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2 years ago

What is the number nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8?.

1 answer:

3 0

The nearest number is 10080 if the number is nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8.

It is defined as the common number of two integers, which is the lowest number that is a multiple of two or more numbers. The full name of LCM is the least common multiple.

It is given that:

The number nearest to 10000 is exactly divisible by 3, 4, 5, 6, 7, and 8.

TO find the number:

Find the LCM of 3, 4, 5, 6, 7, and 8:

LCM = 760

= 10000/760

There are two possibilities:

= 10000 - 760 = 9240

or 10000 + (840 - 760) = 10080

The nearest number = 10080

Thus, the nearest number is 10080 if the number is nearest to 10000 which is exactly divisible by 3, 4, 5, 6, 7, and 8.

Learn more about the LCM here:

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3 years ago

Use the Factor Theorem and synthetic division to show x + 5 is a factor of f(x) = 2x3 + 7x2 − 14x + 5

olga2289 [7]

Answer:

Factor theorem: $f(-5) = 0$ .

Synthetic division: $f(x) = (x + 5)\, (2\, x^2 -3\, x + 1)$ .

Step-by-step explanation:

<h3>Factor Theorem</h3>

By the factor theorem, a monomial of the form $(x - a)$ (where $a$ is a constant) is a factor of polynomial $f(x)$ if and only if $f(a) = 0$ .

In this question, the monomial is $(x + 5)$ , which is equivalently $(x - (-5))$ . $a = -5$ .

$\begin{aligned}& f(-5) \\ &= 2 \times (-5)^{3} + 7 \times (-5)^{2}- 14\times (-5) + 5\\ &= -250 + 175 + 70 + 5 \\ &= 0\end{aligned}$ .

Hence, by the factor theorem, $(x + 5)$ , which is equivalent to $(x - (-5))$ , is a factor of $f(x)$ .

<h3>Synthetic Division</h3>

$\begin{aligned}& f(x) \\ &= 2\, x^{3} + 7\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} - 3\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} + \underbrace{(x + 5)\, (-3\, x)}_{-3\, x^{2} - 15\, x} + (x + 5) \\ &= (x + 5)\, (2\, x^{2} - 3\, x + 1)\end{aligned}$ .

The remainder is $0$ when dividing $f(x)$ by $(x + 5)$ . Hence, $(x + 5)\!$ is a factor of $f(x)\!$ .

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Answer
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2 years ago

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vekshin1

Answer:

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Step-by-step explanation:

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3 years ago