a) I will be substituting for x and simplifying.

\(\frac{4x^3+2x^2+6x+7}{2x+1} = 2x^2 + 3 + \frac{4}{2x+1}\)

\(\frac{4(1000)+2(100)+6(10)+7}{2(10)+1} = 2(100) + 3 + \frac{4}{2(10)+1}\)

\(\frac{4267}{21} = \frac{4267}{21}\)

(b) When plugging in \(x=-\frac{1}{2}\) to the denominators \(2x+1\) on both sides, we will end up dividing by zero, an undefined value.

(c) There's really two ways to do this. One is multiplying both sides by \(2x+1\) and showing the results. The other is dividing \(4x^3+2x^2+6x+7\) by \(2x+1\) through long division to show that it equals \(2x^2+3+\frac{3}{2x+1}\). However, I happen to hate long division with polynomials. So Ima do it tha otha way.

\(\frac{4x^3+2x^2+6x+7}{2x+1} = 2x^2 + 3 + \frac{4}{2x+1}\)

\(4x^3+2x^2+6x+7 = (2x+1) (2x^2 + 3 + \frac{4}{2x+1})\)

\(4x^3+2x^2+6x+7 = 4x^3+2x^2+6x+7\)

The right side multiplication looks intimidating, but it actually multiplies out that easily in one step.

Oh, and, see (b) for why it doesn't work for \(x = -\frac{1}{2}\)

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