<span>(5+2 i)(4-3i) - (5-2yi)(4-3i)
Factorize out (4 -3i)
(4 -3i)( (5 +2i) - (5 -2yi) )
= </span><span><span>(4 -3i)(5 +2i - 5 + 2yi)</span>
= </span><span><span>(4 -3i)(5 - 5 + 2i + 2yi)</span>
= (4 -3i)(2i + 2yi)
= (4 - 3i)(2 + 2y)i. Let's multiply the first two.
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(4 - 3i)(2 + 2y) = 2*(4 -3i) + 2y*(4 - 3i)
= 8 - 6i + 8y - 6yi
= 8 + 8y - 6i - 6yi
(4 - 3i)(2 + 2y)i = (8 + 8y - 6i - 6yi)i Note: i² = -1
= 8i + 8yi - 6i² - 6yi²
= 8i + 8yi - 6(-1) - 6y(-1)
= 8i + 8yi + 6 + 6y
= 6 + 6y + 8i + 8yi
= (6 + 6y) + (8 + 8y)i In the form a + bi
Do you know that multiplicity means the number of times any factor appears in the factored result? Just checking. For example The graph of y = x^2 - 2x + 1 has a multiplicity of 2. they are y = (x - 1) * ( x - 1)
y = x^2 - 3x - 4 has 2 factors.
y = (x - 4)(x + 1) each of the factors has a multiplicity of 1.
So the answer to your question is there are 5 real zeros and 2 complex zeros.
The student because the student depends on the pizza
255-45-10=200
1 week
200/10=20
20 weeks plus one week equals 21 weeks
Thank you have a day!