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1 year ago

Find the rational roots f(x) =3x3+ 2x2 + 3x + 6

Mathematics

1 answer:

Ann [662]1 year ago

4 0

The rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)

<h3>How to determine the rational root of the function f(x)?</h3>

The function is given as:

f(x) = 3x^3 + 2x^2 + 3x + 6

For a function P(x) such that

P(x) = ax^n +...... + b

The rational roots of the function p(x) are

Rational roots = ± Possible factors of b/Possible factors of a

In the function f(x), we have:

a = 3

b = 6

The factors of 3 and 6 are

a = 1 and 3

b = 1, 2, 3 and 6

So, we have:

Rational roots = ±(1, 2, 3, 6)/(1, 3)

Split the expression

Rational roots = ±(1, 2, 3, 6)/1 and ±(1, 2, 3, 6)/3

Evaluate the quotient

Rational roots = ±(1, 2, 3, 6, 1/3, 2/3, 1, 2)

Remove the repetition

Rational roots = ±(1, 2, 3, 6, 1/3, 2/3)

Hence, the rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)

The complete parameters are:

The function is given as:

f(x) = 3x^3 + 2x^2 + 3x + 6

The rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)

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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

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Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$