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Butoxors [25]
3 years ago
5

What are the possible numbers of positive, negative, and complex zeros of f(x) = x6 − x5 − x4 + 4x3 − 12x2 + 12?

Mathematics
2 answers:
FrozenT [24]3 years ago
3 0
Using <span>Descartes' rule of signs.
There are 4 changes of signs of coefficients:
+ to - (+x^6-x^5)
- to + (-x^4+4x^3)
+ to - (+4x^3-12x^2)
- to + (-12x^2+12)
Therefore, there are 4,2 or 0 positive roots </span><span><span>at most</span>. So it must be the third answer.
</span>
Akimi4 [234]3 years ago
3 0

Third option

Positive:4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0

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Which equation describes a function that is NOT linear?
natka813 [3]

Answer:

Step-by-step explanation:

it’s the third one

5 0
3 years ago
Which of the following is a solution of x2 + 6x = −18? x = 3 − 3i x = −3 + 3i x = −6 + 3i x = 6 − 3i
Marina86 [1]

Step 1 ) Move all terms right side of the equation.

\displaystyle\ x^{2} + 6x = -18

\displaystyle\ x^{2} + 6x + 18 = -18 + 18

\displaystyle\ x^{2} +6x + 18 = 0

Step 2 ) Apply quadratic formula. (Note: There are 2 solutions)

\displaystyle\ x^{2} +6x + 18 = 0

\displaystyle\ x = \frac{-b\sqrt{b^{2}-4ac}}{2a}, \displaystyle\ x = \frac{-b-\sqrt{b^{2}-4ac}}{2a}

\displaystyle\ x = \frac{-6+\sqrt{6^{2}-4 * 18}}{2} , \displaystyle\ x = \frac{-6-\sqrt{6^{2}-4*18}}{2}

\displaystyle\ x = \frac{-6+6i}{2} ,

\displaystyle\ x = \frac{-6-6i}{2}

Step 3 ) Simplify.

\displaystyle\ x = \frac{-6+6i}{2} ,

\displaystyle\ x = \frac{-6-6i}{2}

\displaystyle\ x = -3+ 3i,

\displaystyle\ x = -3 - 3i

Since the options only provide one of the answer we found, the answer is...

\displaystyle\ x = -3 - 3i

•

•

- <em>Marlon Nunez</em>

6 0
3 years ago
Read 2 more answers
Can any one help me solve this ?
Neko [114]
The answer to your question is 12b^2
Because 3-12 is -9 pal :)
3 0
3 years ago
Helppppppoppoppppppppp
vampirchik [111]

Answer:

im pretty sure blue

Step-by-step explanation:

.

8 0
3 years ago
In exercises 21 through 24 is it possible to construct a triangle if the given side lengths? If not explain why.
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