How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{2 x - 6}$ $( x^4-2x^3-2x^2+11x+12)/(2x-6)$ ?

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How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{2 x - 6}$ $( x^4-2x^3-2x^2+11x+12)/(2x-6)$ ?

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Answer 1 · 2016-01-07T00:18:40

Long divide coefficients to find:

$\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{2 x - 6} = \frac{1}{2} x^{3} + \frac{1}{2} x^{2} + \frac{1}{2} x + 7$ $(x^4-2x^3-2x^2+11x+12)/(2x-6) = 1/2x^3+1/2x^2+1/2x+7$

with remainder $27$ $27$

Explanation:

I like to long divide coefficients, but first note that $2 x - 6 = 2 (x - 3)$ $2x-6 = 2(x-3)$ , so to simplify the calculation, let's divide by $x - 3$ $x-3$ , then divide the result (and any remainder) by $2$ $2$ at the end:

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We find:

$\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{x - 3} = x^{3} + x^{2} + x + 14$ $(x^4-2x^3-2x^2+11x+12)/(x-3) = x^3+x^2+x+14$

with remainder $54$ $54$ .

Hence (dividing by $2$ $2$ ):

$\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{2 x - 6} = \frac{1}{2} x^{3} + \frac{1}{2} x^{2} + \frac{1}{2} x + 7$ $(x^4-2x^3-2x^2+11x+12)/(2x-6) = 1/2x^3+1/2x^2+1/2x+7$

with remainder $27$ $27$

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How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{2 x - 6}$ $( x^4-2x^3-2x^2+11x+12)/(2x-6)$ ?

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How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 11 x + 12}{2 x - 6}$ $( x^4-2x^3-2x^2+11x+12)/(2x-6)$ ?