if f(x) = ax^6 + bx^4 + cx^3 where a,b, and c are integers, how many distinct rational zeros could f(x) have

Mathematics

1 answer:

natita [175]3 years ago

8 0

You can rewrite your function as

$f(x) = x^3(ax^3+bx+c)$

This implies that

$f(x)=0 \iff x^3=0\quad\lor\quad ax^3+bx+c=0$

Now, we have $x^3=0\iff x=0$ , so it counts as a solution.

On the other hand, depending on the coefficient a, b and c, the cubic equation

$ax^2+bx+c=0$

can have either one or three solutions.

So, we have the solution x=0, and then one or three solutions coming from the cubic part. The equation as a whole thus have either two or four solutions, depending on the coefficients.

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Juli2301 [7.4K]

Answer:

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2 years ago

Sue initially has 5 hours of pop music and 4 hours of classical music in her collection. Every month onwards, the hours of pop m

Snezhnost [94]

<h2>Answer</h2>

f(x) = 5(1.25)x + 4

<h2>Explanation</h2>

To solve this, we are going to use the standard exponential equation:

$f(x)=a(1+b)^x$

where

$a$ is the initial amount

$b$ is the growth rate in decimal form

$x$ is the time (in months for our case)

Since the hours of classic music remain constant, we just need to add them at the end. We know form our problem that Sue initially has 5 hours of pop, so $a=5$ ; we also know that every month onward, the hours of pop music in her collection is 25% more than what she had the previous month, so $b=\frac{25}{100} =0.25$ . Now let's replace the values in our function: