How do you divide (x^4-2x^3-2x^2+9x+3)/(x^2-3) ?

1 Answer
George C.
May 2, 2016

(x^4-2x^3-2x^2+9x+3)/(x^2-3) = (x^2-2x+1) + (3x+6)/(x^2-3)

Explanation:

I like to long divide the coefficients, not forgetting to include 0's for any missing powers of x. In our example, that means the missing x term in the divisor, which is therefore repesented as 1, 0, -3.

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The process is similar to long division of numbers.

We find a quotient 1, -2, 1, meaning x^2-2x+1 and remainder 3, 6, meaning 3x+6

So: (x^4-2x^3-2x^2+9x+3)/(x^2-3) = (x^2-2x+1) + (3x+6)/(x^2-3)

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