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3 years ago

The excluded values of a rational expression are 2 and 5. Which of the following could be this expression?

2 answers:

muminat3 years ago

6 0

You have not added the options, therefore, I cannot provide an exact answer. However, I can help you with the concept.

In any fraction, the denominator cannot be equal to zero because this would make the fraction undefined.
Excluded values are the values that would make the denominator equal to zero.

We are given that the excluded values are 2 and 5, this means that the factors of the denominator are (x-2) and (x-5)
This means that the denominator is x² - 7x + 10

Pick the fraction which has this denominator.

Hope this helps :)

lisabon 2012 [21]3 years ago

6 0

D. x+5/x^2-7x+10 is the correct one.

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3 years ago

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

Step-by-step explanation:

According to the Factor Theorem, if (x - k) is a factor of a polynomial P(x), then P(k) must equal zero.

We are given that a polynomial function has the zeros 2, √3, and -√3. So, we can let k = 2, √3, -√3.

So, according to the Factor Theorem, P(2), P(√3) and P(-√3) must equal 0.

Testing each choice, we can see that only A is true:

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Testing all three values yields that:

$\displaystyle \begin{aligned} f(2) &= (2)^3 - 2(2)^2 -3(2) + 6 \\ &= (8) - (8) -(6) + (6) \\ &= 0\stackrel{\checkmark}{=}0 \\ \displaystyle f(\sqrt{3}) &= (\sqrt{3})^3 - 2(\sqrt{3})^2 - 3(\sqrt{3}) + 6 \\ &=(3\sqrt{3}) -(6)-(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \\ f(-\sqrt{3}) &= (-\sqrt{3})^3 - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 \\ &=(-3\sqrt{3}) -(6)+(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \end{aligned}$

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