Factor –8x3 – 2x2 – 12x – 3 by grouping. What is the resulting expression?

Mathematics

2 answers:

KATRIN_1 [288]3 years ago

7 0

Answer: $(4x+1)(-2x^2-3)$

Step-by-step explanation:

To factor the given polynomial $-8x^3-2x^2-12x-3$

First group the first two terms and the last two terms together respectively.

i.e. $(-8x^3-2x^2)+(-12x-3)$

Take out GCF from each binomial, we get

$-2x^2(4x+1)-3(4x+1)$

Now, factor out the common binomial, we get the resulting expression as

$=(4x+1)(-2x^2-3)$

xenn [34]3 years ago

3 0

Easier if you first factor out the " - " sign. Then you have:

–8x3 – 2x2 – 12x – 3 = -1(<span>8x3 + 2x2 + 12x +3).
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Start with 12 + 3. This factors to 3(4x+1). Next, focus on 8x^3+2x^2. This factors to 2x^2*(4x+1). So, the binomial factor 4x+1 is common to the first two terms and the second two terms. We get: (4x+1) ( 2x^2 + 3).

Now put the "-" sign back in, and we get - (4x+1) ( 2x^2 + 3) (answer)

Please note: Use "^" to denote exponentiation. 2x2 is meaningless; it should be 2x^2.

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3√x -2/x^2<br><br>please show step by step of differentiation before combining the terms.

EastWind [94]

Answer:

Step-by-step explanation:

Before we differentiate, let us assign a variable to the function. Let y be equal to the function i.e let y = 3√x -2/x²

In differentiation if $y = ax^{n}$ , then $\frac{dy}{dx} = nax^{n-1}$ where n is a constant and dy/dx means we are differentiating the function y with respect to x.

Applying the formula o the question given;

$y= 3\sqrt{x} -2/x^2\\y = 3{x}^\frac{1}{2} - 2x^{-2} \\\\$

On differentiating the resulting function;

$\frac{dy}{dx} = \frac{1}{2}*3x^{\frac{1}{2}-1 } - (-2)x^{-2-1} \\\\\frac{dy}{dx} = \frac{1}{2}*3x^{-\frac{1}{2}} + 2x^{-3}\\ \\\frac{dy}{dx} = \frac{1}{2}*{\frac{3}{x^{\frac{1}{2} } }} + \frac{2}{x^{3} } \\\\\frac{dy}{dx} = {\frac{3}{2x^{\frac{1}{2} } }} + \frac{2}{x^{3} }\\\\\frac{dy}{dx} = {\frac{3}{2\sqrt{x} }} + \frac{2}{x^{3} }$