How do you divide $(7x^4 - 4x^2 – 36x+ 81 )/((x + 9) )$ ?

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Answer 1 · 2016-01-03T23:11:59

$7x^4-4x^2-36+81$

$= (x+9)(7x^3-63x^2+563x-5103)+46008$

I like to long divide the coefficients, not forgetting to include a $0$ for any missing power of $x$ (in this case $x^3$ )...

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Long division of coefficients is similar to long division of numbers.

We find:

$7x^4-4x^2-36+81$

$= (x+9)(7x^3-63x^2+563x-5103)+46008$

That is $(7x^4-4x^2-36+81)$ divided by $(x+9)$ results in a quotient $(7x^3-63x^2+563x-5103)$ with remainder $46008$ .

How do you divide (7x^4 - 4x^2 – 36x+ 81 )/((x + 9) )(7x^4 - 4x^2 – 36x+ 81 )/((x + 9) )?