Divide. (x^2+2x+1) /(x-2) /(x^2-1) /(x^2-4)

2 answers:

8 0

( x² + 2 x + 1 ) / ( x - 2 ) / ( x² - 1 ) / ( x² - 4 ) =
= ( x² + 2 x + 1 )·( x² - 4 ) / ( x - 2 )·( x² - 1 ) =
= ( x + 1 )²·( x - 2 )·( x + 2 ) / ( x - 2 )·( x - 1 )·( x + 1 ) =
= ( x + 1 )·( x + 2 ) / ( x - 1 )

sasho [114]3 years ago

4 0

Answer: The required answer is $\dfrac{x^2+3x+2}{x-1}.$

Step-by-step explanation: We are given to divide the following algebraic expression :

$E=\dfrac{(x^2+2x+1)/(x-2)}{(x^2-1)/(x^2-4)}.$

We know that

$\dfrac{a/b}{c/d}=\dfrac{a}{b}\times\dfrac{d}{c}.$

So, for the given expression E, we have

$E\\\\\\=\dfrac{(x^2+2x+1)/(x-2)}{(x^2-1)/(x^2-4)}\\\\\\=\dfrac{x^2+2x+1}{x-2}\times\dfrac{x^2-4}{x^2-1}\\\\\\=\dfrac{(x+1)^2}{(x-2)}\times\dfrac{(x+2)(x-2)}{(x+1)(x-1)}\\\\\\=\dfrac{(x+1)(x+2)}{x-1}\\\\\\=\dfrac{x^2+3x+2}{x-1}.$

Thus, the required answer is $\dfrac{x^2+3x+2}{x-1}.$

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Travka [436]

Answer:

See below

Step-by-step explanation:

For the model function $f(x)=a|x-h|+k$ , $(h,k)$ is the vertex, and $a$ is the horizontal growth or shrink factor.

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The domain of a function is the set of all existing real x-values, thus, our domain for the given function is $(-\infty,\infty)$ , or all real x-values.

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