₯ + 1)
(π‘₯ βˆ’ 1)(π‘₯ + 1) = π‘₯ π‘₯ + 1 βˆ’ 1 π‘₯ + 1
= π‘₯2 + π‘₯ βˆ’ π‘₯ βˆ’ 1
= π‘₯2
βˆ’ 1
π‘₯2 βˆ’ 1
π‘₯ βˆ’ 1
= π‘₯ + 1
π‘₯2 βˆ’ 1
π‘₯ + 1
= π‘₯ βˆ’ 1and
Lehman College, Department of Mathematics
Polynomial Long Division (4 of 8)
Example 5: Perform polynomial long division:
Solution. Since the numerator polynomial is of higher
degree than the denominator, we will perform
polynomial long-division:
π‘₯2
βˆ’ 1
π‘₯ + 1
π‘₯2
+ 0π‘₯ βˆ’ 1π‘₯ + 1
βˆ’(π‘₯2
+ π‘₯)
π‘₯
βˆ’(βˆ’ π‘₯ βˆ’ 1)
0
βˆ’ 1
βˆ’ 1βˆ’π‘₯
Quotient
Remainder
Lehman College, Department of Mathematics
ErdΕ‘s and Tao (1 of 2)
Terence Tao (b. 1975)
- Australian-American Mathematician
Paul ErdΕ‘s (1913-1996)
- Hungarian Mathematician
Lehman College, Department of Mathematics
ErdΕ‘s and Tao (2 of 2)
A 10-year-old Terence Tao hard at work with Paul
ErdΕ‘s in 1985. Courtesy of Wikimedia Commons.
Lehman College, Department of Mathematics
Polynomial Long Division (5 of 8)
Example 6: Perform polynomial long division:
Solution. Since the numerator polynomial is of higher
degree than the denominator, we will perform
polynomial long-division:
π‘₯3
βˆ’ π‘₯ + 3
π‘₯2 + π‘₯ βˆ’ 2
π‘₯3 + 0π‘₯2 βˆ’ π‘₯ + 3π‘₯2
+ π‘₯ βˆ’ 2
βˆ’(π‘₯3
+ π‘₯2
βˆ’ 2π‘₯)
βˆ’π‘₯2 + π‘₯
π‘₯
βˆ’(βˆ’π‘₯2 βˆ’ π‘₯ + 2)
2π‘₯ + 1
βˆ’ 1
+ 3
Quotient
Remainder
Lehman College, Department of Mathematics
Polynomial Long Division (6 of 8)
Solution (cont’d). It follows that:
π‘₯3
βˆ’ π‘₯ + 3
π‘₯2 + π‘₯ βˆ’ 2
= π‘₯ βˆ’ 1 +
2π‘₯ + 1
π‘₯2 + π‘₯ βˆ’ 2
Lehman College, Department of Mathematics
Polynomial Long Division (7 of 8)
Example 6: Perform polynomial long division:
Solution. :
2π‘₯4
+ 3π‘₯3
+ 5π‘₯ βˆ’ 1
π‘₯2 + 3π‘₯ + 2
2π‘₯4
+ 3π‘₯3
+ 0π‘₯2
+ 5π‘₯ βˆ’ 1π‘₯2
+ 3π‘₯ + 2
βˆ’(2π‘₯4
+ 6π‘₯3
+ 4π‘₯2
)
βˆ’3π‘₯3
βˆ’ 4π‘₯2
2π‘₯2
βˆ’(βˆ’3π‘₯3
βˆ’ 9π‘₯2
βˆ’ 6π‘₯)
5π‘₯2 + 11π‘₯
βˆ’ 3π‘₯
+ 5π‘₯
Quotient
Remainder
βˆ’ 1
+ 5
βˆ’ 5π‘₯2
+ 15π‘₯ + 10
βˆ’4π‘₯ βˆ’ 11
Lehman College, Department of Mathematics
Polynomial Long Division (8 of 8)
Solution (cont’d). It follows that:
2π‘₯4
+ 3π‘₯3
+ 5π‘₯ βˆ’ 1
π‘₯2 + 3π‘₯ + 2
= 2π‘₯2
βˆ’ 3π‘₯ + 5 +
βˆ’4π‘₯ βˆ’ 11
π‘₯2 + 3π‘₯ + 2
Lehman College, Department of Mathematics
Polynomial Long Division Tricks (1 of 3)
Example 7. Perform polynomial long division:
Solution: Create multiples of the denominator:
π‘₯2
π‘₯2 + 1
π‘₯2
π‘₯2 + 1
=
π‘₯2
+ 1 βˆ’ 1
π‘₯2 + 1
=
π‘₯2
+ 1
π‘₯2 + 1
βˆ’
1
π‘₯2 + 1
= 1 +
1
π‘₯2 + 1
Lehman College, Department of Mathematics
Polynomial Long Division Tricks (2 of 3)
Example 8. Perform polynomial long division:
Solution. Create multiples of the denominator:
π‘₯2
+ 2π‘₯ + 3
π‘₯ + 1
π‘₯2 + 2π‘₯ + 3
π‘₯ + 1
=
π‘₯2
+ π‘₯ + π‘₯ + 3
π‘₯ + 1
=
π‘₯2 + π‘₯
π‘₯ + 1
+
π‘₯ + 3
π‘₯ + 1
= π‘₯ +
π‘₯ + 1 + 2
π‘₯ + 1
= π‘₯ + 1 +
2
π‘₯ + 1

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Lesson 22: Polynomial Long Division

  • 1. 𝐏𝐓𝐒 πŸ‘ Bridge to Calculus Workshop Summer 2020 Lesson 22 Polynomial Long Division "Perfect numbers like perfect men are very rare.β€œ – Descartes -
  • 2. Lehman College, Department of Mathematics Factoring Binomials (1 of 1) Example 1. Factor the following expression completely: Solution. Note that the greatest common integer factor is 6 and the highest common power of π‘₯ is: 24π‘₯4 βˆ’ 54π‘₯8 π‘₯4 24π‘₯4 βˆ’ 54π‘₯8 = 6π‘₯4 4 βˆ’ 9π‘₯4 = 6π‘₯4 22 βˆ’ 3π‘₯2 2 = 6π‘₯4 2 βˆ’ 3π‘₯2 2 + 3π‘₯2 = 6π‘₯4 2 βˆ’ π‘₯ 3 2 + π‘₯ 3 2 + 3π‘₯2
  • 3. Lehman College, Department of Mathematics Factoring Binomials (1 of 1) Example 2. Factor the following expression completely: Solution. Note that the greatest common integer factor is 1 and the highest common power of π‘₯ is: Here, π‘₯2 + 1 is an irreducible quadratic. Is π‘₯4 + 1 irreducible? No, but it is the product of two irreducible quadratics. π‘₯9 βˆ’ π‘₯ π‘₯ π‘₯9 βˆ’ π‘₯ = π‘₯ π‘₯8 βˆ’ 1 = π‘₯ π‘₯4 βˆ’ 1 π‘₯4 + 1 = π‘₯ π‘₯2 βˆ’ 1 π‘₯2 + 1 π‘₯4 + 1 = π‘₯ π‘₯ βˆ’ 1 π‘₯ + 1 π‘₯2 + 1 π‘₯4 + 1
  • 4. Lehman College, Department of Mathematics Rational Numbers (1 of 2) Let us look at integer multiplication. For example, if: then we say 2 and 3 are integer factors of 6. Now, we introduce a new operation called integer division. If we have the product 2 β‹… 3 = 6, then we define the quotients: Let us extend the concept to quotients, such as: Since there are no integers π‘š and 𝑛, such that: Then the above quotients are not integers. 2 β‹… 3 = 6 6 3 = 2 6 2 = 3and 5 3 6 4 and 3 β‹… 𝑛 = 5 4 β‹… π‘š = 6and
  • 5. Lehman College, Department of Mathematics Rational Numbers (2 of 2) Let 𝑝 and π‘ž be integers with π‘ž β‰  0, then the quotient: Is called a rational number. The adjective rational comes from the word ratio, meaning quotient. Why is division by zero not allowed? Suppose we have: Where π‘Ÿ is a rational number. Then π‘Ÿ β‹… 0 = 2, but the product of any number with zero is zero, so no such number π‘Ÿ can exist, and the quotient is thus undefined. How about the quotient 0 0 ? Suppose 0 0 = π‘Ÿ, so π‘Ÿ β‹… 0 = 0. In this case, π‘Ÿ could be any real, and is thus undefined. 𝑝 π‘ž 2 0 = π‘Ÿ
  • 6. Lehman College, Department of Mathematics Integer Division (1 of 6) Example 1. Perform the following operation: Solution. Since: Then: The number 1 is called the quotient, and 2 is the remainder of integer division. 5 3 3 β‹… 1 = 3 3 β‹… 2 = 6and< 5 > 5 5 3 = 3 + 2 3 = 3 3 + 2 3 = 1 + 2 3
  • 7. Lehman College, Department of Mathematics Rational Function Definition (1 of 1) Let 𝑝(π‘₯) and π‘ž(π‘₯) be polynomials with π‘ž(π‘₯) β‰  0, then the quotient: is called a rational function. For example, the following are all rational functions: 𝑝(π‘₯) π‘ž(π‘₯) 1 π‘₯ π‘₯ + 1 π‘₯2 + 3π‘₯ βˆ’ 4 5π‘₯3 + 2π‘₯ + 3 3π‘₯2 + 2π‘₯ βˆ’ 1 (a) (b) (c) 5(d) π‘₯ + 1 π‘₯ βˆ’ 1 (e) 4π‘₯2 + 2π‘₯ + 5(f)
  • 8. Lehman College, Department of Mathematics Polynomial Long Division (1 of 8) Example 2. Perform the following operation: Solution. We distribute the product: Similar to integer division, it follows that: (3π‘₯ + 5)(π‘₯ + 2) 3π‘₯ + 5 π‘₯ + 2 = 3π‘₯ π‘₯ + 2 + 5 π‘₯ + 2 = 3π‘₯2 + 6π‘₯ + 5π‘₯ + 10 = 3π‘₯2 + 11π‘₯ + 10 3π‘₯2 + 11π‘₯ + 10 π‘₯ + 2 = 3π‘₯ + 5 3π‘₯2 + 11π‘₯ + 10 3π‘₯ + 5 = π‘₯ + 2
  • 9. Lehman College, Department of Mathematics Polynomial Long Division (2 of 8) Example 3: In a rational function, if the numerator polynomial is the same or of higher degree than the denominator, we can perform polynomial long-division: It follows that: 3π‘₯2 + 11π‘₯ + 10π‘₯ + 2 βˆ’(3π‘₯2 + 6π‘₯) 5π‘₯ 3π‘₯ βˆ’ (5π‘₯ + 10) 0 + 5 + 10 3π‘₯2 + 11π‘₯ + 10 π‘₯ + 2 = 3π‘₯ + 5 Quotient Remainder
  • 10. Lehman College, Department of Mathematics Polynomial Long Division (3 of 8) Example 4. Perform the following operation: Solution. We distribute the product: Similar to integer division, it follows that: (π‘₯ βˆ’ 1)(π‘₯ + 1) (π‘₯ βˆ’ 1)(π‘₯ + 1) = π‘₯ π‘₯ + 1 βˆ’ 1 π‘₯ + 1 = π‘₯2 + π‘₯ βˆ’ π‘₯ βˆ’ 1 = π‘₯2 βˆ’ 1 π‘₯2 βˆ’ 1 π‘₯ βˆ’ 1 = π‘₯ + 1 π‘₯2 βˆ’ 1 π‘₯ + 1 = π‘₯ βˆ’ 1and
  • 11. Lehman College, Department of Mathematics Polynomial Long Division (4 of 8) Example 5: Perform polynomial long division: Solution. Since the numerator polynomial is of higher degree than the denominator, we will perform polynomial long-division: π‘₯2 βˆ’ 1 π‘₯ + 1 π‘₯2 + 0π‘₯ βˆ’ 1π‘₯ + 1 βˆ’(π‘₯2 + π‘₯) π‘₯ βˆ’(βˆ’ π‘₯ βˆ’ 1) 0 βˆ’ 1 βˆ’ 1βˆ’π‘₯ Quotient Remainder
  • 12. Lehman College, Department of Mathematics ErdΕ‘s and Tao (1 of 2) Terence Tao (b. 1975) - Australian-American Mathematician Paul ErdΕ‘s (1913-1996) - Hungarian Mathematician
  • 13. Lehman College, Department of Mathematics ErdΕ‘s and Tao (2 of 2) A 10-year-old Terence Tao hard at work with Paul ErdΕ‘s in 1985. Courtesy of Wikimedia Commons.
  • 14. Lehman College, Department of Mathematics Polynomial Long Division (5 of 8) Example 6: Perform polynomial long division: Solution. Since the numerator polynomial is of higher degree than the denominator, we will perform polynomial long-division: π‘₯3 βˆ’ π‘₯ + 3 π‘₯2 + π‘₯ βˆ’ 2 π‘₯3 + 0π‘₯2 βˆ’ π‘₯ + 3π‘₯2 + π‘₯ βˆ’ 2 βˆ’(π‘₯3 + π‘₯2 βˆ’ 2π‘₯) βˆ’π‘₯2 + π‘₯ π‘₯ βˆ’(βˆ’π‘₯2 βˆ’ π‘₯ + 2) 2π‘₯ + 1 βˆ’ 1 + 3 Quotient Remainder
  • 15. Lehman College, Department of Mathematics Polynomial Long Division (6 of 8) Solution (cont’d). It follows that: π‘₯3 βˆ’ π‘₯ + 3 π‘₯2 + π‘₯ βˆ’ 2 = π‘₯ βˆ’ 1 + 2π‘₯ + 1 π‘₯2 + π‘₯ βˆ’ 2
  • 16. Lehman College, Department of Mathematics Polynomial Long Division (7 of 8) Example 6: Perform polynomial long division: Solution. : 2π‘₯4 + 3π‘₯3 + 5π‘₯ βˆ’ 1 π‘₯2 + 3π‘₯ + 2 2π‘₯4 + 3π‘₯3 + 0π‘₯2 + 5π‘₯ βˆ’ 1π‘₯2 + 3π‘₯ + 2 βˆ’(2π‘₯4 + 6π‘₯3 + 4π‘₯2 ) βˆ’3π‘₯3 βˆ’ 4π‘₯2 2π‘₯2 βˆ’(βˆ’3π‘₯3 βˆ’ 9π‘₯2 βˆ’ 6π‘₯) 5π‘₯2 + 11π‘₯ βˆ’ 3π‘₯ + 5π‘₯ Quotient Remainder βˆ’ 1 + 5 βˆ’ 5π‘₯2 + 15π‘₯ + 10 βˆ’4π‘₯ βˆ’ 11
  • 17. Lehman College, Department of Mathematics Polynomial Long Division (8 of 8) Solution (cont’d). It follows that: 2π‘₯4 + 3π‘₯3 + 5π‘₯ βˆ’ 1 π‘₯2 + 3π‘₯ + 2 = 2π‘₯2 βˆ’ 3π‘₯ + 5 + βˆ’4π‘₯ βˆ’ 11 π‘₯2 + 3π‘₯ + 2
  • 18. Lehman College, Department of Mathematics Polynomial Long Division Tricks (1 of 3) Example 7. Perform polynomial long division: Solution: Create multiples of the denominator: π‘₯2 π‘₯2 + 1 π‘₯2 π‘₯2 + 1 = π‘₯2 + 1 βˆ’ 1 π‘₯2 + 1 = π‘₯2 + 1 π‘₯2 + 1 βˆ’ 1 π‘₯2 + 1 = 1 + 1 π‘₯2 + 1
  • 19. Lehman College, Department of Mathematics Polynomial Long Division Tricks (2 of 3) Example 8. Perform polynomial long division: Solution. Create multiples of the denominator: π‘₯2 + 2π‘₯ + 3 π‘₯ + 1 π‘₯2 + 2π‘₯ + 3 π‘₯ + 1 = π‘₯2 + π‘₯ + π‘₯ + 3 π‘₯ + 1 = π‘₯2 + π‘₯ π‘₯ + 1 + π‘₯ + 3 π‘₯ + 1 = π‘₯ + π‘₯ + 1 + 2 π‘₯ + 1 = π‘₯ + 1 + 2 π‘₯ + 1