The polynomial has three real zeros.
x1 = -2
x2 = 2+√5
x3 = 2-√5
x3-2x2-9x-2
=(x+2)(x2-4x-1)
=(x+2)(x-2-√5)(x-2+√5).
You can use synthetic division to factor out x+2.
-2 | 1 -2 -9 -2
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|
| -2 +8 +2
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| 1 -4 -1 0
Stephen C.
x+2=0 and x^2-4x-1=0
right?
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05/29/17
Stephen C.
-2, 2+SQ.RT.5, 2-SQ.RT.5
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05/29/17
Sava D.
tutor
The first part is correct. When you factor x^2-4x-1, you get
(x-2-√5)(x-2+√5).
Each of the factors needs to be equaled to zero, to find the zero.
x-2-√5 = 0
and
x-2+√5 = 0.
Note, the first term in your factoring is x^2. The power 2 means that you need to keep factoring. The linear factors must have a power of 1. This is the requirement. The way the question is worded, the answer is
x3-2x2-9x-2 = (x+2)(x-2-√5)(x-2+√).
In the expression on the right hand side, all factors are linear. In some cases, you get a quadratic trinomial with complex or imaginary zeros. These are called prime trinomials, because their zeros are complex numbers and have no real roots. The graph of these is either above the x-axis, or below the x-axis.
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05/29/17
Sava D.
tutor
I am missing 5 in the last expression. The corrected answer needs to read above:
x3-2x2-9x-2 = (x+2)(x-2-√5)(x-2+√).
x3-2x2-9x-2 = (x+2)(x-2-√5)(x-2+√5).
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05/29/17
Sava D.
tutor
While I was typing, you got it! Good work.
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05/29/17
Stephen C.
05/29/17