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3 years ago

What is the quotient when −3x3 + 5x + 14 is divided by x − 2?

Mathematics

2 answers:

ASHA 777 [7]3 years ago

6 0

The answer is A. -3x2 - 6x - 7

Anika [276]3 years ago

6 0

Answer:

$-3x^{2} -6x-7$

Step-by-step explanation:

The given expression is :

$\frac{-3x^{3}+5x+14}{x-2}$

Factoring $-3x^{3}+5x+14$

Factor out common term -1

$3x^{3}-5x-14$

Now dividing leading coefficients of numerator and divisor.

$\frac{3x^{3}}{x}=3x^{2}$

Multiplying x-2 with $3x^{2}$ = $3x^{3}-6x^{2}$

Subtracting $3x^{3}-6x^{2}$ from $3x^{3}-5x-14$

we get $6x^{2} -5x-4$

Therefore, $\frac{-3x^{3}+5x+14}{x-2}$ = $3x^{2} +\frac{6x^{2}-5x-14}{x-2}$

Now repeating these same steps until (x-2) is factored out, we get the quotient as :

$-3x^{2} -6x-7$

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Step-by-step explanation:

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3 years ago

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elena-14-01-66 [18.8K]

If you're just starting calculus, perhaps you're asking about using the definition of the derivative to differentiate $x^4$ .

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$\dfrac{d}{dx} x^4 = \displaystyle \lim_{h\to0} \frac{(x+h)^4 - x^4}h$

Expand the numerator using the binomial theorem, then simplify and compute the limit.

$\dfrac{d}{dx} x^4 = \displaystyle \lim_{h\to0} \frac{(x^4+4hx^3 + 6h^2x^2 + 4h^3x + h^4) - x^4}h \\\\ ~~~~~~~~ = \lim_{h\to0} \frac{4hx^3 + 6h^2x^2 + 4h^3x + h^4}h \\\\ ~~~~~~~~ = \lim_{h\to0} (4x^3 + 6hx^2 + 4h^2x + h^3) = \boxed{4x^3}$

In general, the derivative of a power function $f(x) = x^n$ is $\frac{df}{dx} = nx^{n-1}$ . (This is the aptly-named "power rule" for differentiation.)