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2 years ago

F(x)=15x³+22x²-15x+2 Write f(x) as a product of linear factors.

Mathematics

1 answer:

Goryan [66]2 years ago

3 0

Answer:

$(5x - 1)(3x - 1)(x + 2)$

Step-by-step explanation:

$(15 {x}^{3} + 22 {x}^{2} - 15x + 2)$

Apply Rational Root Theorem, our possible roots will be

plus or minus( 2/15, 2/5,2/3,2, 1/15,1/5,1/3,1).

I tried root -2 and it work so

If we apply synthetic dividon, we would be left with

$15 {x}^{2} - 8x + 1$

We can factor this regularly.

Apply AC method that a number

AC will multiply to 15 but add to -8.

The answer are -5 and -3 so we write this as

$15 {x}^{2} - 5x - 3x + 1$

Factor by grouping

$(15x {}^{2} - 5x) - (3x + 1)$

$5x(3x - 1) - 1(3x - 1)$

So our factor are

$(5x - 1)(3x - 1)(x + 2)$

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$