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Pavel [41]
2 years ago
11

Please help me with these two :/

Mathematics
1 answer:
serg [7]2 years ago
3 0
The first answer is 80, and the second answer is still 65.5. Hope this helps!
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1. If a polynomial of Degree 4 has an imaginary zero at -2i and a real zero at 5, how many imaginary zeros does it have?
Vadim26 [7]

Answer:

Complex roots for polynomials equations will always come in conjugate pairs.  Since 1-2i is a root, 1+2i will also be a root.  (Just switch the sign of the imaginary part to its opposite.)

Step-by-step explanation:

So far we have 1-2i and 1+2i as roots.  We need two more to make a 4th degree polynomial.

x=1 with multiplicity of 2 means that we count the root twice when we write down the polynomial: (x - 1)(x - 1) = (x - 1)2

Putting it all together, we can construct our polynomial of degree 4 as

y = f(x) = (x - 1)(x - 1)[x - (1-2i)][x - (1+2i)]

Multiply the factors (FOIL):

[x - (1-2i)][x - (1+2i)] =

x2 - (1+2i)x - (1-2i)x + (1-2i)(1+2i) =

x2 - x - 2ix - x + 2ix + (1 + 2i - 2i + 4) =

x2 - 2x + 5

(x - 1)(x - 1) = x2 - x - x + 1 = x2 - 2x + 1

Now multiply the two underlined expressions:

y = (x2 - 2x + 5)(x2 - 2x + 1)

= x4 - 2x3 + 5x2 - 2x3 + 4x2 - 10x + x2 - 2x + 5  by multiplying term by term

= x4 - 4x3 + 10x2 -12x + 5 by collecting like terms.

4 0
2 years ago
Linear equations
yarga [219]
1. x = 2
2. x = 9
3. a = 3
4. x = 13
5. c = 3
6. s = 3
7. x = 7
8. c = 0
9. b = 1
10. c = 4
11. x = 4
12. x = 8
13. x = 12
14. y = 10
15. x = 11
16. x = 10
17. x = 6
18. x = 11
8 0
2 years ago
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