Polynomial Functions help

Mathematics

1 answer:

Studentka2010 [4]3 years ago

6 0

The key piece of information for these questions is the Fundamental Theorem of Algebra, which states that a degree n polynomial has n complex roots. A complex root can be either real or imaginary.

First question, regarding the polynomial y = x^3 - 3x^2 + 16x - 48:
We know there is one real root, the x-intercept.
Since it's a third degree polynomial, there are three complex roots in total.
Therefore, there is one real root and two imaginary roots.
Answer is B

Second question:
You probably can guess the answer, now that you know the Fundamental Theorem of Alegebra:
There are 3 real zeros, each with multiplicity one, meaning each root only happens once. It's a 5th degree polynomial, so there are a total of 5 roots, implying 2 imaginary roots.

Answer is C) 3 real and 2 imaginary zeroes.

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So its 8p = 56 is the frist bit and the second is 7. BOOM

Mice21 [21]

p=7, I don't know if this is a question or not

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3 years ago

For the love of God help me !! I'm desperate for it tomorrow

Eduardwww [97]

Try to relax. Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.

Consider this: (2)^a negative power = (1/2)^the same power but positive.

So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.

What I just said in that paragraph was: log₂ of(N) = - log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.

Now let's look at the problem:

log₂(x-1) + log(base 1/2) (x-2) = log₂(x)

Subtract log₂(x) from each side:

log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0

Subtract log(base 1/2) (x-2) from each side:

log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.

The left side is the same as log₂[ (x-1)/x ]

==> The right side is the same as +log₂(x-2)

Now you have: log₂[ (x-1)/x ] = +log₂(x-2)

And that ugly [ log to the base of 1/2 ] is gone.

Take the antilog of each side:

(x-1)/x = x-2

Multiply each side by 'x' : x - 1 = x² - 2x

Subtract (x-1) from each side:

x² - 2x - (x-1) = 0

x² - 3x + 1 = 0

Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .

I think you have to say that x=2.618 is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.

There,now. Doesn't that feel better.

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Answer:

Step-by-step explanation:

if solving for y

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