1 warm-up divide the following using long division: (6x 3 - 16x 2 + 17x - 6) (3x –2 ) divide the...
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Warm-up• Divide the following using Long Division:• (6x3 - 16x2 + 17x - 6) (3x –2 )
• Divide the following with Synthetic Division• (5x3 – 6x2 + 8) (x – 4)
• Given the following polynomial and one of its factors, Find the remaining factors
• (3x3 + 2x2 –19x + 6) : (x + 3) is a factor
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Warm-up• Divide the following using Long Division:
• (6x3 - 16x2 + 17x - 6) (3x –2 )
• 2x2 – 4x + 3
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Warm-up• Divide the following with Synthetic Division
• (5x3 – 6x2 + 8) (x – 4)
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23256145 2
xxx
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Warm-up• Given the following polynomial and one of its
factors, Find the remaining factors
• (3x3 + 2x2 –19x + 6) : (x + 3) is a factor
• (x – 2)(3x – 1)
Complex NumbersSection 2-4
Digital Lesson
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Objectives
• I can use “i” to write complex numbers
• I can add, subtract, and multiply complex numbers
• I can simplify Negative Square Roots
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Applications
• Impedance readings for electronics and electrical circuits are all measured in complex units
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Complex Numbers
Real Numbers Imaginary Numbers
Rational Irrational
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Complex Numbers
The set of all numbers that can be written
in the format: a + bi ;
“a” is the real number part
“bi’ is the imaginary part
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The Imaginary Unit
11 2 iwherei
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Negative Radicals
25 25 5i i
20 20 2 5i i
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Negative Radicals
18 32
18 4 2i
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To add or subtract complex numbers:
1. Write each complex number in the form a + bi.
2. Add or subtract the real parts of the complex numbers.
3. Add or subtract the imaginary parts of the complex numbers.
(a + bi ) + (c + di ) = (a + c) + (b + d )i
(a + bi ) – (c + di ) = (a – c) + (b – d )i
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Example: Add (11 + 5i) + (8 – 2i )
= 19 + 3i
Group real and imaginary terms.
a + bi form
= (11 + 8) + (5i – 2i )
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Examples: Subtract: (– 21 + 3i ) – (7 – 9i)
= (– 21 – 7) + [(3 – (– 9)]i
= (– 21 – 7) + (3i + 9i)
= –28 + 12i
Group real and
imaginary terms.
a + bi form
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The product of two complex numbers is defined as:
1. Use the FOIL method to find the product.
2. Replace i2 by – 1.
3. Write the answer in the form a + bi.
(a + bi)(c + di ) = (ac – bd ) + (ad + bc)i
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1. 7i (11– 5i) = 77i – 35i2
= 35 + 77i
2. (2 + 3i)(6 – 7i ) = 12 – 14i + 18i – 21i2
= 12 + 4i – 21i2
= 12 + 4i – 21(–1)
= 12 + 4i + 21
= 33 + 4i
= 77i – 35 (– 1)
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2(3 4 )i
(3 4 )(3 4 )i i
29 12 12 16i i i
7 24i
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Homework
• WS 3-7