Question

How do you divide `(x^4 + 3x^3 + 28x + 15) /( x + 5)`?

Answer 1

`(x^4+3x^3+28x+15)/(x+5)=x^3-2x^2+10x-22+125/(x+5)`

Explanation:

You can separate out multiples of `(x+5)` from the numerator progressively as follows:

`(x^4+3x^3+28x+15)/(x+5)`

`=(x^4+5x^3-2x^3+28x+15)/(x+5)`

`=(x^3(x+5)-2x^3+28x+15)/(x+5)`

`=x^3+(-2x^3-10x^2+10x^2+28x+15)/(x+5)`

`=x^3-2x^2+(10x^2+28x+15)/(x+5)`

`=x^3-2x^2+(10x^2+50x-22x+15)/(x+5)`

`=x^3-2x^2+10x+(-22x+15)/(x+5)`

`=x^3-2x^2+10x+(-22x-110+125)/(x+5)`

`=x^3-2x^2+10x-22+125/(x+5)`

This is equivalent to long division of polynomials.

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To check that the remainder is correct, substitute `x=-5` in the original numerator expression:

`x^4+3x^3+28x+15`

`=5^4-3*5^3-28*5+15=625-375-140+15 = 125`

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If you prefer (as I do), you can long divide the coefficients - not forgetting to include a zero for the 'missing' `x^2` term in the dividend...